# Find the MLE of $\hat{\gamma}$ of $\gamma$ based on $X_1, ... , X_n$

new user here self-studying some mathematical statistics.

I have a problem that has been tripping me up for a few days now. The problem is as follows:

For $$1 \leq i \leq n$$ and letting $$X_1, ... , X_n$$ be a random sample from a $$N(\phi, 1)$$ density, we define $$Y_i = 1$$ if $$X_i > 0$$, and we define $$Y_i = 0$$ if $$X_i \leq 0$$. Additionally, we let $$\gamma(\phi) = P_{\phi}[Y_i = 1]$$.

(i) Find the MLE of $$\hat{\gamma}$$ of $$\gamma$$ based on $$X_1, ... , X_n$$, and write the MLE in terms of the CDF of the standard normal distribution.

(ii) Find an approximate large sample $$90$$% confidence interval for $$\gamma$$ based on $$X_1, ... , X_n$$.

My issues are that I'm having trouble finding the MLE in the first part, and then I'm unsure of how to find the confidence interval for $$\gamma$$ based on $$X_1, ..., X_n$$.

I know in general for finding MLEs, you take the joint so you have your likelihood function, $$L(\phi, x)$$, then maybe if need be you take the log-likelihood function $$\mathscr{L}(\phi | x) = ln(L(\phi | x))$$, take this derivative with respect to the parameter of interest, and set it equal to zero, then solve for your parameter of interest and make sure to put a little hat on it at the end so you know it's your MLE.

So for this problem in particular, I know what the likelihood function looks like:

$$L(\phi | x) = ({\frac{1}{2\pi}})^{-\frac{n}{2}}e^{\frac{-1}{2}\sum_{i = 1}^{n}(x_i - \theta)^2}$$ and the log-likelihood function comes out to be $$\mathscr{L}(\phi|x) = \frac{n}{2}ln(2\pi) - \frac{1}{2}\Sigma_{i = 1}^{n}x_{i}^{2} + \phi\Sigma_{i = 1}^{n}x_i - \frac{n}{2}\phi^2$$.

But from here I don't know what to do. Any help would be greatly appreciated. Thanks for taking the time to read and consider my question(s).

• stats.stackexchange.com/q/407344/119261 Commented May 4, 2020 at 14:55
• The last term of the log-likelihood should be $-\frac{n}{2}\phi^2$.
– Ale
Commented May 4, 2020 at 14:57
• @StubbornAtom, thanks for the link! Commented May 4, 2020 at 15:18
• Do you mean based the MLE on $Y_1,\ldots,Y_n$ since otherwise it appears like a standard Normal problem. Commented May 4, 2020 at 17:39
• @Xi'an, nope, though I guess the person who originally made this problem could have made a typo. Commented May 4, 2020 at 18:22

For a random sample of iid normal RV's the MLE of the $$E[X_1]$$ is the sample mean. Hence, the MLE of $$\hat{\phi}$$ is simply the sample mean, $$\bar{X}$$. Then $$\gamma = P(X>0) = 1-\Phi(0-\phi),$$ where $$\Phi(\cdot)$$ is the standard normal cdf. By invariance of MLE, it follows that the MLE of $$\gamma$$ is $$\hat{\gamma} = 1-\Phi(-\hat{\phi}) = 1-\Phi(-\bar{X}).$$
Denote the 90% confidence interval for $$\phi$$ as $$[\hat{\phi}_{10},\hat{\phi}_{90}].$$ It follows that the 90% confidence interval for $$\gamma$$ is $$[1-\Phi(-\hat{\phi}_{10}),1-\Phi(-\hat{\phi}_{90})].$$