conditional expectation $E[{X_1}^2+{X_2}^2|X_1+X_2=t] $ of normal distributed variables

Question

to find the conditional expectation $E[{X_1}^2+{X_2}^2|X_1+X_2=t] $

if $X_i$'s are independent and both are std. normal distributed.

My attempt: as given is $X_1+X_2=t$ , take $X_2= t-X_1$ and calculate

$E[{X_1}^2+(t-X_1)^2]$

$=E[2{X_1}^2-2tX_1+t^2]=2E[{X_1}^2]-2tE[X_1]+t^2$

$=2+t^2$

this gives a multiple of correct answer $(=1+\frac{t^2}{2})$ altough I'm not sure what is wrong in this method. Can smn point out mistakes if present?

Edit: I got a correct and easier solution other than the one mentioned so now I am just looking for a mistake in the one suggested Please.

Answer 1

One possible way is to note that for $T=X_1+X_2$, both $(X_1,T)$ and $(X_2,T)$ have the same bivariate normal distribution. In fact, $$(X_1,T)\sim N_2\left[ 0, \begin{pmatrix} 1& 1/\sqrt 2 \\ 1/\sqrt 2 & 2 \end{pmatrix}\right ]$$

This gives the conditional distribution $$X_1\mid T\sim N\left(\frac{T}{2},\frac{1}{2}\right)$$

Now use linearity of expectation to find $$E\left[X_1^2+X_2^2\mid T\right]=E\left[X_1^2\mid T\right]+E\left[X_2^2\mid T\right]$$

---

Another approach to find the conditional mean is shown in this answer.

And @Maxim points out your mistake correctly.

You cannot simply say that $E\left[X_1^2+(t-X_1)^2\mid X_1+X_2=t\right]=E\left[X_1^2+(t-X_1)^2\right]$. This is not true because $X_1^2+(t-X_1)^2$ is not independent of $\{X_1+X_2=t\}$. When you use linearity of expectation in the next step, the condition $\{X_1+X_2=t\}$ should remain.

Answer 2

Let $H = (X_1 + X_2 = t)$. $X_1 - X_2$ and $X_1 + X_2$ are independent as uncorrelated components of a jointly normal distribution, therefore $$\operatorname{E}((X_1 - X_2)^2 \mid H) = \operatorname{E}((X_1 - X_2)^2), \\ 2 \operatorname{E}(X_1^2 + X_2^2 \mid H) = \operatorname{E}((X_1 + X_2)^2 \mid H) + \operatorname{E}((X_1 - X_2)^2 \mid H) = t^2 + 2.$$