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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.

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  • $\begingroup$ Given that $\hat{d}$ is a function of $T$, it is evident that both expectations are (non-trivial) functions of $T$. If you did not condition on $T$ and took an expectation, it would be nonsensical for $T$ to enter in the formula, wouldn't it? I realize this might not help one's intuition, but surely it shows a clear difference between conditioning on $T$ and not conditioning on it. $\endgroup$ – whuber Oct 23 '12 at 16:19
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    $\begingroup$ As a hint, one powerful (and historically early) approach uses geometrical reasoning. Upon recognizing that, conditional on $T$, the vector $(X_1,X_2,X_3)$ is uniformly distributed on a sphere of radius $\sqrt{T}$, it's easy to see that the two parts of the question are very closely related (the second part immediately answers the first part) and the $1/2$ in the second part is the average height of a random point on the upper hemisphere of a unit sphere. This is illustrated at stats.stackexchange.com/questions/7977 and further discussed at stats.stackexchange.com/questions/22764 $\endgroup$ – whuber Oct 23 '12 at 16:35
  • $\begingroup$ I'm unable to edit my original question. I'm mostly confused about how to write out E[d~ | T] or E[|X1| | T] and begin to work through the algebra $\endgroup$ – xardox Oct 23 '12 at 17:15
  • $\begingroup$ @xardox Of course you cannot edit the question: you are not klingzon! The two accounts use different identifiers. If both you and klingzon verify you are the same person, a moderator can merge your accounts and then you'll be fine. $\endgroup$ – whuber Oct 23 '12 at 19:11
  • $\begingroup$ @gung Oh yes, totally!! Good observation. $\endgroup$ – gui11aume Oct 23 '12 at 20:52
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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 $f_{X_1|T}(x_1 | t) \propto \exp\left\{-\frac{x_1^2}{2d^2} \right\} \exp\left\{-\frac{t-x_1^2}{2d^2}\right\}$.

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  • $\begingroup$ Umm... I'm pretty sure that $T$ has three degrees of freedom, not two. Not that it matters--everything is conditioned on $T$. But maybe I misunderstand the notation. All the work depends on justifying that assertion about the joint distribution. $\endgroup$ – whuber Oct 23 '12 at 20:58
  • $\begingroup$ @whuber absolutely. My bad. By the way, I like your geometric answer (to the duplicate of that question) better :D $\endgroup$ – gui11aume Oct 23 '12 at 21:01
  • $\begingroup$ @whuber actually thinking about it, I stick to my point (see the edit). Did I get something wrong? $\endgroup$ – gui11aume Oct 23 '12 at 22:12
  • $\begingroup$ Now I'm convinced :-). $\endgroup$ – whuber Oct 23 '12 at 22:59

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