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198 Cards in this Set
- Front
- Back
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P(AUB)
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P(A)+P(B)-P(AB)
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P(AUB) if A and B are disjoint
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P(A) + P(B)
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example: Draw one card at random from a deck of ordinary set of playing cards
1) P(Ace or Club) 2) P(Heart or Diamond) |
1) P(ace U club)
= P(A)+P(C)-P(AC) = 4/52 + 13/52 - 1/52 = 16/52 2) P(H U C) = P(H) + P(C) = 13/52 + 13/52 |
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P(A')
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1-P(A)
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P(A-B)
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P(A-B)=P(AB')
= P(A) - P(AB) |
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conditional probability
P(A|B) |
P(A|B)
=P(AB)/P(B) |
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joint probability
P(AB) |
P(AB)
=P(A|B)P(B) =P(B|A)P(A) |
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total or marginal probability
P(A) |
sum(P(ABi),i,1,n)
=sum(P(A|Bi)P(Bi),i,1,n) |
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Baye's Theorem
P(Bj|A) |
P(Bj|A)
=P(BjA)/P(A) = P(A|Bj)P(Bj)/sum(P(A|Bi)P(Bi),i,1,n) |
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Example: Pick a card
1) P(spade) 2) P(spade|black) 3) P(spade|red) |
1) P(spade)=13/52
2) P(spade|black) = P(black and spade)/P(black) 3) P(spade|red) = P(spade and red)/P(red) =0/(26/52) = 0 |
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P(A'|B)
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P(A'|B)
= 1-P(A|B) |
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P((AUC)|B)
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P((AUC)|B)
= P(A|B) + P(C|B) - P(AC|B) |
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disjoint
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AnC=0
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P(AB)
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P(AB)
=P(A|B)P(B) =P(B|A)P(B) |
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Example:
Choosetwo balls without replacement at random from a box which has 3 white and 2 red balls: 1) prob first ball removed white and second ball red? |
1) P(W1R2)
=P(R2|W1)P(W1) =(2/4)*(3/5) |
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example:
combined experiment toss a coin and die what is the total sample space? |
S=S1 * S2
={(H,1),(H,2), ..., (H,6),(T,1), ... (T,6)} |
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cartesian product:
S1 has n elements S2 has m elements what is the total sample space of S? |
m*n pairs ofelements
S=S1 * S2 S is called a cartesian product |
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geometric interpolation:
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table to see possible outcomes of combined experiment
a1 a2 ... an b1 a1b1 a2b2 ... b2 a1b2 ... . . . bm a1bm a2bm ... anbm |
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example:
number of outcomes if toss a coin 3 times |
2*2*2=8
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example:
number of outcomes if roll a six headed dice three times |
6*6*6
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marginal/total probability
P(A)= |
P(A)
= sum(P(ABi),i,1,n) = sum(P(A|Bi)P(Bi),i,1,n) with BiBj=0 ie disjoint |
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example:
f=factory, D=defective given 30% of xistors from f1 50% of xistors from f2 20% of xistors from f3 and that 2% of xistors from f1 are D 4% of xistors from f2 are D 5% of xistors from f3 are D what is 1) the probability we have a defective transistor? 2) probability the defective transistor is from factory 2? |
1) P(D)
= P(Df1)+P(Df2)+P(Df3) = P(D|f1)P(f1) +P(D|f2)P(f2) +P(D|f3)P(f3) = 0.02*0.3 + 0.04*0.5 +0.05*0.2 = 3.6% 2) P(f2|D) = P(f2D)/P(D) = P(D|f2)*P(f2)/P(D) |
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example:
choose a fruit (apple or orange) at random from a box chosen at random. P(apple) = ? |
P(apple)
= P(AB1)+P(AB2)+P(AB3) = P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3) = (3/6)(1/3) + (2/3)(1/3) + (2/6)(1/3) = 37/90 |
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Bayes' Theorem
P(Bj|A) |
P(Bj|A)
=[P(ABj)/P(Bj)]/P(A) |
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example:
communication channel (binary) -- consider binary communication channel transmit data in form of zero or ones. define: T1={# 1 transmitted} T0={# 0 transmitted} R1={# 1 received} R0={# 0 received} given: P(R1|T0)=0.91 P(R0|T0)=0.94 P(T0)=0.45 find: 1) P(R1) 2) P(R0) 3) P(T1|R1) 4) P(T0|R0) 5) P(error) |
1) P(R1)
=P(R1T1)+P(R1T0) =P(R1|T1)P(T1)+P(R1|T0)P(T0) =0.5275 2) P(R0) = 1-P(R1) = 0.4725 3) P(T1|R1) = P(T1R1)/P(R1) = [P(R1|T1)P(T1)/P(R1)] = 0.9488 4) P(T0|R0) =[P(R0|T0)P(T0)]/P(R0) = 0.8952 5) P(error) = P(R1T0) + P(R0T1) = P(R1|T0)P(T0) + P(R0|T1)P(T1) = 0.0765 |
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example:
a system consists of two subsystems A1 and A2. Define events: Wi = {Ai working} Wi'= {Ai not working} for i = 1,2 given: P(W1|W2)=0.9 P(W2)=0.8 P(W1)=0.6 find: P(W2|W1') |
P(W2|W1')
=P(W2 W1')/P(W1') =[P(W1'|W2)P(W2)/P(W1')] ={[1-P(W1|W2)]P(W2)} /(1-P(W1)) |
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P(A'|B)
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P(A'|B)
= 1-P(A|B) |
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Tree Diagrams:
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Useful graphical tool for defining the basic outcomes of a combined experiment. Particularly when events are dependent.
