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Let $$Y=\sum_{i=1}^{15}X_i\text{ , } Z=\sum_{i=6}^{20}X_i,$$ where $X_1,...,X_{20}$ are $iid$ with normal distribution $N(\mu,\sigma^2)$. Find $$\Bbb E(Z|Y=y).$$ My doubtful solution goes as following: $$\Bbb E(Z|Y=y)=\Bbb E(y-(X_1+...+X_5)+X_{16}+...+X_{20})=y.$$ I'm worried about the first equality, since I'm not sure if that's how the conditional expected value works. However, the second equality comes from the properties of the expected value. May I get any tips on how to solve this problem properly?

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  • $\begingroup$ Perhaps an argument by symmetry works. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 2 '18 at 21:29
  • $\begingroup$ You are right to be worried. Your approach is as follows: $Z = Y + W$, where $W$ is a linear combination of some of the $Xi$, so $\mathbb{E}[Z|Y=y] = \mathbb{E}[Y+W|Y=y] = y + \mathbb{E}[W|Y=y] = y + \mathbb{E}[W] = y$. The reason this breaks down is that $W$ is not independent of $Y$, implying $\mathbb{E}[W|Y=y] \neq \mathbb{E}[W]$. The reason $W$ is not independent of $Y$ is that they both contain contributions from $X_1, \;..., \; X_5$. $\endgroup$ – Aditya Dua Nov 2 '18 at 23:54
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That is correct because $\Bbb E[y]=y$ and $\Bbb E[X_1+\cdots+X_5]=\Bbb E[X_{16}+\cdots+X_{20}]=5\mu$ due to $X_i$ being $\textsf{iid}$ so you have that $$\Bbb E[Z\mid Y=y]=y-5\mu+5\mu=y.$$

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It is evident that since $X_i \sim N(\mu, \sigma^2)$ and are independent, $Y \sim N(15\mu, 15\sigma^2)$ and $Z \sim N(15\mu, 15\sigma^2)$. You will have to convince yourself that $Y$ and $Z$ are not just individually Gaussian, but also jointly Gaussian.

Now we can use a standard result. If $Y$ and $Z$ are jointly Gaussian,

$\mathbb{E}[Z | Y=y] = \mathbb{E}[Z] + \frac{Cov(Y,Z)}{Var(Y)}(y-\mathbb{E}[Y])$. We know that:

  • $\mathbb{E}[Y] = \mathbb{E}[Z] = 15 \mu$
  • $Var(Y) = 15 \sigma^2$
  • $Cov(Y,Z) = \mathbb{E}[YZ] - \mathbb{E}[Y]\mathbb{E}[Z] = 225 \mu^2 + 10 \sigma^2 - 225\mu^2 = 10 \sigma^2$ (skipped some algebra here)

Putting this all together:

$\mathbb{E}[Z | Y=y] = 15\mu + 10\sigma^2/15\sigma^2(y-15\mu) = 2y/3 + 5\mu$

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  • $\begingroup$ could you please give me any hint how to prove that they are jointly Gaussian? $\endgroup$ – R.K. Nov 5 '18 at 10:44
  • $\begingroup$ @R.K.There is a theorem which says that random variables $Y$ and $Z$ are jointly Gaussian iff any linear combination of Y and Z is Gaussian. Since both $Y$ and $Z$ are sums of Gaussian RVs ${X_i}$, you can easily prove that the condition holds and hence conclude that $Y$ and $Z$ are jointly Gaussian. You can also try it from first principles, though that could be a bit tedious. $\endgroup$ – Aditya Dua Nov 5 '18 at 17:15

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