Question

I’ve spent hours trying to logic my way through this, please help and show step by step.

To thè value of p you found above, are X and Y independent? Problem 6 Consider two random variables X and Y. Assume that Y a bX for some unknown numbers a and b. Assume that we know E(X-#x, EM-lly, Var(X- , Var(Y- (a) Find values of variances of X and Y such that you can not find a and b. (b) Find values of variances of X and Y such that there are continuum values of a and b that solve the equations. (c) Find values of variances of X and Y such that there is only one soluti on of a and b. And find such a and b. (c) Find values of variances of X and Y such that there are two solutions of a and b. And find such a and b 3

Answer 1

$\mu_Y=E(Y)=a+bE(X)=a+b\mu_X......(1)\\ \sigma_Y^2=Var(Y)=b^2Var(X)=b^2\sigma_X^2~or,~b^2=\frac{\sigma_Y^2}{\sigma_X^2}......(2)\\ (a)~If~\sigma_X^2=0~then~from~(2)~we~see~that~b^2~is~undefined~and~since~value~of\\ a~depends~on~b~(see~(1))~so~we~can't~find~a~also.\\ (b)~If~we~take~value~of~any~one~of~\sigma_X^2~and~\sigma_Y^2~in~a~subset~of~\mathbb{R}^+,~for~example,\\ \sigma_X^2\in (0,1)~and~\sigma_Y^2=1~(say)~then~we~have~infinitely~many~values~of~b~and\\ hence~a~also~have~infinitely~many~values.\\ (c)~Take,~\sigma_Y^2=0~then~b^2=0~or,~b=0~hence~a=\mu_Y.\\ (d)~Take,~\sigma_X^2=\sigma_Y^2=1~then~b=\pm 1~and~then~a=\mu_Y-\mu_X~or~\mu_Y+\mu_X.$