# Proof that two iid Gaussian random variables are conditionally independent of their mean given their sum

My question is a simplification of a statement in this book that i.i.d. Gaussian random variables $$X_1, X_2, ..., X_n \sim \mathcal{N}(\Theta, 1)$$ are conditionally independent of $$\Theta$$ given their sum $$X_1 + ... + X_n$$.

I started working on the statement with $$n=2$$, aiming to show that the conditional distribution of $$X_1,X_2$$ does not involve the parameter $$\Theta$$. Immediately, I run into the problem of having the region $$X_1 + X_2 = c$$ having zero area in $$\mathbb{R}^2$$. Because of this, I cannot talk about

$$\int f_{X_1,X_2|Z}(x,y|z) dxdy$$

where $$Z=X_1+X_2$$ because any neighborhood of any point in $$X_1+X_2=z$$ for a fixed $$z$$ will contain points outside of the restricted region. Working with a conditional distribution does not look promising to me.

My question is: is there an alternative way to show conditional independence of $$X_1,X_2$$ with the mean $$\Theta$$ given their sum $$X_1+X_2$$?

Yes, there is! It is enough to prove that $$\sum_i X_i=T$$ is the sufficient estimator of $$\theta$$

(you can prove it in many ways, e.g. factorization theorem)

More, as Gaussian belongs to the Exponential family, T is not only sufficient but also complete and minimal.

Now what you are looking to prove is exactly the definition of sufficiency of T

• If I use the definition of sufficient statistics in the book, I need to show that $I(X_1,X_2; \Theta) = I(X_1+X_2; \theta)$. From my understanding, the mutual information needs a distribution to work with, which is why I am finding it difficult to proceed. May 26, 2020 at 5:49

The easiest way to proceed is to exprerss your Gaussian $$N(\theta;1)$$ in term of Exponential form

then the proof you are looking for is immediate

• The second way is to use the factorization theorem and the third but very dirty way (I do hope that no Statistician will read what I am writing) is, for $n=2$ calculate the covariance between $X$ and $U=X+Y$, then write the joint density and factorize it in $f_{XU}=f_{U}\times f_{X|U}$ May 26, 2020 at 6:50