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example:
select a box at random and draw two balls without replacement. given box1 = 4 red, 2 white box2 = 3 red, 2 white find probability both balls are white. |
P(W1W2)
=P(B1W1W2 U B2W1W2) = P(B1W1W2) + P(B2W1W2) = P(W2|W1B1)P(W1|B1)P(B1) + P(W2|W1B2)P(W1|B2)P(B2) =(1/5)(2/6)(1/2) +(1/4)(2/5)(1/2) =1/12 |
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P(A|BC)
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P(A|BC)
= P(ABC)/P(BC) |
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example:
combined experiment toss a coin and die what is the total sample space? |
S=S1 * S2
={(H,1),(H,2), ..., (H,6),(T,1), ... (T,6)} |
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cartesian product:
S1 has n elements S2 has m elements what is the total sample space of S? |
m*n pairs ofelements
S=S1 * S2 S is called a cartesian product |
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geometric interpolation:
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table to see possible outcomes of combined experiment
a1 a2 ... an b1 a1b1 a2b2 ... b2 a1b2 ... . . . bm a1bm a2bm ... anbm |
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example:
number of outcomes if toss a coin 3 times |
2*2*2=8
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example:
number of outcomes if roll a six headed dice three times |
6*6*6
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marginal/total probability
P(A)= |
P(A)
= sum(P(ABi),i,1,n) = sum(P(A|Bi)P(Bi),i,1,n) with BiBj=0 ie disjoint |
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example:
f=factory, D=defective given 30% of xistors from f1 50% of xistors from f2 20% of xistors from f3 and that 2% of xistors from f1 are D 4% of xistors from f2 are D 5% of xistors from f3 are D what is 1) the probability we have a defective transistor? 2) probability the defective transistor is from factory 2? |
1) P(D)
= P(Df1)+P(Df2)+P(Df3) = P(D|f1)P(f1) +P(D|f2)P(f2) +P(D|f3)P(f3) = 0.02*0.3 + 0.04*0.5 +0.05*0.2 = 3.6% 2) P(f2|D) = P(f2D)/P(D) = P(D|f2)*P(f2)/P(D) |
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example:
choose a fruit (apple or orange) at random from a box chosen at random. P(apple) = ? |
P(apple)
= P(AB1)+P(AB2)+P(AB3) = P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3) = (3/6)(1/3) + (2/3)(1/3) + (2/6)(1/3) = 37/90 |
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Bayes' Theorem
P(Bj|A) |
P(Bj|A)
=[P(ABj)/P(Bj)]/P(A) |
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example:
communication channel (binary) -- consider binary communication channel transmit data in form of zero or ones. define: T1={# 1 transmitted} T0={# 0 transmitted} R1={# 1 received} R0={# 0 received} given: P(R1|T0)=0.91 P(R0|T0)=0.94 P(T0)=0.45 find: 1) P(R1) 2) P(R0) 3) P(T1|R1) 4) P(T0|R0) 5) P(error) |
1) P(R1)
=P(R1T1)+P(R1T0) =P(R1|T1)P(T1)+P(R1|T0)P(T0) =0.5275 2) P(R0) = 1-P(R1) = 0.4725 3) P(T1|R1) = P(T1R1)/P(R1) = [P(R1|T1)P(T1)/P(R1)] = 0.9488 4) P(T0|R0) =[P(R0|T0)P(T0)]/P(R0) = 0.8952 5) P(error) = P(R1T0) + P(R0T1) = P(R1|T0)P(T0) + P(R0|T1)P(T1) = 0.0765 |
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example:
a system consists of two subsystems A1 and A2. Define events: Wi = {Ai working} Wi'= {Ai not working} for i = 1,2 given: P(W1|W2)=0.9 P(W2)=0.8 P(W1)=0.6 find: P(W2|W1') |
P(W2|W1')
=P(W2 W1')/P(W1') =[P(W1'|W2)P(W2)/P(W1')] ={[1-P(W1|W2)]P(W2)} /(1-P(W1)) |
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P(A'|B)
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P(A'|B)
= 1-P(A|B) |
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Tree Diagrams:
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Useful graphical tool for defining the basic outcomes of a combined experiment. Particularly when events are dependent.
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example:
select a box at random and draw two balls without replacement. given box1 = 4 red, 2 white box2 = 3 red, 2 white find probability both balls are white. |
P(W1W2)
=P(B1W1W2 U B2W1W2) = P(B1W1W2) + P(B2W1W2) = P(W2|W1B1)P(W1|B1)P(B1) + P(W2|W1B2)P(W1|B2)P(B2) =(1/5)(2/6)(1/2) +(1/4)(2/5)(1/2) =1/12 |
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P(A|BC)
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P(A|BC)
= P(ABC)/P(BC) |
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P(ABC)
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P(ABC)
=P(A|BC)P(BC) =P(A|BC)P(B|C)P(C) |
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equations to test independence
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1) P(A|B) = P(A)
2) P(AB) = P(A)P(B) |
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example:
toss a coin twice P(H2|H1)=? |
P(H2|H1)=P(H2)
independent events |
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conditionally independent
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P(AB|C)
= P(A|C)P(B|C) |
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independent events: definition
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two events are called statistically independent if the probability of occurance of one event is not affected by occurance or non-occurance of the other one.
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Three events A, B, and C are independent iff:
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P(AB) = P(A)P(B)
P(AC) = P(A)P(C) P(BC) = P(B)P(C) P(ABC) = P(A)P(B)P(C) |
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total # eqns required to establish n events are independent
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2^n - (n+1)
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example:
toss a die events -- A ={odd number} B ={even number} C ={one or two} 1) are A and B statistically independent events? 2) are A and C statistically independent events? |
1) P(AB) = P(A)P(B)?
P(B)=P(A)=1/2 P(AB)=P(0)=0 P(AB)is not equal to P(A)P(B) thus, A and B are not independent 2) (a) P(AC) = P(A)P(C)? or (b) P(C|A) = P(C)? P(C)=2/6=1/3 P(A)=1/2 P(C|A)=1/3 P(AC)=P(1)=1/6 (a) and (b) are true so A and C are statistically independent |
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example:
two relay contacts are activated by a single armature where: A={First contact closed} B={Second contact closed} C={armature is activated} given P(C)=0.5, P(A|C)=0.9, P(B|C)=0.9, P(A|C')=0.2, P(B|C')=0.1 and A and B are CONDITIONALLY independent find: 1) P(AB|C) 2) P(B) 3) P(AB) |
1) P(AB|C)
=P(A|C)P(B|C) =0.81 2) P(B) =P(B|C)P(C)+P(B|C')P(C') =0.5 3) P(AB) =P(AB|C)P(C)+P(AB|C')P(C') =P(AB|C)P(C) + P(A|C')P(B|C')P(C') =0.415 |
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if n events are independent from eachother, what else can we assume they are independent from?
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any one event Ai is independent of any event formed by unions, intersections, and complements of the other events.
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Cartesian Product of sets A and B
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C=A*B
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Number of Elements in Cartesian Product of sets A and B (A has m elements, B has n elements)
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# elements of aibj = m*n
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Permutation: definition
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Permutation:
An ORDERED arrangement of k distinct objects taken from n distinct objects without replacement |
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Combination: definition
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NON-ORDERED combination of n objects taken k at a time
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Difference between Permutation and Combination
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Permutation order matters.
Combination order does not matter. |
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P(n,k)
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P(n,k)
=n!/(n-k)! |
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P(n,n)
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P(n,n)
=n! |
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C(n,k)
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C(n,k)=
(n k) =n!/(k!*(n-k)!) |
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Combination: definition
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Selection of k distinct objects from total of n distinct objects when order of selection is not needed. ie number of ways to divide n objects into two groups of k and n-k elements.
