Skip to main content
Cross Validated

Return to Question

Added the "exchangeability" tag and removed the "mathematical statistics" tag, since that is a more informative tag for future reference.
Link
edit approved Jan 9, 2021 at 0:51
passerby51
edit approved Jan 9, 2021 at 0:51
passerby51
  • 1.8k
  • 10
  • 11
Tweeted twitter.com/StackStats/status/1058463829229756416
edited body; edited tags
Source Link
whuber ♦
whuber ♦
  • 333.5k
  • 63
  • 792
  • 1.3k

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,\cdots,X_n$$X_1,\ldots,X_n$ are IID.

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

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 $X_1,\cdots,X_n$ are IID.

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

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?

Source Link
learning
learning
  • 293
  • 1
  • 6

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 $X_1,\cdots,X_n$ are IID.

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