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For my textbook, Introduction to Probability by Blitzstein and Hwang, I have the problem where I have the random variables $X$, $Y$, and $Z$ such that $X \sim N(0, 1)$. I am also told that, conditional on $X = x$, $Y$ and $Z$ are independent and identically distribution $N(x, 1)$.

I found the joint PDF of $X$, $Y$, and $Z$ to be

$$P[Y = y, Z = z, X = x] = P[Y = y, Z = z | X = x] P[X = x] = \dfrac{1}{(\sqrt{2 \pi})^3} e^{-1/2 [y^2 + z^2 + 3x^2 - 2x(y + z)]}$$

I now want to find the expected values and variances for each of $X$, $Y$, and $Z$. However, I am unsure of how to do this. I was thinking that, since I have the joint distribution $P[Y = y, Z = z, X = x]$, would I somehow marginalize out each of the other 2 random variables, which would then some how lead to the results I am looking for?

I would greatly appreciate it if someone could please take the time to show and explain how this is done.

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You could integrate out two of the variables from the joint pdf to find the mean and variance of each variable, but that would be unnecessarily complicated.

Are you familiar with the double rules of expectation and variance?

It happens to be true that for any two stochastic variables $X$ and $Y$ that

\begin{align*} \mathbb{E}(X) = \mathbb{E}(\mathbb{E}(X\vert Y)), \text{ and that} \end{align*}

\begin{align*} \mathbb{V}(X) = \mathbb{E}(\mathbb{V}(X\vert Y)) + \mathbb{V}(\mathbb{E}(X\vert Y)). \end{align*}

Using these, and the information given in the problem about the conditional distributions of $Y$ and $Z$ given $X$, you should be able to find their marginal means and variances.

(Added after comment below) For instance, if you want to find the mean of $Y$, use that $Y\sim \mathcal{N}(x, 1),$ so that $\mathbb{E}(Y\vert X=x) = x,$ and the double rule to see that

\begin{align*} \mathbb{E}(Y) = \mathbb{E}(\mathbb{E}(Y\vert X)) = \mathbb{E}(X) = 0. \end{align*}

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  • $\begingroup$ Thanks for the answer. This is the first time I am hearing of this concept. Can you please elaborate on how it would be applied to this problem? $\endgroup$ – The Pointer Oct 14 '19 at 12:54
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    $\begingroup$ I'll edit the post to show you how to do it for the mean for one of $Y$ or $Z$. $\endgroup$ – Simon Boge Brant Oct 14 '19 at 14:21
  • $\begingroup$ Ok. Thank you for the help. $\endgroup$ – The Pointer Oct 14 '19 at 14:22

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