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example:
number of ways we can choose two applicants from five different applicants of a1,a2,a3,a4,a5 |
C(5,2)
=(5 2) =5!/(2!3!) =10 |
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binomial coefficient
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(n
k)=(n,k) |
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binomial expansion
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(p+q)^n
=sum((n,k)p^k*q^(n-k),k,0,n) |
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generalized combination:
definition |
number of ways to partition n distinct objects into k distinct groups containing n1,n2,...,nk objects respectively where
sum(ni,i,1,k)=1 |
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generalized combination:
equation |
C(n,(n1,n2,...,nk))
= n!/(n1!n2!...nk!) |
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Bernoulli Trials (aka Binomial Experiment): definition
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*n identical subexperiments
*Each sub experiment has two outcomes: success or failure *subexperiments are independent *each subexperiment has identical probabilities |
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Bernoulli Trials (aka Binomial Experiment):
equation |
P(success any trial)=p
P(success any trial)=q p+q=1 P(number of successes = k) =(n,k)p^k*q^(n-k) |
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Bernoulli Trial:
P(k1<=k<=k2) |
Bernoulli Trial:
P(k1<=k<=k2) = sum((n,k)p^k *q^(n-k),k,k1,k2) |
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example:
toss a coin three times where P(H)=p and P(T)=q find: 1) P(3H) 2) P(2H and 1T) 3) P(2H at most) 4) P(1H at least) |
1) P(3H)
=(3,3)p^3*q^0 =p^3 2) P(2H and 1T) =P(2H) =(3,2)p^2*q^1 =3p^2*q 3) P(2H at most) =P(k<=2) =P(0H)+P(1H)+P(2H) =1-P(3H) =1-p^3 4) P(1H at least) =P(k>=1) =P(1H)+P(2H)+P(3H) =1-P(0H) =1-P(k=0) =1-(3,0)p^0*q^3 =1-q^3 |
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example:
consider a relay is operating 98% of the time under certain conditions. for 10 trials under the same conditions find: 1) P{all trials successful} 2) P{1 failure and 9 success} 3) P{1st 9 success & last fails} |
1) P{all trials successful}
=(10,10)p^10 =0.98^10 =0.8171 2) P{1 failure and 9 success} =(10,9)p^9*q =0.1667 3) P{1st 9 success & last fails} =(0.98)^9*(0.02)^1 =0.0167 |
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reliability: definition
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reliability: definition --
reliability of a system is probability of success of that system. |
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elementery systems:
FAi= FAi'= pAi= qAi= R= Q= |
elementery systems:
FAi={Ai open/fails} FAi'={Ai closed/works} pAi=P(FAi') qAi=P(FAi) R=P(success) Q=P(fail) |
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elementery systems:
series --[A1]--[A2]--[A3]-- R=? Q=? |
R
=P(FA1'FA2'FA3') =pA1*pA2*pA3 Q =P(FA1 U FA2 U FA3) =1-R =1-pA1*pA2*pA3 |
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elementery systems:
parallel |---[A1]---| ---|---[A2]---|--- |---[A3]---| R=? Q=? |
R
=1-Q =1-qA1*qA2*qA3 Q =qA1*qA2*qA3 |
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Real Random Variable: definition
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Real Random Variable: definition --
A real single valued function defined over a sample space and mapped the sample space into a set of real numbers. We represent a random variable by capital letters and any particular values of r.v. with lower case letters. |
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example:
tos a die and define the random variable X as: X(fi)=10i with i=1,2,3,4,5,6 1)P(X=20) 2)P(X=25) 3)P(X<=25) 4)P(15<=X<=35) |
1)P(X=20)
=1/6 2)P(X=25) =0 3)P(X<=25) = P(X=10 U X=20) = 1/6 + 1/6 = 2/6 4)P(15<=X<=35) =P(X=20 U X=30) =2/6 |
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Discrete Random Variable: definition
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Discrete Random Variable: definition --
Can take a discrete (countable) numbers of values. |
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example:
toss 2 coins and let X define the number of heads observed. 1)P(X=1) 2)P(X=0) 3)P(X=2) |
events:
H1H2 (Two heads) H1T2,T1H2 (One head) T1T2 (No heads) 1)P(X=1) =P(H1T2 U T1H2) =1/4 + 1/4 =1/2 2)P(X=0) =P(T1T2) =1/4 3)P(X=2) =P(H1H2) =1/4 |
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Continuous Random Variable: definition
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Continuous Random Variable: definition --
can take all values within a specified interval. Can NOT result from discrete sample space. |
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Probability Mass Function: discrete random variables:
equation |
PX(x)
=P(X=x) =P({Si:X(Si)=x}) |
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example:
tos a die and define the random variable X as: X(fi)=10i with i=1,2,3,4,5,6 find probability mass function |
f(x)=PX(x)
=1/6 delta(x-10) +1/6 delta(x-20) +1/6 delta(x-30) +1/6 delta(x-40) +1/6 delta(x-50) +1/6 delta(x-60) |
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Probability Distribution Function (P.D.F)
(aka Cumulative Distribution Function) |
Px of the random variable X is defined as:
FX(x) = P(X<=x) = P({Si:X(Si)<=x}) for continuous random variables, probability of one specific point is zero |
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example:
toss a die. X(fi)=10i with i=1,2,3,4,5,6 0) FX(x) 1)FX(0) 2)FX(10) 3)FX(20) 4)FX(30) 5)FX(100) 6)P(x<40) |
0) FX(x)=
1/6*U(X-10) +1/6*U(X-20) +1/6*U(X-30) +1/6*U(X-40) +1/6*U(X-50) +1/6*U(X-60) 1)FX(0)=0 2)FX(10) =P(X<=10) =1/6 3)FX(20) =P(X<=20) =2/6 4)FX(30) =P(X<=30) =3/6 5)FX(100) =P(X<=100) =6/6 = 1 6) P(x<40) = 3/6 |
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impulse function/delta function:
1) delta(x) 2) delta(x-a) |
1) delta(x)=
1 when x=0 0 when x!=0 2) delta(x-a)= 1 when x=a 0 when x!=a |
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unit step function:
1) u(x) 2) u(x-a) |
1) u(x)
1 when x>=0 0 when x<0 2) u(x-a)= 1 when x>=a 0 when x<a |
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relation between delta(x) and u(x)
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delta(x)=du(x)/dx
and u(x) = integral(delta(x),x) |
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relation between PDF and probability mass function
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F(x)=integral(f(x),x)
f(x)=(d/dx)F(x) probability mass function: f(x) probability distribution fu: F(x) |
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example:
Toss a coin two times and let X be the number of heads observed 1)find the probability mass fu. 2) find the probability distribution fu |
1) f(x)
=PX(x) =(1/4)delta(x) +(1/4+1/4)delta(x-1) +(1/4)delta(x) 2)FX(x) =(1/4)u(x) +(1/4+1/4)u(x-1) +(1/4)u(x-2) |
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integral(
f(x)delta(x-x0) ,x,-infinity,infinity) |
integral(
f(x)delta(x-x0) ,x,-infinity,infinity) =f(x0) |
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short note notation for P(x) and F(x)
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PX(x)
=f(x) =sum (P(X=xi)delta(x-xi),all xi) FX(x) =sum (P(X=xi)u(x-xi),all xi<=x) |
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Four properties of PDF
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1)FX(-infinite)
=P(X<-infinite) =P(0)=0 2)FX(infinite) =P(X<infinite) =P(S) =1 3)FX(x) is an increasing fu. of x (positive slope) 4)FX(x)=FX(x+) at the point of discontinuity where: F(x+)=lim(F(x+epsilon),epsilon-->0) {epsilon>0} |
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Probability Measurements from PDF
1)P(x1<X<=x2) 2)P(X=x1) 3)P(x1<=X<=x2) |
1)P(x1<X<=x2)
=FX(x2)-FX(x1) for x1<x2 2)P(X=x1) =F(x1+)-F(x1-) when X discontinuous at point x1 =0 when X is continuous at point x1 3)P(x1<=X<=x2) =P(X=x1)+P(x1<=X<=x2) =FX(x2+)-FX(x1-) when X is discontinuous at point x1 =FX(x2)-FX(x1) when X is continuous at x1 |
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Probability Measurements from PDF (for discrete r.v.)
