# Explain conditional distribution given sufficient statistics.

I understood proof of sufficient statistics by factorization theorem. But I can't solve fllowing two questions about conditional distribution.

1. When $X_1, X_2, ..., X_n \sim N(0, \sigma^2)$ are i.i.d., expain conditional distribution of $X=(X_1, X_2, ..., X_n)$ given $T=\sum_{i=1}^{n}X_i^2$.

2. When $X_1, X_2, ..., X_n \sim N(\mu, \sigma^2)$ are i.i.d., expain conditional distribution of $X=(X_1, X_2, ..., X_n)$ given $T=(\bar{X}, \sum_{i=1}^{n}(X_i-\bar{X})^2)$.

I was given a hint of Q1, "Consider uniformaly distribution on a hyper-sphere $\sum x_i^2=t$."

• @TheBluegrassMathematician I tried to use definition of conditional distribution. $X\sim N(0, \sigma^2I_n)$, $\sum(X_i/\sigma)^2\sim\chi^2(n)$, $f_{X, T}(x, t) = \frac{1}{{(2\pi\sigma^2)}^{n/2}}\exp(-\frac{1}{2\sigma^2}\sum x_i^2)I(\sum x_i^2=t)$, $f_T(t)=\frac{1}{(2\sigma^2)^{n/2}}\frac{1}{\Gamma(n/2)}t^{n/2-1}\exp(-\frac{1}{2\sigma^2}t)$. And I used $f_{X|T=t}(x) = f_{X,T}(x, t)/f_T(t)$, I got my answer "If $\sum x_i^2=t$, then $f_{X|T=t}(x)=\Gamma(n/2)/(\pi^{n/2}t^{n/2-1})$". But I can't explain this distribution.
– user476802
Nov 13, 2017 at 13:46

The point in $\#1$ is that the conditional density of $(X_1,\ldots,X_n)$ given the value of $X_1^2+\cdots+X_n^2$ does not depend on $\sigma.$

Suppose you're given that $X_1^2+\cdots+X_n^2 = r^2.$ The joint density of $(X_1,\ldots,X_n)$ is proportional (as a function of $(x_1,\ldots,x_n)$ to $$e^{-(x_1^2+\cdots+x_n^2)/(2\sigma^2)}.$$ When restricted to the set on which that sum of squares is $r^2,$ that becomes $e^{-r^2/(2\sigma^2)}.$ Now the crucial point is that that value, $e^{-r^2/(2\sigma^2)}$, does not change as the point $(x_1,\ldots,x_n)$ moves about on the sphere $x_1^2+\cdots+x_n^2=r^2.$ So we have $$\int\limits_\text{sphere} e^{-r^2/(2\sigma^2)} \, dA = e^{-r^2/(2\sigma^2)} \times \text{surface area}.$$ For this to be a density, we need to multiply it by a normalizing constant $c,$ so that $$\big(c e^{-r^2/(2\sigma^2)} \times \text{surface area} \big) = 1.$$ Then the density is $$ce^{-r^2/(2\sigma^2)} = \left( \frac 1 {e^{-r^2/(2\sigma^2)} \times \text{surface area}} \right) e^{-r^2/(2\sigma^2)} = \cdots$$ and this does not depend on $\sigma^2.$

I.e. the conditional distribution of $(X_1,\ldots,X_n)$ given $T$ does not depend on $\sigma^2.$

• Thanks. All I had to do was explain dependency of $\sigma^2$. By the way I have a question. According to your answer, density is 1/surface area $= \frac{\Gamma(n/2)}{2\pi^{n/2}t^{n-1}}$. But my result is $\frac{\Gamma(n/2)}{\pi^{n/2}t^{n/2-1}}$. What's wrong my idea?
– user476802
Nov 14, 2017 at 12:44
• @tanio : As far as the question you asked is concerned, all one needs is that you end up with something that doesn't depend on $\sigma.$ Maybe your last comment is worth a separate question. Nov 15, 2017 at 5:07