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198 Cards in this Set

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P(AUB)
P(A)+P(B)-P(AB)
P(AUB) if A and B are disjoint
P(A) + P(B)
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
P(A')
1-P(A)
P(A-B)
P(A-B)=P(AB')
= P(A) - P(AB)
conditional probability
P(A|B)
P(A|B)
=P(AB)/P(B)
joint probability
P(AB)
P(AB)
=P(A|B)P(B)
=P(B|A)P(A)
total or marginal probability
P(A)
sum(P(ABi),i,1,n)
=sum(P(A|Bi)P(Bi),i,1,n)
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)
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
P(A'|B)
P(A'|B)
= 1-P(A|B)
P((AUC)|B)
P((AUC)|B)
= P(A|B) + P(C|B) - P(AC|B)
disjoint
AnC=0
P(AB)
P(AB)
=P(A|B)P(B)
=P(B|A)P(B)
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)
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)}
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
geometric interpolation:
table to see possible outcomes of combined experiment

a1 a2 ... an
b1 a1b1 a2b2 ...
b2 a1b2 ...
.
.
.
bm a1bm a2bm ... anbm
example:
number of outcomes if toss a coin 3 times
2*2*2=8
example:
number of outcomes if roll a six headed dice three times
6*6*6
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
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)
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
Bayes' Theorem
P(Bj|A)
P(Bj|A)
=[P(ABj)/P(Bj)]/P(A)
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
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))
P(A'|B)
P(A'|B)
= 1-P(A|B)
Tree Diagrams:
Useful graphical tool for defining the basic outcomes of a combined experiment. Particularly when events are dependent.
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
P(A|BC)
P(A|BC)
= P(ABC)/P(BC)
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)}
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
geometric interpolation:
table to see possible outcomes of combined experiment

a1 a2 ... an
b1 a1b1 a2b2 ...
b2 a1b2 ...
.
.
.
bm a1bm a2bm ... anbm
example:
number of outcomes if toss a coin 3 times
2*2*2=8
example:
number of outcomes if roll a six headed dice three times
6*6*6
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
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)
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
Bayes' Theorem
P(Bj|A)
P(Bj|A)
=[P(ABj)/P(Bj)]/P(A)
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
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))
P(A'|B)
P(A'|B)
= 1-P(A|B)
Tree Diagrams:
Useful graphical tool for defining the basic outcomes of a combined experiment. Particularly when events are dependent.
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
P(A|BC)
P(A|BC)
= P(ABC)/P(BC)
P(ABC)
P(ABC)
=P(A|BC)P(BC)
=P(A|BC)P(B|C)P(C)
equations to test independence
1) P(A|B) = P(A)
2) P(AB) = P(A)P(B)
example:
toss a coin twice
P(H2|H1)=?
P(H2|H1)=P(H2)
independent events
conditionally independent
P(AB|C)
= P(A|C)P(B|C)
independent events: definition
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.
Three events A, B, and C are independent iff:
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)
total # eqns required to establish n events are independent
2^n - (n+1)
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
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
if n events are independent from eachother, what else can we assume they are independent from?
any one event Ai is independent of any event formed by unions, intersections, and complements of the other events.
Cartesian Product of sets A and B
C=A*B
Number of Elements in Cartesian Product of sets A and B (A has m elements, B has n elements)
# elements of aibj = m*n
Permutation: definition
Permutation:
An ORDERED arrangement of k distinct objects taken from n distinct objects without replacement
Combination: definition
NON-ORDERED combination of n objects taken k at a time
Difference between Permutation and Combination
Permutation order matters.
Combination order does not matter.
P(n,k)
P(n,k)
=n!/(n-k)!
P(n,n)
P(n,n)
=n!
C(n,k)
C(n,k)=
(n
k)
=n!/(k!*(n-k)!)
Combination: definition
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.
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
binomial coefficient
(n
k)=(n,k)
binomial expansion
(p+q)^n
=sum((n,k)p^k*q^(n-k),k,0,n)
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
generalized combination:
equation
C(n,(n1,n2,...,nk))
= n!/(n1!n2!...nk!)
Bernoulli Trials (aka Binomial Experiment): definition
*n identical subexperiments
*Each sub experiment has two outcomes: success or failure
*subexperiments are independent
*each subexperiment has identical probabilities
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)
Bernoulli Trial:
P(k1<=k<=k2)
Bernoulli Trial:
P(k1<=k<=k2)
= sum((n,k)p^k
*q^(n-k),k,k1,k2)
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
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
reliability: definition
reliability: definition --
reliability of a system is probability of success of that system.
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)
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
elementery systems:
parallel
|---[A1]---|
---|---[A2]---|---
|---[A3]---|