1)P(x1<X<=x2) 2)P(X=x1) 3)P(x1<=X<=x2) |
1)P(x1<X<=x2)
=F(x2)-F(x1) 2)P(X=x1) =FX(x1+)-F(x1') 3)P(x1<=X<=x2) =FX(x2)-FX(x1') |
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example:
toss a die where X(fi)=10i where i=1,2,3,4,5,6 1)P(X=30) 2)P(X=25) 3)P(20<X<=40) 4)P(20<=X<=40) |
1)P(X=30)
=FX(30+)-FX(30-) =3/6 - 2/6 =1/6 2)P(X=25)=0 since x is continuous at 25 3)P(20<X<=40) =FX(40)-FX(20) =4/6-2/6 =2/6 4)P(20<=X<=40) =FX(40)-FX(20-) =4/6-1/6 =3/6 |
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probability density function (pdf): eqn
|
pdf
=fX(x) =dFX(x)/dx =sum(PX(xi)delta(x-xi),i,1,n) |
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four properties of pdf
|
four properties of pdf
1) fX(x)>=0 always positive fu since has positive slope 2)integral(fX(x) ,x,-infinity,infinity) =FX(x),-infinity,infinity =FX(infinity)-FX(-infinity) =1 3)FX(x) = integral(fX(z),z,-infinity,x) 4)P(x1<=X<=x2) =integral(fX(x),x,x1,x2) =FX(x2)-FX(x1) |
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example:
let X be a continuous r.v. with pdf =c*x^2 when 0<=X<=1 =0 elsewhere 1)P(0.5<=X<=0.75) 2)FX(x) |
First find c:
integral(c*x^2,x,0,1)=1 -->c=3 1)P(0.5<=X<=0.75) integral(3x^2,x,0.5,0.75) =0.75^3 - 0.5^3 2)FX(x) = 0 when x<0 =integral(3z^2,z,0,x) when 0<=x<=1 =1 when x>1 |
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Binomial distribution used by
1)probability mass fu 2)PDF 3)pdf |
X=# successes
n=# trials 1)probability mass fu P(X=k)=pX(k) =(n,k)p^k*q^(n-k) 2)PDF = FX(l) =sum(PX(k)u(x-k),k,0,l) =sum((n,k)p^k*q^(n-k)*u(x-k) ,k,0,l) 3)pdf = fX(x) =sum(PX(k)*delta(x-k),k,0,n) |
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example:
toss a coin 10 times and find pdf and PDF for #heads |
X=# heads, n=10
f(x)=sum((10,k)*p^k*q^(10-k) *delta(x-k),k,0,10) F(x) =P(X<=2) =sum((10,k)*p^k*q^(10-k) *u(x-k),k,0,10) |
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Poisson: definition & eqn
|
If X defines number of events happening in an interval of length T then X has a poisson distribution where:
P(X=k) =PX(k) =e^-b*b^k/k! when k=0,1,2,... =0 otherwise b=avg number events in interval T |
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relation binomial and poisson
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binomial prob fu converges to poisson when n-->infinity and p-->0 in such a way that np=b
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example:
let X be the # of accidents in one intersection duringone week where avg # of accidants in one week is reported as 2 1)write PDF and pdf of X 2)FX(2) 3)P(2<=X<=4) |
1)write PDF and pdf of X
b=2 fX(x) =sum (PX(xi)delta(x-xi),all xi) =sum(e^-b*b^k/k!*delta(x-k),k,0,infinity) PDF=FX(l) =sum(e^-b*b^k/k!*u(l-k),k,0,l) =sum(e^-2*2^k/k!*u(l-k),k,0,l) 2)FX(2) =sum(e^-2*2^k/k!,k,0,2) =0.675 3)P(2<=X<=4) =sum(e^-2*2^k/k!*,k,2,4) |
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uniform density function: definition and pdf and PDF eqns
|
X is equally likely between a andb
pdf: fX(x) =1/(b-a) a<=x<=b =0 elsewhere PDF: FX(x) =0 x<=a =1 x>=b =(x-a)/(b-a) a<=x<=b |
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example:
randomX uniformly distributed between 2 and 4 1)pdf 2)PDF |
1)fX(x) = 1/(4-2) = 1/2
2)FX(x) = (x-2)/(4-2) =(x-2)/2 |
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example:
R is a random variable for which all values between 80 and 100 are equally likely. All otehr values are impossible. Find: 1) P(R between 90 and 95) 2) P(R between 90 and 95|between 85 and 95) given f(x)=0.05 when 80<R<100 |
1) P(90<R<95)
=(0.05)*(90-95)=0.25 2) P(90<R<95|85<R<95) =P(90<R<95)/P(85<R<95) =[(90-95)*0.05]/[(95-85)*0.05] |
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exponential pdf
|
fx(X)
= (1/b)e^(-x/b) when x>=0 =0 when x<0 FX(x) =1-e^(-x/b) when x>=0 =0 when x<0 |
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applications of exponential pdf
|
*time between arrival of plane
*fluctuations in signal strength received by radar from certain types of aircraft |
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Normal or Gaussian: eqn
|
f(x)
= (1/(sqrt(2Pi)*sigma)) *e^(-(x-mu)^2/(2*sigma^2)) |
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Normal or Gaussian: def
|
1) f(x) is bell shaped and symmetrical around mu
2) f(x) has two parameters: mu=mean sigma=std deviation sigma^2=variance |
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Normal or Gaussian: short notation
|
N(mu,sigma)
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Application of Normal or Gaussian:
|
noise in communication system
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Probability Measurements from PDF
1)P(x1<X<=x2) 2)P(X=x1) 3)P(x1<=X<=x2) |
1)P(x1<X<=x2)
=FX(x2)-FX(x1) for x1<x2 2)P(X=x1) =F(x1+)-F(x1-) when X discontinuous at point x1 =0 when X is continuous at point x1 3)P(x1<=X<=x2) =P(X=x1)+P(x1<=X<=x2) =FX(x2+)-FX(x1-) when X is discontinuous at point x1 =FX(x2)-FX(x1) when X is continuous at x1 |
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Probability Measurements from PDF (for discrete r.v.)
