Problem Statement

Let $X_1, X_2, ..., X_n$ be i.i.d. random variables from a normal distribution with mean $\mu$ and variance $1$. Find an unbiased estimator for $\tau(\mu):=P(X_1>0)$.

Attempt at a Solution

We can manipulate $\tau(\mu)$ to obtain that it is the cdf $\Phi$ of a standard normal by noting that \begin{aligned} P(X_1>0)&=P\left(\frac{X_1-\mu}{\sigma}>-\frac{\mu}{\sigma}\right)\\ &=1-\Phi(-\mu/\sigma) \\ &=\Phi(\mu/\sigma) \\ &=\Phi(\mu) \end{aligned} since $\sigma = 1$. We therefore want a statistic $S\big(\vec X\big)$ so that $E[S] = \Phi(\mu)$. Because $\overline X$ is the UMVUE for $\mu$, I am inclined to search for a statistic that is a function of $\overline X$; i.e., an $S$ such that $$ \int_{-\infty}^\infty S(t)\frac{n}{\sqrt{2\pi}}e^{-n^2(t-\mu)^2/2}dt = \int_{-\infty}^\mu\frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt $$ where the LHS is $E\big[S\big(\overline X\big)\big]$ in terms of the pdf for $\overline X$, which is a normal with mean $\mu$ and variance $\sigma^2/n=1/n$, and the RHS is the integral of the pdf of a standard normal. From here, I have considered:

  • A change of variables to get the limits of integration to agree.
  • Choosing $S=\Phi\big(\overline X \big)$, although there isn't any reason to expect this to be unbiased for $\Phi(\mu)$ just because $\overline X$ is unbiased for $\mu$.
  • Expressing the integral on the LHS in different forms; for example, in terms of the joint pdf of the $X_i$.

The ad-hoc nature of my approach here is representative of my endeavors to find unbiased estimators in general. I hold out hope for a more systematic approach to doing so than blindly manipulating half-baked guesses informed by complete and sufficient statistics for exponential family distributions.

  • 2
    $\begingroup$ Please consider adding the self-study tag. Have you considered en.wikipedia.org/wiki/Empirical_distribution_function ? $\endgroup$ – Christoph Hanck Dec 22 '16 at 12:39
  • $\begingroup$ Have you thought about looking at the likelihood to get the MLE for mu? Then any function of the MLE is an MLE.. This means you can get the MLE for tau(mu). But is it unbiased? I find the logic of your approach impeccible. $\endgroup$ – Michael R. Chernick Dec 22 '16 at 16:26
  • $\begingroup$ @AlecosPapadopoulos I assume $\mu$ is unknown; if it were known, I could just compute $P(X_1>0)$ directly without needing an estimator, I think, since $\sigma$ is known also. $\endgroup$ – Eric Kightley Dec 22 '16 at 19:38
  • $\begingroup$ @MichaelChernick I have indeed computed the MLE for $\mu$ (it's $\overline X$). I agree that $\tau(\overline X)$ is then the MLE for $\tau(\mu)$, butI can't figure out how to compute the integrals to check if this is unbiased (this is how I wound up trying to change variables to swap the order of integration in the OP). I'm updating the question now to include more details about why I am stuck on this integral. $\endgroup$ – Eric Kightley Dec 22 '16 at 19:44
  • 1
    $\begingroup$ @ChristophHanck I have added the self-study tag. I had not considered the EDF, which is indeed an unbiased estimator for $\Phi(\mu)$. However, finding the unbiased estimator is only one part of the problem, which also asks for the UMVUE. I suspect this ought to be given in terms of the complete and sufficient statistics for a normal distribution. I'll update the question to reflect my recent efforts using the EDF. $\endgroup$ – Eric Kightley Dec 22 '16 at 19:47

As a comment suggested, an unbiased estimator is (one minus) the empirical distribution function

$$\hat P(X_1 > 0) = 1-\hat F_X(0) = 1-\frac 1n \sum_{i=1}^n I\{x_i \leq 0\}$$

where $I\{\}$ is the indicator function, because

$$E[\hat P(X_1 > 0)]=1- E[\hat F_X(0)] = 1-\frac 1n \sum_{i=1}^n E[I\{x_i \leq 0\}] $$

$$= 1-\frac 1nn P(X_1\leq0) = P(X_1 > 0)$$

To the degree that we are estimating a probability with respect to a known threshold (in our case $0$), it doesn't matter whether we know the parameters of the distribution ($\mu$) or not.

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