Finding the conditional expected value of the sum of independent random variables.

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?

• Perhaps an argument by symmetry works. – GNUSupporter 8964民主女神 地下教會 Nov 2 '18 at 21:29
• 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 $X_1, \;..., \; X_5$. – Aditya Dua Nov 2 '18 at 23:54

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 $$\Bbb E[Z\mid Y=y]=y-5\mu+5\mu=y.$$

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:

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

• could you please give me any hint how to prove that they are jointly Gaussian? – R.K. Nov 5 '18 at 10:44
• @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. – Aditya Dua Nov 5 '18 at 17:15