1)P(x1<X<=x2) 2)P(X=x1) 3)P(x1<=X<=x2) |
1)P(x1<X<=x2)
=F(x2)-F(x1) 2)P(X=x1) =FX(x1+)-F(x1') 3)P(x1<=X<=x2) =FX(x2)-FX(x1') |
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example:
toss a die where X(fi)=10i where i=1,2,3,4,5,6 1)P(X=30) 2)P(X=25) 3)P(20<X<=40) 4)P(20<=X<=40) |
1)P(X=30)
=FX(30+)-FX(30-) =3/6 - 2/6 =1/6 2)P(X=25)=0 since x is continuous at 25 3)P(20<X<=40) =FX(40)-FX(20) =4/6-2/6 =2/6 4)P(20<=X<=40) =FX(40)-FX(20-) =4/6-1/6 =3/6 |
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probability density function (pdf): eqn
|
pdf
=fX(x) =dFX(x)/dx =sum(PX(xi)delta(x-xi),i,1,n) |
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four properties of pdf
|
four properties of pdf
1) fX(x)>=0 always positive fu since has positive slope 2)integral(fX(x) ,x,-infinity,infinity) =FX(x),-infinity,infinity =FX(infinity)-FX(-infinity) =1 3)FX(x) = integral(fX(z),z,-infinity,x) 4)P(x1<=X<=x2) =integral(fX(x),x,x1,x2) =FX(x2)-FX(x1) |
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example:
let X be a continuous r.v. with pdf =c*x^2 when 0<=X<=1 =0 elsewhere 1)P(0.5<=X<=0.75) 2)FX(x) |
First find c:
integral(c*x^2,x,0,1)=1 -->c=3 1)P(0.5<=X<=0.75) integral(3x^2,x,0.5,0.75) =0.75^3 - 0.5^3 2)FX(x) = 0 when x<0 =integral(3z^2,z,0,x) when 0<=x<=1 =1 when x>1 |
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Binomial distribution used by
1)probability mass fu 2)PDF 3)pdf |
X=# successes
n=# trials 1)probability mass fu P(X=k)=pX(k) =(n,k)p^k*q^(n-k) 2)PDF = FX(l) =sum(PX(k)u(x-k),k,0,l) =sum((n,k)p^k*q^(n-k)*u(x-k) ,k,0,l) 3)pdf = fX(x) =sum(PX(k)*delta(x-k),k,0,n) |
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example:
toss a coin 10 times and find pdf and PDF for #heads |
X=# heads, n=10
f(x)=sum((10,k)*p^k*q^(10-k) *delta(x-k),k,0,10) F(x) =P(X<=2) =sum((10,k)*p^k*q^(10-k) *u(x-k),k,0,10) |
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Poisson: definition & eqn
|
If X defines number of events happening in an interval of length T then X has a poisson distribution where:
P(X=k) =PX(k) =e^-b*b^k/k! when k=0,1,2,... =0 otherwise b=avg number events in interval T |
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relation binomial and poisson
|
binomial prob fu converges to poisson when n-->infinity and p-->0 in such a way that np=b
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Probability Distribution Function (P.D.F)
(aka Cumulative Distribution Function) |
Px of the random variable X is defined as:
FX(x) = P(X<=x) = P({Si:X(Si)<=x}) for continuous random variables, probability of one specific point is zero |
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example:
toss a die. X(fi)=10i with i=1,2,3,4,5,6 0) FX(x) 1)FX(0) 2)FX(10) 3)FX(20) 4)FX(30) 5)FX(100) 6)P(x<40) |
0) FX(x)=
1/6*U(X-10) +1/6*U(X-20) +1/6*U(X-30) +1/6*U(X-40) +1/6*U(X-50) +1/6*U(X-60) 1)FX(0)=0 2)FX(10) =P(X<=10) =1/6 3)FX(20) =P(X<=20) =2/6 4)FX(30) =P(X<=30) =3/6 5)FX(100) =P(X<=100) =6/6 = 1 6) P(x<40) = 3/6 |
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impulse function/delta function:
1) delta(x) 2) delta(x-a) |
1) delta(x)=
1 when x=0 0 when x!=0 2) delta(x-a)= 1 when x=a 0 when x!=a |
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unit step function:
1) u(x) 2) u(x-a) |
1) u(x)
1 when x>=0 0 when x<0 2) u(x-a)= 1 when x>=a 0 when x<a |
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relation between delta(x) and u(x)
|
delta(x)=du(x)/dx
and u(x) = integral(delta(x),x) |
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relation between PDF and probability mass function
|
F(x)=integral(f(x),x)
f(x)=(d/dx)F(x) probability mass function: f(x) probability distribution fu: F(x) |
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example:
Toss a coin two times and let X be the number of heads observed 1)find the probability mass fu. 2) find the probability distribution fu |
1) f(x)
=PX(x) =(1/4)delta(x) +(1/4+1/4)delta(x-1) +(1/4)delta(x) 2)FX(x) =(1/4)u(x) +(1/4+1/4)u(x-1) +(1/4)u(x-2) |
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integral(
f(x)delta(x-x0) ,x,-infinity,infinity) |
integral(
f(x)delta(x-x0) ,x,-infinity,infinity) =f(x0) |
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short note notation for P(x) and F(x)
|
PX(x)
=f(x) =sum (P(X=xi)delta(x-xi),all xi) FX(x) =sum (P(X=xi)u(x-xi),all xi<=x) |
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Four properties of PDF
|
1)FX(-infinite)
=P(X<-infinite) =P(0)=0 2)FX(infinite) =P(X<infinite) =P(S) =1 3)FX(x) is an increasing fu. of x (positive slope) 4)FX(x)=FX(x+) at the point of discontinuity where: F(x+)=lim(F(x+epsilon),epsilon-->0) {epsilon>0} |
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example:
let X be the # of accidents in one intersection duringone week where avg # of accidants in one week is reported as 2 1)write PDF and pdf of X 2)FX(2) 3)P(2<=X<=4) |
1)write PDF and pdf of X
b=2 fX(x) =sum (PX(xi)delta(x-xi),all xi) =sum(e^-b*b^k/k!*delta(x-k),k,0,infinity) PDF=FX(l) =sum(e^-b*b^k/k!*u(l-k),k,0,l) =sum(e^-2*2^k/k!*u(l-k),k,0,l) 2)FX(2) =sum(e^-2*2^k/k!,k,0,2) =0.675 3)P(2<=X<=4) =sum(e^-2*2^k/k!*,k,2,4) |
