만약

질문

경우 $X_1,\cdots,X_n \sim \mathcal{N}(\mu, 1)$ IID 후 계산되어 $\mathbb{E}\left( X_1 \mid T \right)$ 여기서는 $T = \sum_i X_i$ .

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시도 : 아래가 올바른지 확인하십시오.

$$\begin{align} \sum_i \mathbb{E}\left( X_i \mid T \right) = \mathbb{E}\left( \sum_i X_i \mid T \right) = T . \end{align}$$ 와 같은 조건부 기대 값의 합을합시다 . 그것은 각 수단 $\mathbb{E}\left( X_i \mid T \right) = \frac{T}{n}$ 이후$X_1,\ldots,X_n$IID이다.

따라서 $\mathbb{E}\left( X_1 \mid T \right) = \frac{T}{n}$ . 맞습니까?

아이디어는 옳지 만 좀 더 엄격하게 표현해야한다는 의문이 있습니다. 그러므로 나는 표기법과 아이디어의 본질을 노출시키는 데 중점을 둘 것입니다.

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교환 성 이라는 아이디어로 시작합시다 .

확률 변수 $\mathbf X=(X_1, X_2, \ldots, X_n)$ 인 교환 시 순열 변수의 분포 $\mathbf{X}^\sigma=(X_{\sigma(1)}, X_{\sigma(2)}, \ldots, X_{\sigma(n)})$ 모든 가능한 순열 $\sigma$ 대해 모두 동일합니다 .

분명히 iid는 교환 가능함을 의미합니다.

표기의 문제로서, 물품 $X^\sigma_i = X_{\sigma(i)}$ 위한 $i^\text{th}$ 성분 $\mathbf{X}^\sigma$ 및하자

$$T^\sigma = \sum_{i=1}^n X^\sigma_i = \sum_{i=1}^n X_i = T.$$

하자 $j$ BE 모든 인덱스 및하자 $\sigma$ 전송 인덱스 임의 치환 될 $1$ 에 $j = \sigma(1).$ (이러한 $\sigma$ 는 항상 $1$ 과 $j.$ 교환 할 수 있기 때문에 존재합니다 . ) $\mathbf X$ 교환 가능성은 암시합니다

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

(첫 번째 불평등에서) 우리는 $\mathbf X$ 를 동일하게 분포 된 벡터 로 대체했을뿐입니다.$\mathbf X^\sigma.$ 이것이 문제의 핵심입니다.

따라서

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

어떻게

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

$\newcommand{\one}{\mathbf 1}$This is not a proof (and +1 to @whuber's answer), but it's a geometric way to build some intuition as to why $E(X_1 | T) = T/n$ is a sensible answer.

Let $X = (X_1,\dots,X_n)^T$ and $\one = (1,\dots,1)^T$ so $T = \one^TX$. We're then conditioning on the event that $\one^TX = t$ for some $t \in \mathbb R$, so this is like drawing multivariate Gaussians supported on $\mathbb R^n$ but only looking at the ones that end up in the affine space $\{x \in \mathbb R^n : \one^Tx = t\}$ . 그런 다음이 좁은 공간에있는 점의$x_1$ 좌표의평균을 알고 싶습니다.

우리는

$$X \sim \mathcal N(\mu \one, I)$$ 을 알고 있으므로 일정한 평균 벡터를 가진 구형 가우시안을 얻었으며 평균 벡터 $\mu\one$ 은 초평면 $x^T\one = 0$ 의 법선 벡터와 같은 줄에 있습니다.

이것은 아래 그림과 같은 상황을 제공합니다.

핵심 아이디어 : 먼저 아핀 부분 공간 $H_t := \{x : x^T\one = t\}$ 대한 밀도를 상상해보십시오 . E ( X ) ∈ 스팬  1 이므로 $X$ 의 밀도는 $x_1 = x_2$ 주위에서 대칭 입니다. 밀도는 대칭에 대한 것이다 H를 t 로 시간 t가 동일한 라인을 통해 대칭이며 대칭 인 주위의 포인트는 선의 교점 인 X 1 + X 2$E(X) \in \text{span } \one$$H_t$$H_t$$x_1 + x_2 = t$$x_1 = x_2$$x = (t/2, t/2)$

$E(X_1 | T)$$H_t$$x_1$$H_t$ the distribution of the $x_1$ coordinates will also be symmetric, and it'll have the same center point of $t/2$. The mean of a symmetric distribution is the central point of symmetry so this means $E(X_1 | T) = T/2$, and that $E(X_1| T) = E(X_2 | T)$ since $X_1$ and $X_2$ can be excahnged without affecting anything.

In higher dimensions this gets hard (or impossible) to exactly visualize, but the same idea applies: we've got a spherical Gaussian with a mean in the span of $\one$, and we're looking at an affine subspace that's perpendicular to that. The balance point of the distribution on the subspace will still be the intersection of $\text{span }\one$ and $\{x : x^T\one = t\}$ which is at $x=(t/n, \dots, t/n)$, and the density is still symmetric so this balance point is again the mean.

Again, that's not a proof, but I think it gives a decent idea of why you'd expect this behavior in the first place.

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Beyond this, as some such as @StubbornAtom have noted, this doesn't actually require $X$ to be Gaussian. In 2-D, note that if $X$ is exchangeable then $f(x_1, x_2) = f(x_2, x_1)$ (more generally, $f(x) = f(x^\sigma)$) so $f$ must be symmetric over the line $x_1 = x_2$. We also have $E(X) \in \text{span }\one$ so everything I said regarding the "key idea" in the first picture still exactly holds. Here's an example where the $X_i$ are iid from a Gaussian mixture model. All the lines have the same meaning as before.

I think your answer is right, although I'm not entirely sure about the killer line in your proof, about it being true "because they are i.i.d". A more wordy way to the same solution is as follows:

Think about what $\mathbb{E}(x_{i}|T)$ actually means. You know that you have a sample with N readings and that their mean is T. What this actually means, is that now, the underlying distribution they were sampled from no longer matters (you'll notice you at no point used the fact it was sampled from a Gaussian in your proof).

$\mathbb{E}(x_{i}|T)$ is the answer to the question, if you sampled from your sample, with replacement many times, what would be the average you obtained. This is the sum over all the possible values, multiplied by their probability, or $\sum_{i=1}^{N}\frac{1}{N}x_{i}$ which equals T.