0
$\begingroup$

Let $Y$ be a random variable with $E[Y^2]=3.$ Let $Y_1$, $Y_2$, $Y$ be independent and identically distributed. Assume that Y has the same distribution as $\frac{Y_1 + Y_2}{\sqrt{2}}.$ Find Var[Y].

Attempt:

$E[Y^2] = E[(\frac{Y_1 + Y_2}{\sqrt{2}})^2] = \frac{1}{2}(E[Y_1]^2 + 2E[Y_1]E[Y_2]+ E[Y_2]^2) = 3$

$\Longrightarrow E[Y_1]^2 + 2E[Y_1]E[Y_2]+ E[Y_2]^2 = 6$

$Var[Y] = E[Y^2] - E[Y]^2$

$\Longrightarrow Var[Y] = 3 - E[Y]^2$

$\Longrightarrow Var[Y] = 3 - E[\frac{Y_1 + Y_2}{\sqrt{2}}]^2$

$\Longrightarrow Var[Y] = 3 - \frac{E[Y_1 + Y_2]^2}{2}$

$\Longrightarrow Var[Y] = 3 - \frac{(E[Y_1] + E[Y_2])^2}{2}$

$\Longrightarrow Var[Y] = 3 - \frac{E[Y_1]^2 + 2E[Y_1]E[Y_2]+ E[Y_2]^2}{2}$

$\Longrightarrow Var[Y] = 3 - \frac{6}{2} = 0$

I'm not sure if this is right, it seems too easy. May someone check it.

$\endgroup$
  • $\begingroup$ How did you compute $E(Y)$? $\endgroup$ – Mauricio G Tec Mar 7 '16 at 23:10
  • $\begingroup$ The first line is wrong already. You interchanged the square with the expectation. $\endgroup$ – Mauricio G Tec Mar 7 '16 at 23:12
  • $\begingroup$ $\mathsf E\big((Y_1+Y_2)^2\big) = \mathsf E(Y_1^2+2Y_1Y_2+Y_2^2) \mathop{=}^{iid} 2 \big(\mathsf E(Y_1^2)+ \mathsf E(Y_1)^2\big)$ $\endgroup$ – Graham Kemp Mar 7 '16 at 23:32
  • $\begingroup$ @TiffanyButterfly It is wrong because: $\mathsf E(Y_1^2) \neq \mathsf E(Y_1) ^2$ $\endgroup$ – Graham Kemp Mar 8 '16 at 0:09
1
$\begingroup$

The first line contains an error since $\mathsf E(Y_1^2)$ is not the same thing as $\mathsf E(Y_1)^2$

$$\begin{align} \mathsf E(Y^2) = & ~ \mathsf E(\tfrac 1 2(Y_1+Y_2)^2) & \textsf{Identical Distribution} \\[1ex] = & ~ \tfrac 12\mathsf E(Y_1^2+Y_1Y_2+Y_2^2) \\[1ex] = & ~ \tfrac 1 2\big(\mathsf E(Y_1^2)+\mathsf E(Y_1Y_2)+\mathsf E(Y_2^2)\Big) & \textsf{Linearity of Expectation} \\[1ex] = & ~ \tfrac 1 2\big(\mathsf E(Y_1^2)+\mathsf E(Y_1)\mathsf E(Y_2)+\mathsf E(Y_2^2)\Big) & \textsf{Independence} \\[1ex] = & ~ \tfrac 1 2\big(\mathsf E(Y^2)+\mathsf E(Y)\mathsf E(Y)+\mathsf E(Y^2)\Big) & \textsf{Identical Distribution} \\[2ex] \therefore \mathsf E(Y) = & ~ 0 \end{align}$$

John Dawkins points out a quicker way to obtain this from $\mathsf E(Y)=\mathsf E((Y_1+Y_2)/\sqrt{2}) = \sqrt 2~\mathsf E(Y)$

$\endgroup$
3
$\begingroup$

Because $Y$ has the same distribution as $(Y_1+Y_2)/\sqrt{2}$, we have $E[Y]=(E[Y_1]+E[Y_2])/\sqrt{2}=2E[Y]/\sqrt{2}$, which forces $E[Y]=0$.

$\endgroup$

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.