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

Question

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,
\begin{align} \sum_i \mathbb{E}\left( X_i \mid T \right) = \mathbb{E}\left( \sum_i X_i \mid T \right) = T . \end{align}
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.

Consequently

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

whence

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