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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 $X_1+X_2$?

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2 Answers 2

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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

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  • $\begingroup$ 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. $\endgroup$
    – owovrokfop
    May 26, 2020 at 5:49
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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

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  • $\begingroup$ 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 $f_{XU}=f_{U}\times f_{X|U}$ $\endgroup$
    – tommik
    May 26, 2020 at 6:50

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