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### 88 Cards in this Set

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 Expected Value (definition) Process of averaging when random variable is involved Expected Value EX[g(x)] (continuous equation) EX[g(x)] = integral(g(x)f(x)dx,-inf.,inf.) Expected Value EX[g(x)] (discrete equation) EX[g(x)] = sum(g(xi)P(xi),all i) Mean same as expected value but when g(x)=X. 1)mean is weighted average value of all xi values 2) the mean is center of gravity of the probability density function different notations for mean E[X] =muX _ =X Mean (discrete equation) sum(xi*PX(xi),all i) Mean (continuous equation) integral(x*fX(x)dx,-inf.,inf.) Variance (definition) spread of r.v. around mean values. average values for error. Larger variance = larger spread. Smaller variance means closer around the center. Variance (notation) var[X] =sigmaX^2 Variance (discrete equation) =EX[(X-muX)^2] =sum((xi-muX)^2*P(xi),all i) 4 properties of variance 1) variance is a measure of the spread about the mean value 2) variance is a positive value 3) variance is moment of inertia of the P.d.f. about the mean 4) if σX²=0 then P(X=μX)=1 (ie X is a constant) If probability density fu is even symmetric about x=a then μX=a since f(a-x)=f(a+x) example: random variable X is uniformly distributed on interva [a,b], findmean and variance of random variable X find: mean and variance mean=μX=E[X]= ∫(x*(1/(b-a)),x,a,b)=(a+b)/2 <-- midpoint var[X]=σX² = E[(x-μX)²] =∫((x-μX)²*(1/(b-a)),x,+a,b) =(1/12)*(b-a)² example *** you are playinga game with a friend and throwing a pair of dice, if the sum of the up faces is a prime number you win corresponding number of dollars. and if it is not a prime number you lose corresponding number of dollars. how much do you expect to win or to lose? X={2,3,4,...,12} E[X]=∑(xi*P(xi),all i) = (1/36)*2+(2/36)*3-(3/36)*4+(4/36)*5-(5/36)*6+(6/36)*7-(5/36)*8-(4/36)*9-(3/36)*10+(2/36)*11-(1/36)*12 =-1.89 LOSE example: given X={2,3,5,6} f(x)=0.2*∂(x-2)+0.1*∂(x-3)+0.4*∂(x-5)+0.3*∂(x-6) find: mean and variance _ X =∑(xi*P(xi),all i) =2*0.2+3*0.1+5*0.4+6*0.3 =4.5 σ²=E[(x-4.5)²] =∑(xi*P(xi),all i) =(2-4.5)²*0.2+(3-4.5)²*0.1+(5-4.5)²*0.4+(6-4.5)²*0.3 =2.25 mode (definition) value of X where f(x) is maximum. for normal distribution function mode and mean are the same quantities median (definition) the value of X such as xm where: FX(xm)=P(X≤xm) =∫(f(x),x,-∞,xm) =1/2 Standard Deviation (def and eqn) positive square root of variance of random variable sdv[X]=σX=√(σX²) moments (around origin) mk =E[X^k] =∫(x^k*f(x),x,-∞,∞) (continuous) =∑(xi^k*P(xi),all i) (discrete) central moments (definition) moment around mean values Central Moments (equation) λk =E[(x-μX)^k] =∫((x-μX)^k*f(x),x,-∞,∞) (continuous) =∑((x-μX)^k*P(xi),all i) (discrete) m1 μ=mean first central moment is the mean m2 E[X²] =average power =mean squared value λ2 =σX² =variance λ3 =E[(x-μX)³] which is a measure of the asymmetry of f(x) about mean. λ3/σX³ normalized third central moment is called skewness or coefficient of skewness λ3 if p.d.f. is symmetric about x=μX λ3=0 EX[c] where c is a constant EX[c]=c E[X1+X2] E[X1+X2] =E[X1]+E[X2] when does E[X1*X2] =E[X1]*E[X2} for two independent r.v. X1 and X2 E[X1*X2] for two independent r.v. X1 and X2 E[X1*X2]=E[X1]*E[X2] under what conditions does E[(X1+X2)²]=E[X1²]+E[X2²] when they are both independent random variables whose mean values are zero