If $X_1,\cdots,X_n \sim \mathcal{N}(\mu, 1)$ are IID, then compute $\mathbb{E}\left( X_1 \mid T \right)$, where $T = \sum_i X_i$


If $X_1,\cdots,X_n \sim \mathcal{N}(\mu, 1)$ are IID, then compute $\mathbb{E}\left( X_1 \mid T \right)$, where $T = \sum_i X_i$.

Attempt: Please check if the below is correct.

Let say, we take the sum of the those conditional expectations such that,
\sum_i \mathbb{E}\left( X_i \mid T \right) = \mathbb{E}\left( \sum_i X_i \mid T \right) = T .

It means that each $\mathbb{E}\left( X_i \mid T \right) = \frac{T}{n}$ since $X_1,\ldots,X_n$ are IID.

Thus, $\mathbb{E}\left( X_1 \mid T \right) = \frac{T}{n}$. Is it correct?


The idea’s right–but there’s a question of expressing it a little more rigorously. I will therefore focus on notation and on exposing the essence of the idea.

Let’s begin with the idea of exchangeability:

A random variable $\mathbf X=(X_1, X_2, \ldots, X_n)$ is exchangeable when the distributions of the permuted variables $\mathbf{X}^\sigma=(X_{\sigma(1)}, X_{\sigma(2)}, \ldots, X_{\sigma(n)})$ are all the same for every possible permutation $\sigma$.

Clearly iid implies exchangeable.

As a matter of notation, write $X^\sigma_i = X_{\sigma(i)}$ for the $i^\text{th}$ component of $\mathbf{X}^\sigma$ and let $$T^\sigma = \sum_{i=1}^n X^\sigma_i = \sum_{i=1}^n X_i = T.$$

Let $j$ be any index and let $\sigma$ be any permutation of the indices that sends $1$ to $j = \sigma(1).$ (Such a $\sigma$ exists because one can always just swap $1$ and $j.$) Exchangeability of $\mathbf X$ implies

$$E[X_1\mid T] = E[X^\sigma_1\mid T^\sigma] = E[X_j\mid T],$$

because (in the first inequality) we have merely replaced $\mathbf X$ by the identically distributed vector $\mathbf X^\sigma.$ This is the crux of the matter.


$$T = E[T \mid T] = E[\sum_{i=1}^n X_i\mid T] = \sum_{i=1}^n E[X_i\mid T] = \sum_{i=1}^n E[X_1\mid T] = n E[X_1 \mid T],$$


$$E[X_1\mid T] = \frac{1}{n} T.$$

Source : Link , Question Author : learning , Answer Author : whuber

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