# Obtaining the expected value $E[X_{(1)} \mid\overline X = c]$

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 $$E[\min(X_1, \dots, X_n) \mid \overline X] = E[X_1 \leq x, \dots, X_n \leq x\mid \overline X]$$?

• 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.. Sep 23, 2019 at 6:17
• 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.) Sep 23, 2019 at 6:37
• 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. Sep 23, 2019 at 12:46
• 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? Sep 23, 2019 at 19:06
• 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 $x$. Sep 23, 2019 at 19:16

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 $$\theta=0$$.

• 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 $\theta$. May 7, 2021 at 16:17