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

  • $\begingroup$ 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

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

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  • $\begingroup$ 𝖡𝖺𝗋(X)=𝖀(X2)βˆ’π–€(X)2 , 𝖡𝖺𝗋(X)=1, and 𝖀(X)=0, therefore 𝖀(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
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    $\begingroup$ $(X^2)(Y^2)+10(X^2)YZ+25(X^2)(Z^2)= (X^2)(Y+\color{red}5Z)^2$ @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

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