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uniform density function: definition and pdf and PDF eqns
|
X is equally likely between a andb
pdf: fX(x) =1/(b-a) a<=x<=b =0 elsewhere PDF: FX(x) =0 x<=a =1 x>=b =(x-a)/(b-a) a<=x<=b |
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example:
randomX uniformly distributed between 2 and 4 1)pdf 2)PDF |
1)fX(x) = 1/(4-2) = 1/2
2)FX(x) = (x-2)/(4-2) =(x-2)/2 |
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example:
R is a random variable for which all values between 80 and 100 are equally likely. All otehr values are impossible. Find: 1) P(R between 90 and 95) 2) P(R between 90 and 95|between 85 and 95) given f(x)=0.05 when 80<R<100 |
1) P(90<R<95)
=(0.05)*(90-95)=0.25 2) P(90<R<95|85<R<95) =P(90<R<95)/P(85<R<95) =[(90-95)*0.05]/[(95-85)*0.05] |
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exponential pdf
|
fx(X)
= (1/b)e^(-x/b) when x>=0 =0 when x<0 FX(x) =1-e^(-x/b) when x>=0 =0 when x<0 |
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applications of exponential pdf
|
*time between arrival of plane
*fluctuations in signal strength received by radar from certain types of aircraft |
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Normal or Gaussian: eqn
|
f(x)
= (1/(sqrt(2Pi)*sigma)) *e^(-(x-mu)^2/(2*sigma^2)) |
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Normal or Gaussian: def
|
1) f(x) is bell shaped and symmetrical around mu
2) f(x) has two parameters: mu=mean sigma=std deviation sigma^2=variance |
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Normal or Gaussian: short notation
|
N(mu,sigma)
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Application of Normal or Gaussian:
|
noise in communication system
|
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example:
let X be the # of accidents in one intersection duringone week where avg # of accidants in one week is reported as 2 1)write PDF and pdf of X 2)FX(2) 3)P(2<=X<=4) |
1)write PDF and pdf of X
b=2 fX(x) =sum (PX(xi)delta(x-xi),all xi) =sum(e^-b*b^k/k!*delta(x-k),k,0,infinity) PDF=FX(l) =sum(e^-b*b^k/k!*u(l-k),k,0,l) =sum(e^-2*2^k/k!*u(l-k),k,0,l) 2)FX(2) =sum(e^-2*2^k/k!,k,0,2) =0.675 3)P(2<=X<=4) =sum(e^-2*2^k/k!*,k,2,4) |
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uniform density function: definition and pdf and PDF eqns
|
X is equally likely between a andb
pdf: fX(x) =1/(b-a) a<=x<=b =0 elsewhere PDF: FX(x) =0 x<=a =1 x>=b =(x-a)/(b-a) a<=x<=b |
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example:
randomX uniformly distributed between 2 and 4 1)pdf 2)PDF |
1)fX(x) = 1/(4-2) = 1/2
2)FX(x) = (x-2)/(4-2) =(x-2)/2 |
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example:
R is a random variable for which all values between 80 and 100 are equally likely. All otehr values are impossible. Find: 1) P(R between 90 and 95) 2) P(R between 90 and 95|between 85 and 95) given f(x)=0.05 when 80<R<100 |
1) P(90<R<95)
=(0.05)*(90-95)=0.25 2) P(90<R<95|85<R<95) =P(90<R<95)/P(85<R<95) =[(90-95)*0.05]/[(95-85)*0.05] |
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exponential pdf
|
fx(X)
= (1/b)e^(-x/b) when x>=0 =0 when x<0 FX(x) =1-e^(-x/b) when x>=0 =0 when x<0 |
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applications of exponential pdf
|
*time between arrival of plane
*fluctuations in signal strength received by radar from certain types of aircraft |
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|
Normal or Gaussian: eqn
|
f(x)
= (1/(sqrt(2Pi)*sigma)) *e^(-(x-mu)^2/(2*sigma^2)) |
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|
Normal or Gaussian: def
|
1) f(x) is bell shaped and symmetrical around mu
2) f(x) has two parameters: mu=mean sigma=std deviation sigma^2=variance |
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Normal or Gaussian: short notation
|
N(mu,sigma)
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Application of Normal or Gaussian:
|
noise in communication system
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Let X be N(mu,sigma) find:
P(x1<=X<=x2)=? |
P(x1<=X<=x2)
=integral(fX(x),x,x1,x2) = integral([1/(sqrt(2Pi)*sigma)]*e^(-Z^2/2),Z,Z1,Z2) =FSN(Z2)-FSN(Z1) with Z=(x-mu)/(sigma) |
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Let X be N(mu,sigma) find:
P(x1<=X<=x2)=? using the short way and three properties |
P(x1<=X<=x2)
=FSN(Z2)-FSN(Z1) with Z=(x-mu)/(sigma) remember 1)FSN(-z)=1-FSN(z) 2)FSN(infinity)=1 3)FSN(-infinity)=0 |
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example:
consider X is N(2,3) find 1) P(1<=X<=5) 2) P(X>=5) |
1) P(1<=X<=5)
=FSN((5-2)/3)-FSN((1-2)/3) =FSN(1)-FSN(-1/3) and from table =0.84134-(1-0.6293)=0.4707 2)P(X>=5) =P(5<=X<infinity) =FSN(infinity)-FSN((5-2)/3) =1-0.8413=0.1586 |
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Joint Probability Mass Fu (Discrete r.v): notation and 2 properties
|
PXY(x,y)=P(X=x and Y=y)
1)PXY(x,y)>=0 2)sum(sum(PXY(xi,yj),all xi) ,all yj)=1 |
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Joint Distribution Function (JPDF): eqn
|
A={X<=x} B={Y<=y}
P(AB) =FXY(x,y) =P(X<=x and Y<=y) |
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Joint Distribution Function (JPDF): 6 properties
|
1)FXY(-infinity,infinity)=0
