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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 $g(\widehat{\theta})$.

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  • $\begingroup$ 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, 2019 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. $\endgroup$ May 8, 2019 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? $\endgroup$ May 8, 2019 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. $\endgroup$ May 8, 2019 at 22:05

1 Answer 1

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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, 2019 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 $\Phi(X_1)$? $\endgroup$ May 10, 2019 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, 2019 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. $\endgroup$ May 10, 2019 at 18:48

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