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From the figure above,

pr{0<X+Y<z}={

112z2-12(2-z)2 if1if0<zz<2<1.

An alternative derivation uses integration. For example, in the case z<1,

pr{0<X+Y<z}=0<x+y<zf(x,y)dxdy =y=0zx=0z-ydxdy =y=0z(z-y)dy=12z2.

(b) If Z=X+Y, then Z has cumulative distribution function F(z) where, from (a) above,

F(z)=pr{X+Yz}={

112z2-12(2-z)2 if1z<2if0<z<1.

Probability density function for Z is then f(z)=dF(z)dz={

z

if 0<z<1, 2-z if 1z<2.

Z has a triangular distribution on the interval (0,2) with mode at z=1.

Worked Example: Lecture 14

Measurements of stature were made on each member of a large population of pairs of adult brothers. The height of the elder brother was denoted by X and of the younger brother by Y. Both X and Y had the same mean μ and the same standard deviation σ. The correlation coefficient was ρ. Deduce the mean and variance of (i) U=X-Y, and (ii) V=X+Y.

Derive the covariance of U and V.

Answer:

E[U]=E[X-Y]=E[X]-E[Y]=μ-μ=0.

Var [U]= Var [X-Y]= Var [X]+ Var [Y]-2cov(X,Y)=σ2+σ2-2ρσ2=2σ2(1-ρ) , as

corr(X, Y) =cov(X,Y)Var[X]Var[Y] cov (X,Y)=corr(X,Y)Var[X]Var[Y]=ρσ2σ2=ρσ2

E[V]=E[X+Y]=E[X]+E[Y]=μ+μ=2μ.

Var[V] = Var [X+Y]=Var[X]+ Var [Y]+2cov(X,Y)=σ2+σ2+2ρσ2=2σ2(1+ρ) .

cov (U,V)=cov(X-Y,X+Y) = cov (X,X) —cov (Y,Y)+ cov (X,Y)- cov (Y,X) = Var[X] —Var [Y]+cov(X,Y) —cov (X,Y)

= σ2-σ2+cov(X,Y) —cov (X,Y)

= 0.

Worked Example: Lecture 14.

Let T=a1X1+a2X2, where X1 and X2 are uncorrelated random variables with mean μ and variance σ2, and a1 and a2 are constants chosen so that E[T]=μ.

Deduce that the choice a2=1-a1 gives E[T]=μ.

In this case prove that the variance of T is a minimum if a1=a2=12.

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