2)FXY(infinity,infinity)=1 3)FXY(-infinity,y)=0 FXY(x,-infinity)=0 4)FXY(infinity,y)=FY(y) FXY(x,infinity)=FX(x) 5)FXY(x,y)=FXY(x+,y+) continuous toward higher limit for both arguments 6)non-decreasing in each argument |
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example:
findthe joint probability density function and joint distribution fu of X and Y associated with the experiment of tossing one pair of die where X and Y each represent the number appearing on each die |
PXY(X=i,Y=j)
=PXY(i,j) =1/36 1)jpdf =fXY(x,y) =sum(sum(PXY(i,j)*delta(x-i)*delta(y-j)),i,1,6),j,1,6) 2)JPDF =FXY(m,n) =P(X<=m,Y<=n) =sum(sum(PXY(i,j)*u(m-i)*u(n-j),i,1,6),j,1,6) =m*n/36 |
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Joint Probability Density Function: (continuous) eqn
|
fXY(x,y)=[d^2/(dxdy)]FXY(x,y)
where FXY(x,y)= integral(integral( fxy(mu,lambda), mu,-infinity,x), lambda,-infinity,y) |
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Joint Probability Density Function: (continuous) properties {2}
|
1) fXY(x,y)>=0
always positive 2) integral(integral( fXY(x,y) ,x,-infinity,infinity) ,y,-infinity,infinity) =P(S)=1 area under density fu is always equal to one |
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Probabilities in terms of jpdf:
1) P(x1<=X<=x2 and y1<=Y<=y2) 2) fXY(x,y) (discrete) |
1) P(x1<=X<=x2 and y1<=Y<=y2)
=integral(integral( fXY(x,y) ,x,x1,x2) ,y,y1,y2) ie integrating to find volume 2) fXY(x,y)= sum(sum( P(xi,yj) *delta(x-xi)*delta(y-yj) ,i,1,n),j,1,m) |
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example:
consider r.v. X and Y with the following jpdf: fXY(x,y) =kxy when 0<=x<=1 and 0<=y<=1 =0 elsewhere 1)FXY(x,y) 2)P(X<=1/2 and Y<=3/4) 3)P(x<=y) |
first find k:
integral(integral( kxy,x,0,1),y,0,1)=1 -->k=4 1)FXY(x,y) =integral(integral( 4*mu*lambda ,mu,0,x),lambda,0,y) =x^2*y^2 2)P(X<=1/2 and Y<=3/4) =FXY(1/2,3/4) =9/64 3)P(x<=y) =integral(integral( 4xy,x,0,y),y,0,1) =1/2 draw the graph to find integral limitations |
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Marginal Distribution Functions:
FX(x)= FY(y)= |
FX(x)=P(X<=x and Y<infinity)
=FXY(x,infinity) FY(y)=P(X<infinity and Y<=y) =FXY(infinity,y) |
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Marginal Probability Mass Fu:
PX(xi)= PY(yi)= |
PX(xi)=
sum(PXY(xi,yj),all j) PY(yi)= sum(PXY(xi,yj),all i) |
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Marginal Probability Density Fu:
fX(x)= fY(y)= |
fX(x)
=integral( fXY(x,y) ,y,-infinity,infinity) fY(y) =integral( fXY(x,y) ,x,-infinity,infinity) |
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Marginal Probability Density Fu:
find FX from fX |
FX
=integral( fX(mu),mu,-infinity,x) |
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Marginal Probability Density Fu:
find FX from fXY |
FX
=integral(integral( fXY(mu,lambda) ,lambda,-infinity,infinity ,mu,-infinity,x) |
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independence of two random variables:
continuous eqns |
independent iff:
fXY(x,y)=fX(x)*fY(y) |
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independence of two random variables:
discrete eqns |
independent iff:
PXY(x,y)=PX(x)*PY(y) or FXY(x,y)=FX(x)*FY(y) or f(x|y)=f(x) f(y|x)=f(y) |
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example:
fXY(x,y) =8xy when 0<=x<=1 and 0<=y<=x =0 elsewhere 1)P(x<=1/2) 2)f(y) 3)f(x) 4)are X and Y independent? |
1)P(x<=1/2)
integral(f(x),x,0,1/2) =integral(integral( f(x,y),y,0,x),x,0,1/2) =1/16 2)f(y) =integral(fXY(x,y),x,y,1) =integral(8xy,x,y,1) =4y*(1-y^2) 3)f(x) =integral(fXY(x,y),y,0,x) =4x^3 4)are x and y independent? NO b/c fXY(x,y) != f(x)f(y) |
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example:
X and Y are jointly distributed with FXY(x,y)=(1/6)[x^2*y+x*y^2] 0<=x<=1 0<=y<=2 1)FX(x) 2)FY(y) 3)fXY(x,y) 4)fX(x) 5)fY(y) 6)are X and Y independent? |
1)FX(x)
=FXY(x,2) =(1/3)*[x^2+2x] 2)FY(y) =FY(y) =FXY(1,y) =(1/6)[y+y^2] 3)fXY(x,y) =(d^2/(dxdy))FXY(x,y) =(1/3)(x+y) 4)fX(x) =integral(fXY(x,y),y,0,2) =(2/3)(x+1) or =dF(x)/dx=(2/3)(x+1) 5)fY(y) =integral(fXY(x,y),x,0,1) 6)are X and Y independent? NO b/c f(x,y)!= f(x)f(y) |
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example (discrete):
assume the joint sample space XY has only three possible values: P(1,1)=0.2 P(2,1)=0.3 P(3,3)=0.5 1)jpdf 2)JPDF 3)FX(x) 4)FY(y) |
1)jpdf
=f(x,y) =0.2*delta(x-1)*delta(y-1) +0.3*delta(x-2)*delta(y-1) +0.5*delta(x-3)*delta(y-3) 2)JPDF =F(x,y) =0.2*u(x-1)*u(y-1) +0.3*u(x-2)*u(y-1) +0.5*u(x-3)*u(y-3) 3)FX(x) =FXY(x,infinity) =0.2*u(x-1) +0.3*u(x-2) +0.5*u(x-3) 4)FY(y) =FXY(infinity,y) =0.2*u(y-1) +0.3*u(y-1) +0.5*u(y-3) |
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Conditional Distribution Function:
FX(x|B)=? |
FX(x|B)
=P(X<=x and B)/P(B) |
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Conditional Distribution Function:
properties {4} |
1)FX(-infinity|B)=0
2)FX(infinity|B)=0 3)non decreasing fu of X 4)FX(x|B)=FX(x+|B) discontinues at point x |
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Conditional Density Function:
fX(x|B)=? |
fX(x|B)
=dF(x|B)/dx |
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Conditional Distribution Function:
