# Maximum likelihood estimator for Bernoulli parameter based on standard normal

$$X_i \sim Normal(\psi,1), \ \ i = 1, ..., n$$

$$Y_i = 1$$ if $$X_i \ge 0.$$

$$Y_i = 0$$ if $$X_i < 0.$$

Let $$\theta = P(Y_i = 1)$$.

What is the MLE of $$\theta$$?

I know how to find the MLE of a Bernoulli success parameter, but I don't know how to do this question.

• Since $P(Y_i=1)=P(X_i\ge 0)=1/2$, I think you did not quote the question correctly. The distribution of $X_i$'s probably depends on $\theta$, in which case $P(Y_i=1)$ would be a function of $\theta$. Then the MLE can be found using the invariance property. Oct 28, 2018 at 16:53
• Yes, it is a test question. Yes, I did write it correctly. Yes, I agree that $\theta = 1/2$, but I don't know how to show it.
– MSE
Oct 28, 2018 at 17:30
• My guess is that the actual intent of the question is as follows: you don't know the mean of the Normal distribution, so you don't know $\theta$. You observe the $x_i$. What is the MLE of $\theta$? This could lead to the mistaken answer "the observed fraction of $x_i \geq 0$". (That's how I would try to answer it, at any rate.) Oct 28, 2018 at 17:51
• Thanks, @jbowman (and @StubbornAtom). Let's assume that you're right, and I will change the expected value of the normal distribution to be unknown. How would you solve it?
– MSE
Oct 28, 2018 at 18:54
• Hint: The MLE of a function of the unknown parameters is the function of the MLE of the unknown parameters (the "invariance property"). So, what are the unknown parameters? What are their MLEs? What is the function that transforms those parameters to the thing of interest ($P(Y_i=1)$)? Oct 28, 2018 at 18:58

Based on @jbowman's comments, I think that the solution is this:

$$\hat{\psi} = \bar{X}$$

$$\theta = P(X_i > 0) = P(\frac{X_i - \psi}{1} > \frac{0 - \psi}{1}) = P(Z > -\psi) = 1 - \Phi(-\psi) = \Phi(\psi)$$,

where $$\Phi$$ is the cumulative distribution function (CDF) of the standard normal distribution.

By the invariance principle, $$\hat{\theta} = \Phi(\hat{\psi}) = \Phi(\bar{X})$$

• No... what if $\bar{X} = 5$? What would you say $P(Y=1)$ is? Yet your formula would return a very, very small formula. Also, you need to take into account the fact that the variance of $\bar{X} \neq 1$, due to the sample size. But you are on the right track. Oct 30, 2018 at 13:49
• I can't control the value of the data. If the sample mean truly is 5, then $P(Y = 1)$ will be very big. I don't see anything wrong with that. The variance as given in the parent normal distribution is 1, so that's why I divided by 1. Why is this wrong?
– MSE
Oct 30, 2018 at 14:46
• If $\bar{X}=5$, then you have $\theta = \Phi(-5)$, which isn't very big. 2) You're right about the variance, I slightly mis-remembered the problem for a moment. Oct 30, 2018 at 15:55
• Ah - thank you, @jbowman. I wrote $P(Z > -\psi)$, but I interpreted it as $P(Z < \psi)$. I have corrected my solution. Does it look right now?
– MSE
Oct 30, 2018 at 17:34
• Yes, except you are mixing up your estimated parameters ($\hat{\theta}$) with your actual parameters in your second line: it's $\theta$ that equals $P(X_i > 0)$, and $\hat{\theta}$ is your estimate of that. Other than that, fine! Oct 30, 2018 at 18:29