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Suppose we have $X_1,\dots, X_n \overset{\text{iid}}{\sim} N(\mu = 0, \sigma^2 = 1)$, for a known $n$. And we want to calculate $E[X_{(1)} \mid \overline X = c]$, where $c \in \mathbb{R}$ is known, $X_{(1)}$ is the first order statistic of the $X_i$'s and $\overline X$ is the sample mean of the data.

What I can see is that $X_{(1)}$ is an estimator of the data, because is an order statistic. Also, $X_{(1)}$ is a sufficient statistic of the data, for the same reason. Then, this expected value is really similar to the Rao-Blackwell Theorem to me, but how can I get/calculate this conditional expected value? Shouldn't one of the parameters ($\mu$, $\sigma^2$) need to be unknown to apply Rao-Blackwell?

Also, since $X_{(1)} = \min(X_1, \dots, X_n)$, the expected value would be $E[\min(X_1, \dots, X_n) \mid \overline X] = E[X_1 \leq x, \dots, X_n \leq x\mid \overline X]$?

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  • $\begingroup$ Sufficiency has to do with a parameter or parameters. You have a known distribution so there is no parameter to estimate. I don't know what you mean by a sufficient statistic for the data. Since the sample mean equals c under the conditioning you already know that the minimum of the sequence is <= to c. I think you mean to condition on X bar = c We know that the minimum is <=x if & only if all the $X_i$ are <= x.. But you are using it in the expectation incorrectly.. $\endgroup$ Sep 23, 2019 at 6:17
  • $\begingroup$ I think that is where my misconception is, I'm seeing E[X_(1)} | X bar] as defined in the Rao-Blackwell Theorem (θ∗ =E (ˆθ | T), where ˆθ is an estimator and T is a sufficient statistic.) $\endgroup$
    – Juan Chong
    Sep 23, 2019 at 6:37
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    $\begingroup$ The difference is that $X$$_($$_1$$_)$ is not a parameter of the population distribution it's a function of the sample and it's not a sufficient statistic for an unknown parameter of the population distribution. $\endgroup$ Sep 23, 2019 at 12:46
  • $\begingroup$ Then, for example in this case, I could get $P(X_{(1)} \leq x)$, which if I didn't made a mistake should be $1 - (\frac{1}{\sqrt{2 \pi \sigma^2}}) e^{-\frac{1}{2 \sigma^2} \sum_{i=1}^{n} (x_i - \mu)^2}$. And then, I can get the $E[X_{(1)} | \bar{X}]$ from the definition of expectation? $\endgroup$
    – Juan Chong
    Sep 23, 2019 at 19:06
  • $\begingroup$ In the formula you need to replace $\mu$ with 0 & $\sigma^2$ with 1. Also the normal density needs to be integrate from -$\infinity$ to $x$. $\endgroup$ Sep 23, 2019 at 19:16

1 Answer 1

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Suppose $X_1,X_2,\ldots,X_n$ are i.i.d $N(\theta,1)$ where $\theta\in \mathbb R$.

Then $\overline X$ is a complete sufficient statistic for $\theta$.

Note that $X_{(1)}-\overline X=(X_{(1)}-\theta)-(\overline X-\theta)$, so that its distribution is free of $\theta$.

By Basu's theorem, the ancillary statistic $X_{(1)}-\overline X$ is independent of $\overline X$.

Therefore,

$$E\left[X_{(1)}-\overline X\right]=E\left[X_{(1)}-\overline X\mid \overline X\right]=E\left[X_{(1)}\mid \overline X\right]-\overline X$$

So for every $\theta$, we must have

$$E\left[X_{(1)}\mid \overline X\right]=E\left[X_{(1)}-\overline X\right]+ \overline X=E\left[X_{(1)}\right]-\theta+\overline X$$

In particular, the above holds for $\theta=0$.

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  • $\begingroup$ Since $E[X_{(1)}]=c+\theta$ where $c$ is the expectation of the first order statistic from $N(0,1)$, the conditional expectation is always free of $\theta$. $\endgroup$ May 7, 2021 at 16:17

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