properties {2} |
1) fX(x|B)>=0
density fu always positive 2)integral(fX(x|B),x, -infinity,infinity)=1 area under density fu always equal to one |
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FX(x|B)
in terms of fX(x|B) |
FX(x|B)
=integral( fX(mu|B) ,mu,-infinity,X) |
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P(x1<=X<=x2|B)
in terms of fX(x|B) |
P(x1<=X<=x2|B)
=integral( fX(x|B),x,x1,x2) |
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Conditional Distribution and Desnity functions:
5 cases of defining event B |
case 1:
B={x1<=X<=x2} case 2: B={X<=x1} case 3: B={y1<=Y<=y2} case 4: B={Y=y} case 5: B={Y<=y} |
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case 1:
B={x1<=X<=x2} FX(x|B)=? fX(x|B)=? |
case 1:
B={x1<=X<=x2} FX(x|B) =0 when X<x1 =[FX(x)-FX(x1)]/P(B) when x1<=X<=x2 =1 when X>=x2 fX(x|B) =0 when X<x1 =fX(x)/P(B) when x1<=X<=x2 =1 when X>=x2 where P(B) =integral(f(x),x,x1,x2) |
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case 2:
B={X<=x1} FX(x|B)=? fX(x|B)=? |
case 2:
B={X<=x1} FX(x|B) =F(x)/P(B) when X<x1 =1 when X>=x1 fX(x|B) =f(x)/P(B) when x<x1 =0 when x>=x1 |
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case 3: interval conditioning on Y
B={y1<=Y<=y2} FX(x|B)=? fX(x|B)=? |
case 3:
B={y1<=Y<=y2} FX(x|B) =P(X<=x and y1<=Y<=y2)/P(B) =[integral(integral( f(x,y) ,x,-infinity,x) ,y,y1,y2)]/ [integral(f(y),y,y1,y2)] fX(x|B) =[integral( f(x,y),y,y1,y2]/ [P(B)] =[integral( f(x,y),y,y1,y2]/ [integral(f(y),y,y1,y2)] |
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case 4: point conditioning
B={Y=y} FX(x|B)=? fX(x|B)=? |
case 4:
B={Y=y} FX(x|B) =[integral(fXY(mu,lambda) ,mu,-infinity,x)]/fY(y) fX(x|B)=fX|Y(x|y) =f(x|Y=y) =f(x,y)/f(y) |
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case 5:
B={Y<=y} FX(x|B)=? fX(x|B)=? |
F(x|B)
=F(x,y)/F(y) f(x|B) =(1/F(y))*(dF(x,y),x) |
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example:
consider two r.v. with jpdf defined as: fXY(x,y)=k(x+y) with 0<=x<=1 and 0<=y<=2 determine: 1)f(x|y=1) 2)F(x|y=1) 3)f(y|B) where B={1<=Y<=2} 4)f(x|y<=1) 5)F(x|y<=1) |
first find k:
integral(integral( k*(x+y),x,0,1),y,0,1)=1 -->k=1/3 1)f(x|y=1) =f(x,1)/f(y) =[1/3(x+y)]/ [integral(1/3(x+y),y,0,1)] =(x+y)/((1/2)+y) -->f(x|y=1)=(2/3)(x+1) 2)F(x|y=1) =integral(fX|Y(mu|y),mu,0,x) =integral( [mu+y]/[1/2+y],mu,0,x) =[0.5x^2+xy]/[0.5+y] -->F(x|y=1)=[0.5x^2+x]/[1.5] 3)f(y|B) where B={1<=Y<=2} =f(y)/P(1<=Y<=2) =(1/4)+(1/2)y where P(1<=Y<=2) =integral(f(y),y,1,2)=2/3 4)f(x|y<=1) =(d/dx)F(x|y<=1)=x+1/2 5)F(x|y<=1) =FXY(x,1)/FY(1) =[integral(integral( fXY(mu,lambda) ,mu,0,x),lambda,0,y)]/ [FXY(1,y)] -->(x^2+x)/2 |
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how to get marginal from conditional jpdf:
f(x) P(xi) |
X is continuous:
f(x) =integral( f(x|y)f(y) ,y,-infinity,infinity) X is discrete: P(xi)= sum( PX|Y(xi|yj)PY(yj),all j) |
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example: let Y=1,2,3 depend on the number of coins tossed. let X be the number of heads. given:
PY(1)=1/4, PY(2)=1/4, PY(3)=1/2, and that the coins are fair coins, also i=0,1,2,3 find PX(i)= 1)PX(0) 2)PX(2) |
1) step 1: find conditionals
P(0|1)=1/2 P(0|2)=1/4 P(0|3)=1/8 step 2: PX(0)=sum( P(0|j)*PY(j),j,1,3) =(1/2)(1/4)+(1/4)(1/4) +(1/8)(1/2) =1/4 2) step 1: find conditionals P(2|2)=1/4 P(2|3)=3/8 hint: can also do these using binomial --> P(2|3)=(3,2)(1/2)^2*(1/2)^1 step 2: PX(2)=sum( P(2|j)*PY(j),j,1,3) =4/16 |
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random vectors:
(using n=4) |
PDF:
FX1X2X3X4(x1,x2,x3,x4) =P{(X1<=x1,X2<=x2, X3<=x3,X4<=x4)} pdf: fX1...Xn =[d^n/(dx1...dxn)]* [FX1...Xn(x1...xn)] fX1X2(x1,x2)= integral(integral( fX1X2X3X4(x1,x2,x3,x4) ,x3,-infinity,infinity) ,x4,-infinity,infinity) |
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random vectors:
conditional density fX1X2|X3X4(x1x2|x3x4) |
fX1X2|X3X4(x1x2|x3x4)
=fX1X2X3X4(x1x2x3x4)/ fX3X4(x3x4) |
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example:
given F(x1,x2,x3) and f(x1,x2,x3) find marginals F(x1) f(x1) |
F(x1)=F(x1,infinity,infinity)
f(x1)=integral(integral( f(x1,x2,x3) ,x2,-infinity,infinity) ,x3,-infinity,infinity) |
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example:
given f(x1,x2,x3) find f(x1|x2x3) |
f(x1|x2x3)
=f(x1,x2,x3)/f(x2,x3) |
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independent random variables:
if A={X<=x} and B={Y<=y} what eqn determines the two are independent? |
FXY(x,y)=FX(x)FY(y)
or fXY(x,y)=fX(x)fY(y) f(x|y)=f(x) |
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example:
given X and Y are two independent r.v. and f(x)=2x 0<=x<=1 f(y)=2y 0<=y<=1 1)f(x,y) |
1)f(x,y)
=f(x)f(y) =4xy 0<=x<=1 0<=y<=1 |
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example:
given X and Y are two independent r.v. and f(x)=2x 0<=x<=1 f(y)=2y 0<=y<=1 2)P{max(x,y)<=1/2} 3)P{min(x,y)<=1/2} |
2)P{max(x,y)<=1/2}
to find max: graph x=y max(x,y) =x when x>y =y when x<y P(max(x,y)<=1/2) =integral(integral( 4xy,x,0,1/2),y,0,1/2) =1/16 3)P{min(x,y)<=1/2} min(x,y) =y when x>y =x when x<y P(min(x,y)<=1/2) =1-P(min(x,y)>1/2) =1-integral(integral( f(x,y),x),y) =1-integral(interal( 4xy,x,1/2,1),y,1/2,1) =1-9/16 =7/16 |
