-1
$\begingroup$

Let X, Y and Z be three independent random variables such that E(X)=E(Y)=E(Z)=0 and Var(X)=Var(Y)=Var(Z)=1. Calculate E[(X^2)(Y+5Z)^2]

I know that the answer is 26.

Since all of the expected values of X Y and Z are all the same, I have replaced each expected value of X Y and Z with just E[X]. For example, E[(X^2)(Y^2)] is now (E[X])^4.

Doing this, I'm left with 26E[X^4]

Since the variance of each random variable is one, I know I need to somehow turn E[X^4] into the formula for variance.... Thanks guys

$\endgroup$
  • $\begingroup$ You have a basic mistake here. Why do you think that $E[XY]=E[X^2]$? $X$ and $Y$ are independent. $\endgroup$ – Ted Shifrin May 19 '18 at 5:04
  • $\begingroup$ Since they are all independent, I thought that meant E[XY]=E[X]E[Y] and since E[X]=E[Y], E[X]E[Y]=E[X]E[X]=E[X^2]... is that not correct? $\endgroup$ – Altered Beast May 19 '18 at 5:08
  • $\begingroup$ No, $E[X^2] = E[X]^2$ if and only if the variance of $X$ is $0$. Indeed, check out one of the usual formulas for variance. $\endgroup$ – Ted Shifrin May 19 '18 at 5:10
  • $\begingroup$ Hm, I see what you mean. Except then why is it that E[XY]=E[X]E[Y], but if I just replace the Y with another X, that statement is no longer true? Thanks @TedShifrin $\endgroup$ – Altered Beast May 19 '18 at 5:13
  • $\begingroup$ Because $E[XY]=E[X]E[Y]$ holds only when $X$ and $Y$ are independent!! (Go back and derive this in a discrete probability case.) Certainly, if you think of any standard example with $E[X]=0$, you'll have $E[X^2]>0$. $\endgroup$ – Ted Shifrin May 19 '18 at 5:17
0
$\begingroup$

The random variables $X,Y,Z$ are mutually independent, and have identical expectation, of $0$, and variance, of $1$.   This does not mean they are interchangable; they are still three distinct variables.

For instance, $X$ is independent from $Y$, but clearly not independent from $X$.

Because $\mathsf {Cov}(X,Y)=\mathsf E(XY)-\mathsf E(X)\mathsf E(Y)$ , $\mathsf {Cov}(X,Y)=0$, $\mathsf E(X)=0$, and $\mathsf E(Y)=0$, therefore $\mathsf E(XY)=0$.

Because $\mathsf {Var}(X)=\mathsf E(X^2)-\mathsf E(X)^2$ , $\mathsf {Var}(X)=1$, and $\mathsf E(X)=0$, therefore $\mathsf E(X^2)=1$ .

So obviously $\mathsf E(X^2)\neq \mathsf E(XY)$ and so forth.


So we have : $\mathsf E[(X^2)(Y+5Z)^2]~{=\mathsf E(X^2)~\mathsf E(Y^2+10YZ+25Z^2)\\~\vdots\\ = 26}$

$\endgroup$
  • $\begingroup$ 𝖵𝖺𝗋(X)=𝖤(X2)−𝖤(X)2 , 𝖵𝖺𝗋(X)=1, and 𝖤(X)=0, therefore 𝖤(X2)=1 is very useful thank you. Using this, if I expand out (X^2)(Y+Z)^2 first, I get (X^2)(Y^2)+10(X^2)ZY+25(Z^2)(X^2). I can then take the expected value of those three terms separately, and since E(X^2)=E(Y^2)=E(Z^2)=1, you get 1+0+25=26. Would you agree? (I got the answer 26 from the back of the book if that helps...) @grahamkemp $\endgroup$ – Altered Beast May 19 '18 at 7:17
  • 1
    $\begingroup$ $(X^2)(Y^2)+10(X^2)YZ+25(X^2)(Z^2)= (X^2)(Y+\color{red}5Z)^2$ @JuliaV . Have you missed the $5$ in your post? $\endgroup$ – Graham Kemp May 19 '18 at 7:52
  • $\begingroup$ My apologies, you're right. Will edit now. $\endgroup$ – Altered Beast May 20 '18 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.