Joint ProbabilityLecture 3Spring 2002Joint Probability Distribution FunctionFX1X2(x1,x2)=P [(X1≤ x1) ∩ (X2≤ x2)]1. FX1X2(−∞, −∞)=02. FX1X2(−∞,x2)=0for any x23. FX1X2(x1, −∞)=0foranyx14. FX1X2(+∞, +∞)=15. FX1X2(+∞,x2)=FX2(x2)foranyx26. FX1X2(x1, +∞)=FX1(x1)foranyx1Lecture 2 1Joint Probability Distribution FunctionThe probability that an experiment produces a pair (X1,X2)thatfalls in a rectangular region with lower left corner (a, c) and upperright corner (b, d)isP [(a<X1≤ b) ∩ (c<X2≤ d)] = FX1X2(b, d) − FX1X2(a, d) −FX1X2(b, c)+FX1X2(a, c)Lecture 2 2Joint Probability Density FunctionfX1X2(x1,x2)=∂2FX1X2(x1,x2)∂x1∂x2fU,V(u, v) ≥ 0FU,V(u, v)=u−∞v−∞fU,V(ξ, η)dξdη∞−∞fU,V(ξ, η)dξdη =1FU(u)=u−∞∞−∞fU,V(ξ, η)dξdηFV(v)=∞−∞v−∞fU,V(ξ, η)dξdηfU(u)=∞−∞fU,V(u, η)dηfV(v)=∞−∞fU,V(ξ, v)dξLecture 2 3Example;CONSTRUCT X AND Y ARRAYS FOR THE f(X,Y) pdft=FindGen(301)Meshdom,t,t,X,Y;CONSTRUCT AND DISPLAY THE PDFf=6400-((x-150)^2+(y-150)^2 < 6400)f=f/Total(f) Window,/FreeShade_Surf,fWshowLecture 2 4Example (continued);CALCULATE AND DISPLAY THE CDFFC=FltArr(301,301)FC[*,0]=CumSum(f[*,0])FC[0,*]=CumSum(f[0,*])For ky=1,300 DO $For kx=1,300 DO $FC[kx,ky]=FC[kx-1,ky]+ $FC[kx,ky-1]-FC[kx-1,ky-1]+f[kx,ky]Window,/Free Shade_Surf,FC WshowLecture 2 5Example (continued);DISPLAY A CONTOUR PLOT OF THE CDFWindow,/FreeContour,FC,Levels=Findgen(21)/20WshowLecture 2 6Conditional ProbabilitiesP [U ∈A,V ∈B]=ξ∈Aη∈BfU,V(ξ, η)dξdηThe probability that V ∈Bwithout regard to the value of U is givenbyP [V ∈B]=P [U ∈U,V ∈B]=∞ξ=−∞η∈BfU,V(ξ, η)dξdηLecture 2 7Conditional ProbabilitiesP [U ∈A|V ∈B]=P [U∈A,V ∈B]P [V ∈B]=ξ∈Aη∈BfU,V(ξ, η)dξdη∞ξ=−∞η∈BfU,V(ξ, η)dξdηprovided P [V ∈B] > 0.Lecture 2 8Conditional Probability Distribution FunctionFU(u|V ∈B)=P [U ≤ u|V ∈B]whenever P [V ∈B] > 0.The conditional probability distribution function has all of the prop-erties of an ordinary one-dimensional probability distribution func-tion. That is, it is a nondecreasing function with FU(−∞|V ∈B)=0and FU(∞|V ∈B)=1.Lecture 2 9Conditional Probability Density FunctionfU(u|V ∈B)=dFU(u|V ∈B)duwherever the derivative exists.Lecture 2 10Statistically Independent Random VariablesTwo random variables, U and V are statistically independent randomvariables if and only ifP [U ∈A,V ∈B]=P [U ∈A]P [V ∈B]for every possible choice of A and B.Lecture 2 11Statistically Independent Random VariablesIf U and V are statistically independent random variables thenP [U ≤ u, V ≤ v]=P [U ≤ u]P [V ≤ v]and this is equivalent to the following statement in terms of thedistribution functions:FU,V(u, v)=FU(u)FV(v)If the distribution functions are differentiable for almost all u and vthen we also have the resultfU,V(u, v)=fU(u)fV(v)Lecture 2 12Bivariate Normal DistributionA normal probability density function for two random variables Uand V with zero means and unit variance isf(U, V )=12π 1 − ρ2e−U2−2ρUV +V22(1−ρ2)where ρ is the correlation coefficient between U and V .Ifρ =0thenfU,V(u, v)=12πe−u2+v22Lecture 2 13Bivariate Normal DistributionNow,fU(u)=∞−∞fU,V(u, v)dv =1√2πe−u2/2which is easily established by using the fact that1√2π∞−∞e−v2/2dv =1By a similar integration, we also find thatfV(v)=1√2πe−v2/2ThereforefU,V(u, v)=fU(u)fV(v)and uncorrelated normal random variables are statistically indepen-dent.Lecture 2 14Function of a Random VariableLet U be an random variable and V = g(U). Then V is also a rv.P (V ≤ v)=P (U ≤ g−1(v))If the inverse function u = g−1(v)isknownand is single-valued thenthe probability can be expressed asFV(v)=P [u ≤ g−1(v)] =g−1(v)−∞fU(u)dufV(v)=ddvP [u : g(u) ≤ v]Lecture 2