variance (another equation) var[X] =σX² =E[(x-μX)²] =E[X²]-μX² exmaple consider two random variables X and Y with the following relation: Y=aX+b find: mean and variance of Y in terms of mean and variance of X mean: μY=E[Y]=E[aX+b]=E[aX]+E[b]=a*μX+b variance: σY²=E[Y²]-μY² with: E[Y²]=E[(aX+b)²]=a²E[X²]+b²+2abE[X] and μY² defined above thus: σY² =( a²E[X²]+b²+2abE[X])-(a*μX+b)² =a²*σX² consider bernouli r.v. X whihc takes two values of 0 or 1 with: P(x=1)=p and P(x=0)=q find mean and variance mean: μX=E[X]=∑(xi*P(xi),all i)=1*p+0*q=p variance: σX²=E[X²]-μX² with second moment m2=E[X²] =∑(xi²*P(xi),all i)=1²*p+0²*q=p thus σX²=p-p²=p*(1-p)=p*q example: for binomial random variable X calculate the mean and variance [recall P(x=k)=(n,k)p^k*q^(n-k) with k=0,1,...,n] mean: μX=E[X] =∑(xi*P(xi),all i) =∑(k*PX(k),k,0,n) =∑(k*(n,k)p^k*q^(n-k),k,0,n) =∑(k*{n!/[(n-k!)*k!]}p^k*q^(n-k),k,0,n) =n*p*∑({(n-1)!/[(n-k!)*(k-1)!]}*p^(k-1)*q^(n-k),k,1,n) =n*p*∑((n-1,k-1)p^(k-1)*q^(n-k),k,1,n) =n*p*(p+q)^(n-1)=np variance: m2=E[X²] =∑(k²*(n,k)p^k*q^(n-k),k,0,n) =n*(n-1)*p²*+np σX² =m2-m1² =[n*(n-1)*p²*+np]-[np]² =np(1-p) =npq mean and variance of binomial random variable X mean=np variance=npq example: a fair coin is flipped ten times what is the mean and variance? mean=5 variance=2.5 in which conditions would we use moments of normal for zero mean gaussian random variable X or for gaussian random variable X with mean μX moments of normal equation for r.v. X=N(0,σ) (ie zero mean gaussian r.v. X) nth moment mn= when n is even: =(1)(3)…(n-1)*σX^n when n is odd: =0 moments of normal equation for gaussian random variable X with mean μX and variance σX² λn=E[(x-μX)^n] when n is even: =(1)(3)…(n-1)*σX^n when n is odd: =0 Chebychev's Inequality upper band for any arbitrary density function f(x) P(|x-μX|>k)≤σX²/k² which gives the upper band on probability of x deviated from mean by k Chebychev's Inequality lower band for any arbitrary density function f(x) P(|x-μX|≤k)≥ 1-σX²/k² ie 1- P(|x-μX|>k)≥1-σX²/k² which gives the lower band on probability of x deviated from mean by k example consider r.v. X with σX²=3 1)find upper band on the r.v. X deviated from its mean by 2 2)find lower band on probability of X to be in the range of 2 from its mean 1) P(|x-μX|>2)≤σX²/2² -->P(|x-μX|>2)≤ 3/4 2) P(|x-μX|≤2)≥ 1-σX²/k² --> P(|x-μX|≤2)≥ 1/4 characteristic function (symbol and equation) ΦX(ω) =E[exp(jωx)] =∫(exp(jωx)*f(x),x,-∞,∞) (x is continuous) =∑(exp(j*ω*xi)*PX(xi),all xi) (x is discrete) relation of ΦX(ω) to the probability density fu. ΦX(ω) is the "fourier transform" of the probability density function with sign change on ω as: FX(ω)=F.T.(fX(x))=ΦX(-) inversion formula (uses finds pdf from characteristic eqn inversion formula (eqn) fX(x)=1/(2*π) * ∫( ΦX(ω)*exp(-jωx),ω,-∞,∞) Uses of characteristic eqn 1) determination of higher order moments 2)determination of densities of sums of independent ranom ΦX(0) ΦX(0)=1 |ΦX(ω)| |ΦX(ω)|≤1 euler’s eqn |exp(jωx)|=|cos(ωx)+jsin(ωx)|=√(cos²ωx + sin²ωx)=1 if X and Y are two independent r.v. and Z=X+Y then ΦZ(ω)=? ΦZ(ω)=ΦX(ω)*ΦY(ω) if Y=aX+b ΦY(ω)=? ΦY(ω)=exp(jωb)ΦX(aω) under what conditions will ΦZ(ω)=ΦX(ω)*ΦY(ω) Z=X+Y with X and Y as independent random variables moments in terms of characteristic function: mn=? mn=E[X^n]=(1/j^n)*nth_derivative(ΦX(ω),ω)|ω=0 j^n*E[X^n]=? nth_derivative(ΦX(ω),ω)|ω=0 example: calculate the c.f. for poisson r.v. X using c.f. find mean and variance of X [recall