# Find MLE and CI for $\psi = \mathbb{P}_\theta(X_1>0)$ when $X_1,...,X_n \sim \text{IID } \mathcal{N}(\theta,1)$

Question: Let $$X_1,\cdots,X_n \sim \text{IID }\mathcal{N}(\theta,1)$$ where $$\theta\in\mathbb{R}$$ is unknown and let $$\psi = \mathbb{P}_\theta(X_1>0)$$. Find the maximum likelihood estimator $$\hat{\psi}$$ of $$\psi$$ and find a 95% confidence interval for $$\psi$$.

Comments: I think I understand the first part of the two part question. From the log-likelihood function, it can be shown that $$\hat{\theta}_{MLE}=\bar{X}_n=:\sum_{i=1}^n X_i/n.$$ Then, with the invariance property of the MLEs, we have $$\hat{\psi}_{MLE} = \mathbb{P}\left(\mathcal{N}(\bar{X}_n,1)>0\right).$$

My question is how do we construct the confidence interval? I am having difficulty calculating the any of the moments $$\mathbb{E} \hat{\psi}_{MLE}$$ and $$\mathbb{E} \hat{\psi}_{MLE}^2$$ and hence $$\mathbb{V}(\hat{\psi}_{MLE}).$$

Since $$\bar{X}_n\sim \mathcal{N}(\theta,1/n)$$, letting $$f_{\bar{X}_n}(x)$$ be the density function, I have tried using Tonelli's to switch the order of integration on the following expression $$\mathbb{E} \hat{\psi}_{MLE}=\int_{\mathbb{R}} \mathbb{P}\left(\mathcal{N}(x,1)>0\right) f_{\bar{X}_n}(x)dx=\int_{\mathbb{R}}\int_0^\infty \frac{1}{\sqrt{2\pi}}e^{-(y-x)^2/2}dy \frac{1}{\sqrt{2\pi n}}e^{-(x-\theta)^2/(2\sqrt{n})}dx,$$ but Tonelli's seems to not be the correct route. How do I obtain the variance for the confidence interval?

Edit: Further, I am interested in finding the (nondegenerate) asymptotic distribution of $$\hat{\psi}_{MLE}$$.

$$\psi$$ is a smooth function of $$\theta$$, so the delta-method says

$$\sqrt{n}(\psi(\hat\theta)-\psi(\theta))\stackrel{d}{\to} N(0, \sigma^2)$$ where $$\sigma^2$$ is the limiting value of $$n\psi'(\theta)^2\mathrm{var}[\hat\theta]$$

You know $$\mathrm{var}[\hat\theta]=1/n$$, so you just need $$\psi'(\theta)$$, which is $$\phi(0-\theta)$$ by the fundamental theorem of calculus (or something very close to it).

And a check

> theta<-1
>  n<-50
>  barx<-rnorm(100000,m=theta,s=1/sqrt(n))
>  psi<-function(theta) pnorm(0,theta,1)
> psi(1)
[1] 0.1586553
> mean(psi(barx))
[1] 0.1609813
> dnorm(0,1,1)^2
[1] 0.05854983
> var(psi(barx))*50
[1] 0.05881804

• Yes! I keep forgetting to check for the delta method. However, to construct the confidence interval for fixed $n$, do we not need to find the standard error for fixed $n$? I understand that it will converge to the limiting variance from the delta method but how does one get an expression for the confidence interval for fixed n? Jun 11, 2020 at 0:55
• The fixed-$n$ variance is approximately $1/n$ times the asymptotic variance -- that's how my simulation check worked, since I was comparing $n$ times the simulation variance to the asymptotic variance. Jun 11, 2020 at 1:01
• Thanks for the suggestion. Any ideas on getting an exact expression? Jun 11, 2020 at 1:23
• I would be a bit surprised if there were a closed-form expression for the finite-sample variance. Jun 11, 2020 at 1:44