Conditional Expectation / Estimator Confusion

Question

Let $X_1, X_2, X_3 \sim N(0, d^2)$ and $T = X_1^2 + X_2^2 + X_3^2.$

I have an estimator for $d$, $$\hat{d} = \frac{\sqrt{T\ 2\pi}}{4},$$ and another estimator for $d$, $$\tilde{d} = \frac{1}{3} \sqrt{\pi / 2}\ \left(|X_1| + |X_2| + |X_3|\right).$$

I need to show that $\mathbb{E}(\tilde{d} | T) = \hat{d}$ and that $\mathbb{E}(|X_1| | T) = \frac{1}{2} \sqrt{T}$.

I'm not clear how conditioning the expectation on $T$ changes things and how it affects the computation of expectation.

---

(The following edit was attempted by the OP in an answer, I'm updating the question on their behalf.)

I'm mostly confused about how to write out $E[d\sim | T]$ or $E[|X1| | T]$ and begin to work through the algebra.

Answer 1

Since this looks like homework, or exercise I will skim through the solution.

Let's write

$$f_{X_i|T}(x_i | t) = \frac{f_{X_i, T}(x_i, t)}{f_T(t)}.$$

The joint distribution $f_{X_i, T}(x_i, t)$ is the product of a Gaussian $(0, d^2)$ and a $\chi^2(2)$ (the post does not specify that $X_1, X_2, X_3$ are independent, but I will take it for granted):

$$f_{X_i, T}(x_i, t) \propto \exp\left\{-\frac{x_i^2}{2d^2} \right\} \exp\left\{-\frac{t-x_i^2}{2d^2}\right\},$$

which (miraculously) simplifies to

$$f_{X_i, T}(x_i, t) \propto \exp\left\{-\frac{t}{2d^2}\right\}.$$

From here this is quite easy because we realize that the numerator does not depend on $x_i$ so the distribution of $x_i$ given $t$ is uniform. The boundaries for $x_i$ given $t$ are $-\sqrt{t}$ and $+\sqrt{t}$ by construction, so the expected value $E(|X_i| | T=t)$ is

$$ 2\int_0^{\sqrt{t}} \frac{x_i}{2\sqrt{t}} dx_i = \frac{1}{2}\sqrt{t}.$$

The expected value of a sum is the sum of the expected values, so getting $E(\tilde{d} | T=t)$ is piece of cake.

EDIT: Following @whuber's comment, here is how I got $f_{X_i|T}(x_i | t)$. I will assume $i=1$ for the sake of the argument. Let's define $Z = X_2^2 + X_3^2$, which is a scaled $\chi^2(2)$, i.e with 2 degrees of freedom.

The joint density of $(X_1, Z)$ is the product of their densities because they depend on different variables, so $\propto \exp\left\{-\frac{x_1^2}{2d^2} \right\} \exp\left\{-\frac{z}{2d^2}\right\}$. Now we use the change of variable $(X_1, Z) \rightarrow (X_1, X_1^2 + Z = T)$. The Jacobian of the inverse transformation is 1, so $f_{X_1|T}(x_1 | t) \propto \exp\left\{-\frac{x_1^2}{2d^2} \right\} \exp\left\{-\frac{t-x_1^2}{2d^2}\right\}$.