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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.

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  • $\begingroup$ How did you compute $E(Y)$? $\endgroup$ Commented Mar 7, 2016 at 23:10
  • $\begingroup$ The first line is wrong already. You interchanged the square with the expectation. $\endgroup$ Commented Mar 7, 2016 at 23:12
  • $\begingroup$ $\mathsf E\big((Y_1+Y_2)^2\big) = \mathsf E(Y_1^2+2Y_1Y_2+Y_2^2) \mathop{=}^{iid} 2 \big(\mathsf E(Y_1^2)+ \mathsf E(Y_1)^2\big)$ $\endgroup$ Commented Mar 7, 2016 at 23:32
  • $\begingroup$ @TiffanyButterfly It is wrong because: $\mathsf E(Y_1^2) \neq \mathsf E(Y_1) ^2$ $\endgroup$ Commented Mar 8, 2016 at 0:09

2 Answers 2

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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 $E[Y]=0$.

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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 $\mathsf E(Y)=\mathsf E((Y_1+Y_2)/\sqrt{2}) = \sqrt 2~\mathsf E(Y)$

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