# UMVUE for probability of cutoff

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.

• Please add the self-study tag. Jun 16 '19 at 11:39
• 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. Jun 16 '19 at 12:24
• @StubbornAtom, many thanks. Just to check, $X_1 | \bar{X}$ has mean $\bar{X}$ and variance $1+1/n$ and hence we may compute $P(X_1 < u | \bar{X})$.
– Jo'
Jun 16 '19 at 13:03
• If I did correctly, it is the variance of $X_1 - \bar{X} \sim N(0, 1+1/n)$
– Jo'
Jun 16 '19 at 14:14
• 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). Jun 16 '19 at 14:32


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

(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.)

• How to go about it if we have unknown variance? Mar 27 '21 at 6:49

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 $$\mu$$.

• May I ask how it follows $E \phi(c(u-\bar X))=P(Z \le c(u- \bar X))$
– Jo'
Jun 16 '19 at 12:39
• 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 $W$, $E\phi(W)=EE(1\{Z\le W\}|W)=E1\{Z\le W\}=P(Z\le W)$. Jun 16 '19 at 13:04
• 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. Jun 16 '19 at 14:52

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

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