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Consider the following question:

Suppose $X_{i}\sim N(\beta_{1}z_{i},\sigma^2)$. For a fixed $u$ and known $\beta_{1}$ calculate the probability $X_{i}$ is less than u. Derive the MLE of this probability.

Now, the probability in the second sentence can be derived by using a linear transformation to the standard normal. Namely:

$$\Phi\left(\frac{u-\beta_{1}z_{i}}{\sigma}\right) \tag{1}$$

My issue is with deriving the MLE of (1). My calcualtions are as follows:

\begin{align} L(z_{i})&=\prod_{i}^{n} \Phi\left(\frac{u-\beta_{1}z_{i}}{\sigma}\right) \tag{2}\\ \implies \textit{l}(z_{i})&= \sum_{i}^{n}\ln\left(\Phi\left(\frac{u-\beta_{1}z_{i}}{\sigma}\right)\right)\\ \implies \frac{\partial\textit{l}(z_{i})}{\partial z_{i}}&= \sum_{i}^{n} \frac{f(\frac{u-\beta_{1}z_{i}}{\sigma})}{\Phi(\frac{u-\beta_{1}z_{i}}{\sigma})} \tag{3} \end{align}

Set (3) equal to 0 to obtain the MLE of $z_{i}$. Specifically:

\begin{align} \sum_{i}^{n} \frac{f(\frac{u-\beta_{1}z_{i}}{\sigma})}{\Phi(\frac{u-\beta_{1}z_{i}}{\sigma})} =0 \tag{4} \end{align}

It is at this point I am unsure how to proceed as the denominator of equation (4) does not allow me to cancel out any terms with the numerator. Furthermore, I am uncomfortable with step (2) as this doesn't satisfy the definition of the likelihood function.

Mohammed
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    Hint: The easiest way to do this exploits a very nice property of maximum likelihood estimators, ie. one would not have to maximize the transformed likelihood for the probability of interest – Tyrel Stokes Dec 27 '20 at 21:15
  • Would this work? Let $I(x_{i}|z_{i})=\int_{-\infty}^{u}f(x_{i}|z_{i})$. This is equivalent to calculating $P(X_{i} – Mohammed Dec 28 '20 at 03:02
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    Yeah that works. You need far less than what you showed, but good thinking. See this answer about the invariance property of the MLE! https://stats.stackexchange.com/questions/77573/invariance-property-of-mle-what-is-the-mle-of-theta2-of-normal-barx2#:~:text=Invariance%20property%20of%20MLE%3A%20if,is%20f(%CB%86%CE%B8).&text=The%20book%20says%2C%20%22For%20example,can't%20use%20invariance%20property. – Tyrel Stokes Dec 28 '20 at 04:40
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    The question is ill-posed. The problem is not to derive the MLE of $\sigma$ based on $\mathbb I_{X_i – Xi'an Dec 28 '20 at 08:11
  • Oh I see! I had been misinterpreting the question. In this case one could use the invariance property of the MLE to calculate the desired estimator. – Mohammed Dec 28 '20 at 09:04

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