poisson eqn: P(X=k)=exp(-b)*b^k/k! with k=0,1,...] characteristic function: ΦX(ω) =E[exp(jωx)] =∑(exp(jωk)*exp(-b)*b^k/k =exp[b*(exp(jω)-1)] mean: E[X]=(1/j)*∂(ΦX(ω),ω)|(ω=0) =(1/j)∂(exp[b*(exp(jω)-1)]) variance: E[X²]=2nd_derivative(ΦX(ω)|(ω=0)) =b+b² thus σX²=b+b²-b²=b example: find c.f. for standard normal r.v. [ie N(0,1)] ΦX(ω) =E[exp(jωx)] =∫(exp(jωx)*f(x),x,-∞,∞) =(1/√(2*π))*∫(exp(jωx)*exp(-x²/2),x,-∞,∞) =exp(-ω²/2) example: find c.f. for normal r.v. with mean μ and variance σ² by using the following transformaton Y=σX+μ c.f. of Y is: ΦY(ω) =exp(jωμ)*ΦX(σω) =exp(jωμ)*exp((jω)²/2) =exp(jωμ-ω²σ²/2) example: find mean and variance of Y for standard normal r.v. [ie N(0,1)] using ΦY(ω) E[Y] =(1/j)*∂(Φ(ω),ω)|(ω=0) =(1/j)*(jμ-2σ²ω/2)*ΦY(ω)|(ω=0) =μ E[Y²] =(1/j²)*2nd_derivative(Φ(ω),ω)|(ω=0) =σ²-μ² thus σ²=E[Y²]-{E[Y]}² =σ²-μ²+μ² =σ² exercise: find c.f. for geometric r.v.with the following probability mass function calculate mean and variance using c.f. geometric probability: P(X=k)=p*q^(k-1) k=1,2,... ΦX(ω) =∑(exp(jωk)*p*q^(k-1),k,1,∞) =p*exp(jω)/(1-q*exp(jω)) since |exp(jω)*q|<1: E[X] =(1/j)*∂(ΦX(ω),ω)|(ω=0) =1/p σX²=p/q² expected value of a function of random variables E[g(x1,…,xn)] =∫…(∫(g(x1…xn)*f(x1,…,xn),x1,-∞,∞)…,xn,-∞,∞) if g(x1,…,xn)= ∑(ai*xi,i,1,n) then _ g=E[g(x1…xn)] =∑(ai*E[xi],i,1,n) ie the mean value of a weighted sum of r.v.s equals the weighted sum of mean values joint moments of two r.v. mnk=? mnk =E[x^n*y^k] =∫(∫((x^n)*(y^k)*f(x,y),x,-∞,∞),y,-∞,∞) order of joint moment n+k joint moments of two r.v. m01 _ m01=Y joint moments of two r.v. m10 _ m10=X joint moments of two r.v. m11 m11 =E[xy] =Rxy =∫(∫(xy*f(xy),x,-∞,∞),y,-∞,∞) correlation (notation) m11 =E[xy] =Rxy uncorrelated (definition) Rxy=E[X]E[Y] or if two random variables are independent orthogonal r.v. (definition) Rxy=0 if two r.v. are independent they are also ____ if two r.v. are independent they are also uncorrelated. (Vice-versa is not true in general.) joint central moments (notation) λnk =E[(X-μX)^n*(Y-μY)^k] joint central moments (equation) λnk =E[(X-μX)^n*(Y-μY)^k] =∫(∫( [(x-μX)^n*(y-μY)^k]*f(x,y),x,-∞,∞), y,-∞,∞) λ20=? λ02=? λ20=E[(x-μx)²]=σx² λ02=E[(y-μy)²]=σy² covariance (definition) covariance of X and Y is defined by σxy or Cxy and defined as: Cxy =E[(x-μX)*(y-μY)] =E[XY]-μX*μY if Cxy=0 then X and Y are uncorrelated when X and Y are two independent r.v. the covariance is Cxy=0 thus the two variables are uncorrelated λ11 λ11=Cxy correlation coefficient (definition) correlation coefficient ρxy is a measurement of linear dependency of two r.v. X and Y correlation coefficient (eqn) ρxy=Cxy/(σx*σy) properties of correlation coefficient (2) 1)|ρxy|≤1 2)if Y=aX+b with a and b as constants, then ρxy=1 if a>0 ρxy=-1 if a<0 relationship of correlation and independence if two r.v. are independent they are also uncorrelated. Howerver, the converse is not necessarily true. what does the correlation between r.v. X and Y measure? measures their tendency to lie in a straight line example: consider y=x² when X is N(0,1). X and Y are dependent. find covariance. covariance: Cxy =[XY]-μx*μy =E[X³]-0*μy =E[X³] =0 (b/c odd moment) thus X and Y are uncorrelated but certainly not independent conditional expectation E[g(x)|B]=? E[g(x)|B] =∫(g(x)*f(x|B),x,-∞,∞) =∑(g(xi)*P(X=xi|B),all i) conditional expectation (notation) E[g(x)|B] =EX|B[g(x)|B] conditional mean value E[x|B] =∫(x*f(x|B),x,-∞,∞) =∑(xi*P(X=xi|B),all i)