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Consider a number $X$ being selected according to normal distribution $\frac{1}{\sqrt{2 \pi \sigma^2}}exp\left[-\frac{X^2}{2 \sigma^2}\right]$. Suppose the outcome is $X_1$. Then, a new number $Y$ is picked according to probability distribution $\frac{1}{\sqrt{2 \pi \sigma^2}}exp\left[-\frac{(Y-X_1)^2}{2 \sigma^2}\right]$. Lets say it is $Y_1$.

What can be said about the average of the quantity $(Y_1-X_1)^2$?

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    $\begingroup$ Maybe add self-study tag? Your friend is the double expectation theorem! $\endgroup$ – kjetil b halvorsen Mar 26 '18 at 17:24
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    $\begingroup$ @Kjetil Is that a different name for the law of total expectation? $\endgroup$ – Glen_b -Reinstate Monica Mar 27 '18 at 0:37
  • $\begingroup$ @Glen_b: Yes, it is. Seems to have many names ... $\endgroup$ – kjetil b halvorsen Mar 27 '18 at 12:52
  • $\begingroup$ @kjetilbhalvorsen Even after studying the theorem mentioned by you I am still unsure about how to compute basic quantities like E[Y1]. Could you kindly explain in some detail? $\endgroup$ – Mark Robinson Apr 1 '18 at 6:06
  • $\begingroup$ In the second density you wrote $\mu^2$ where I suppose you meant $\sigma^2$. I corrected it. If you really meant $\mu^2$, please explain why. $\endgroup$ – kjetil b halvorsen Jul 11 '18 at 23:24
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So you have $X \sim \cal{N}(0, \sigma^2)$. Then $Y \mid X=X_1 \sim \cal{N}(X_1, \sigma^2)$. Remember the double expectation theorem says that $$ \DeclareMathOperator{\E}{\mathbb{E}} \E \E [Y \mid X] = \E Y $$ Using this we can compute $$ \E (Y-X_1)^2 = \E \E [(Y-X_1)^2 \mid X_1] = \E [\sigma^2 \mid X_1] = \sigma^2 $$

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