R=?
Q=?
R
=1-Q
=1-qA1*qA2*qA3

Q
=qA1*qA2*qA3
Real Random Variable: definition
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.
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
Discrete Random Variable: definition
Discrete Random Variable: definition --
Can take a discrete (countable) numbers of values.
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
Continuous Random Variable: definition
Continuous Random Variable: definition --
can take all values within a specified interval. Can NOT result from discrete sample space.
Probability Mass Function: discrete random variables:
equation
PX(x)
=P(X=x)
=P({Si:X(Si)=x})
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)
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
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
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
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
relation between delta(x) and u(x)
delta(x)=du(x)/dx
and
u(x) = integral(delta(x),x)
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)
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)
integral(
f(x)delta(x-x0)
,x,-infinity,infinity)
integral(
f(x)delta(x-x0)
,x,-infinity,infinity)
=f(x0)
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)
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}
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
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')
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
probability density function (pdf): eqn
pdf
=fX(x)
=dFX(x)/dx
=sum(PX(xi)delta(x-xi),i,1,n)
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)
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
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)
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)
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
relation binomial and poisson
binomial prob fu converges to poisson when n-->infinity and p-->0 in such a way that np=b
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)
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
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
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]
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
applications of exponential pdf
*time between arrival of plane
*fluctuations in signal strength received by radar from certain types of aircraft
Normal or Gaussian: eqn
f(x)
= (1/(sqrt(2Pi)*sigma))
*e^(-(x-mu)^2/(2*sigma^2))
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
Normal or Gaussian: short notation
N(mu,sigma)
Application of Normal or Gaussian:
noise in communication system
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
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')
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
probability density function (pdf): eqn
pdf
=fX(x)
=dFX(x)/dx
=sum(PX(xi)delta(x-xi),i,1,n)
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)
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
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)
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)
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
relation binomial and poisson
binomial prob fu converges to poisson when n-->infinity and p-->0 in such a way that np=b
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
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
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
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
relation between delta(x) and u(x)
delta(x)=du(x)/dx
and
u(x) = integral(delta(x),x)
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)
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)
integral(
f(x)delta(x-x0)
,x,-infinity,infinity)
integral(
f(x)delta(x-x0)
,x,-infinity,infinity)
=f(x0)
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)
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}
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)
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
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
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]
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
applications of exponential pdf
*time between arrival of plane
*fluctuations in signal strength received by radar from certain types of aircraft
Normal or Gaussian: eqn
f(x)
= (1/(sqrt(2Pi)*sigma))
*e^(-(x-mu)^2/(2*sigma^2))
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
Normal or Gaussian: short notation
N(mu,sigma)
Application of Normal or Gaussian:
noise in communication system
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)
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
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
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]
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
applications of exponential pdf
*time between arrival of plane
*fluctuations in signal strength received by radar from certain types of aircraft
Normal or Gaussian: eqn
f(x)
= (1/(sqrt(2Pi)*sigma))
*e^(-(x-mu)^2/(2*sigma^2))
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
Normal or Gaussian: short notation
N(mu,sigma)
Application of Normal or Gaussian:
noise in communication system
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)
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
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
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
Joint Distribution Function (JPDF): eqn
A={X<=x} B={Y<=y}

P(AB)
=FXY(x,y)
=P(X<=x and Y<=y)
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
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
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)
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
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)
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
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)
Marginal Probability Mass Fu:
PX(xi)=
PY(yi)=
PX(xi)=
sum(PXY(xi,yj),all j)

PY(yi)=
sum(PXY(xi,yj),all i)
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)
Marginal Probability Density Fu:
find FX from fX
FX
=integral(
fX(mu),mu,-infinity,x)
Marginal Probability Density Fu:
find FX from fXY
FX
=integral(integral(
fXY(mu,lambda)
,lambda,-infinity,infinity
,mu,-infinity,x)
independence of two random variables:
continuous eqns
independent iff:
fXY(x,y)=fX(x)*fY(y)
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)
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)
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)
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)
Conditional Distribution Function:
FX(x|B)=?
FX(x|B)
=P(X<=x and B)/P(B)
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
Conditional Density Function:
fX(x|B)=?
fX(x|B)
=dF(x|B)/dx
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
FX(x|B)
in terms of fX(x|B)
FX(x|B)
=integral(
fX(mu|B)
,mu,-infinity,X)
P(x1<=X<=x2|B)
in terms of fX(x|B)
P(x1<=X<=x2|B)
=integral(
fX(x|B),x,x1,x2)
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}
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)
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
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)]
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)
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)
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
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)
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
random vectors:
PDF
pdf

(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)
random vectors:
conditional density
fX1X2|X3X4(x1x2|x3x4)
fX1X2|X3X4(x1x2|x3x4)
=fX1X2X3X4(x1x2x3x4)/
fX3X4(x3x4)
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)
example:
given f(x1,x2,x3)
find f(x1|x2x3)
f(x1|x2x3)
=f(x1,x2,x3)/f(x2,x3)
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)
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
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