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?
2026-03-26 02:55:34.1774493734
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Finding the conditional expected value of the sum of independent random variables.
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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:
$\mathbb{E}[Z | Y=y] = 15\mu + 10\sigma^2/15\sigma^2(y-15\mu) = 2y/3 + 5\mu$
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 $$\Bbb E[Z\mid Y=y]=y-5\mu+5\mu=y.$$