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Let $X_i \sim N(\mu,1)$, i.i.d. We aim to find UMVUE for $p(\mu) = P_{\mu}(X_1 \leq u)$ for some fixed $u$.

I have shown that $\bar{X}$ and $X_1 - \bar{X}$ are independent. ($\bar{X}$: sample mean).

Now I want to use this to compute $E_\mu[1\{X_1 < u\}\mid\bar{X}]$ and if I am not mistaken, by Lehmann-Scheffé, it is UMVUE.

I am quite new in statistics, maybe my arguments are wrong, or I do not know how to compute.
Any hint appreciated.

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    $\begingroup$ Please add the self-study tag. $\endgroup$ Jun 16 '19 at 11:39
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    $\begingroup$ Your approach is correct. Since $(X_1,\overline X)$ is jointly normal, use the fact that $X_1\mid \overline X$ has a normal distribution. $\endgroup$ Jun 16 '19 at 12:24
  • $\begingroup$ @StubbornAtom, many thanks. Just to check, $X_1 | \bar{X}$ has mean $\bar{X}$ and variance $1+1/n$ and hence we may compute $P(X_1 < u | \bar{X})$. $\endgroup$
    – Jo'
    Jun 16 '19 at 13:03
  • $\begingroup$ If I did correctly, it is the variance of $X_1 - \bar{X} \sim N(0, 1+1/n)$ $\endgroup$
    – Jo'
    Jun 16 '19 at 14:14
  • $\begingroup$ No I am talking about $X_1\mid \overline X$; for that you need the covariance between $X_1$ and $\overline X$ ( I am using this theory). $\endgroup$ Jun 16 '19 at 14:32
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$\newcommand{\v}{\operatorname{var}}\newcommand{\c}{\operatorname{cov}}\newcommand{\e}{\operatorname{E}}$

Lemma: Suppose $U,V$ are normally distributed, but also jointly normally distributed, i.e. so distributed that no matter which (non-random) scalars $a$ and $b$ are, the random variable $aU+bV$ is normally distributed. Suppose $\e(U) = \alpha,$ $\e(V) = \beta,$ $\v(U) = \sigma^2,$ $\v(V) = \tau^2,$ and $\c(U,V) = \rho\sigma\tau$ (so $\rho$ is the correlation). Then the conditional distribution of $V$ given $U$ is $$ V \mid U \sim N\left( \beta + \rho\tau\cdot\frac{U-\alpha} \sigma, \quad \tau^2(1-\rho^2) \right). $$

Now suppose $V=X_1$ and $U= \overline X.$ Observe that these satisfy the hypotheses of the Lemma. And \begin{align} \alpha & = \mu, \\ \beta & = \mu, \\ \sigma^2 & = 1/n, \\ \tau^2 & = 1, \\ \rho & = 1/\sqrt n. \end{align} Therefore by the Lemma, $$ X_1\mid \overline X \sim N\left( \overline X ,\quad \frac {n-1} n \right). \tag 1 $$

Therefore $$ \Pr\left( X_1<u \mid \overline X \right) = \Phi\left( \frac{u - \overline X}{\sqrt{\frac{n-1} n}} \right). $$

(The lack of dependence of line $(1)$ above upon $\mu$ is what it means to say that $\overline X$ is sufficient for this family of distributions.)

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  • $\begingroup$ How to go about it if we have unknown variance? $\endgroup$
    – userNoOne
    Mar 27 '21 at 6:49
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The parameter of interest $$ p = P(X < u) = P(X-\mu \le u-\mu)=\Phi(u-\mu) $$ so a possible estimator of $p$ is $$ \hat p = \Phi(c(u-\bar X)). $$ where $c$ is a suitably chosen constant. This has expectation \begin{align} E\hat p &=E \Phi(c(u-\bar X)) \\&=P(Z \le c(u- \bar X)) & \text{where $Z\sim N(0,1)$} \\&=P(Z+c\bar X \le cu) \\&=P(\frac{Z+c(\bar X -\mu)}{\sqrt{1+c^2/n}} \le \frac{c(u - \mu)}{\sqrt{1+c^2/n}}) \\&=\Phi(\frac{c(u - \mu)}{\sqrt{1+c^2/n}}) \\&=p \end{align} for \begin{align} \frac{c}{\sqrt{1+c^2/n}}&=1 \\ c^2&=1+c^2/n \\ c&=1/\sqrt{1-1/n}. \end{align} So $$ \hat p = \Phi(\frac{u-\bar X}{\sqrt{1-1/n}}) $$ is unbiased for $p$ and, by the Lehmann-Scheffé theorem, also UMVUE since $\bar X$ is sufficient for $\mu$.

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  • $\begingroup$ May I ask how it follows $E \phi(c(u-\bar X))=P(Z \le c(u- \bar X))$ $\endgroup$
    – Jo'
    Jun 16 '19 at 12:39
  • $\begingroup$ The standard normal cdf can be expressed as $\phi(x)=P(Z\le x)=E 1\{Z\le x\}$ where $Z$ is standard normal. Using the law of total expectation, it then follows that for any random variable $W$, $E\phi(W)=EE(1\{Z\le W\}|W)=E1\{Z\le W\}=P(Z\le W)$. $\endgroup$ Jun 16 '19 at 13:04
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    $\begingroup$ You might consider changing your notation to match common notation for CDF's of the normal distribution. Usually a lower-case phi ($\phi$) is reserved for the PDF of a normal, while $\Phi$ is typically reserved for the CDF. Your non-standard notation may cause some confusion. $\endgroup$ Jun 16 '19 at 14:52
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I'll add another answer as this approach is another simple way to find the UMVUE.

Because $\overline X$ is a complete sufficient statistic for $\mu$ whenever $\mu\in \mathbb R$, the ancillary statistic $X_1-\overline X$ is independent of $\overline X$ by Basu's theorem. This independence can also be seen from the fact that $(X_1-\overline X,\overline X)$ is jointly normal and $\operatorname{Cov}(X_1-\overline X,\overline X)=0$.

This means

\begin{align} P(X_1\le u\mid \overline X)&= P(X_1-\overline X\le u -\overline X\mid \overline X) \\&=P(X_1-\overline X \le u-\overline X) \end{align}

As $X_1-\overline X \sim N \left(0,1-\frac1n \right)$, the last probability equals

$$\Phi\left(\frac{u-\overline X}{\sqrt{1-\frac1n}}\right)$$

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