This is part(a) of exercise 6 of Chapter 9 from Wasserman's All of Statistics.

Let $X_1,\dots,X_n\sim N(\theta,1)$. Define $Y_i=\begin{cases} 1 &\text{ if }X_i>0 \\ 0 &\text{ if }X_i\le 0.\end{cases}$. Let $\psi=P(Y_1=1)$. I must find the maximum likelihood estimator of $\psi$.

What I've tried so far:

I know that $P(Y_1=1)=P(X_1>0)= 1-\Phi_{\theta}(0)$, where $\Phi_{\theta}$ is the c.d.f. of $N(\theta,1)$. I'm wondering if we can use the equivariance property, i.e., if $\tau=g(\theta)$ and $\widehat{\theta}$ is the MLE of $\theta$, then the MLE of $\tau$ is $g(\widehat{\theta})$.

  • $\begingroup$ It's an interesting question (+1) as to whether treating $Y_i$ as a Binomial variate with unknown probability parameter or using the underlying distribution of $X_i$ with unknown mean is better. I'd guess the former, personally, since $Y_i$ is distributed Binomial, but I could certainly be convinced otherwise. $\endgroup$ – jbowman May 8 '19 at 19:24
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    $\begingroup$ $P(Y_1=1)=\Phi(\theta)$ ($\Phi$ is CDF of standard normal), so MLE of $P(Y_1=1)$ is $\Phi(\hat\theta)$ where $\hat\theta$ is MLE of $\theta$, as you guessed. $\endgroup$ – StubbornAtom May 8 '19 at 21:37
  • $\begingroup$ @StubbornAtom Thank you; just to be sure, you use the fact that $\Phi_{\theta}(0)=\Phi(\theta)$, where $\Phi_{\theta}$ is the c.d.f for $N(\theta,1)$ and $\Phi$ is that for $N(0,1)$, right? Further, in Wasserman, it is calculated that the MLE for a normal with unknown mean and unknown standard deviation is the sample mean and sample variance. Since the standard deviation is known to be 1, will the sample mean still be the MLE of $\theta$? That is, do I need to repeat the MLE computations? $\endgroup$ – FakeAnalyst56 May 8 '19 at 21:57
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    $\begingroup$ 1. $\Phi_{\theta}(0)=P(X_1<0)=P(X_1-\theta<-\theta)=\Phi(-\theta)$, 2. That is correct; MLE computation is straightforward in any case. $\endgroup$ – StubbornAtom May 8 '19 at 22:05

First, denoting $\Phi$ to be the standard normal c.d.f., we have \begin{align*}\psi &= P(Y_1=1) \\ &= P(X_1>0) \\ &= 1-P(X_1 \le 0) \\ &= 1-P(X_1-\theta\le -\theta)\\ &=1-\Phi(-\theta)\\ &=\Phi(\theta),\end{align*} where in the third line, we use the fact that $X_1-\theta\sim N(0,1)$. Consequently, by equivariance of the MLE, as $\widehat{\theta}=\overline{X}_n$ is the MLE of $\theta$, $\Phi(\overline{X}_n)$ is the MLE of $\psi$.

Credit to StubbornAtom for the hint to this solution.

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    $\begingroup$ Note that you can only calculate $\bar{X}_n$ if $\{X_i\}$ are observable. $\endgroup$ – Taylor May 10 '19 at 17:21
  • $\begingroup$ @Taylor Does that mean if we are forced to compute the MLE of $\psi$ only using the information of $X_1$, the answer is that the MLE is $\Phi(X_1)$? $\endgroup$ – FakeAnalyst56 May 10 '19 at 18:05
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    $\begingroup$ If by "forced" you mean "allowed" because that data is able to be read into your machine, then yes, equivariance is the way to go. But if you only have the $Y$ data, then you have to use that different likelihood. $\endgroup$ – Taylor May 10 '19 at 18:07
  • $\begingroup$ If you observe $Y_i$ instead of $X_i$, then mle of $\psi$ is $\overline Y$ (assuming $Y_i$ is defined as such for every $i$). This is because $Y_i\sim\mathsf{Ber}(\psi)$ and we know that mle of the 'success' probability is the sample mean. If the question doesn't mention anything regarding observed variables then either answer could be correct. $\endgroup$ – StubbornAtom May 10 '19 at 18:48

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