Based on a random sample of size n from a normal distribution, $X$~$N(\mu, \sigma^2)$ find the MLE (maximum likelihood estimator) of the following:

$P[X>c]$ for arbitrary $c$.

This seems to be strange question, and the provided solution is even more troubling:

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Would that not be the method of moments estimate? Surely that solution isn't correct. But if it's not... then how exactly do I calculate the MLE? I'm just downright confused.


The question, as you wrote it, is worded in an unclear way. My interpretation is the following:

Suppose you have a sample of $X_i\sim \mathcal{N}(\mu,\sigma)$, which are i.i.d. Find the Maximum likelihood estimate for the two parameters, $\mu$ and $\sigma$.


$$\mathcal{L}(\mu,\sigma)=(\frac{1}{\sqrt{2\pi\sigma^2}})^n\cdot \prod_{i=1}^ne^{-\frac{(X_i-\mu)^2}{2\sigma^2}}$$ which is the likelihood function, that we seek to maximize. To that end, we take its logarithm; it will make the calculation easier, while preserving the extrema, being a monotonically increasing function. Thus

$$\mathcal{F}=\ln(\mathcal{L}(\mu,\sigma))=-\frac{n}{2}ln(2\pi)-\frac{n}{2}ln(\sigma^2)+\sum_{i=1}^n-\frac{(X_i-\mu)^2}{2\sigma^2}$$ Now, you simply differentiate with respect to the two parametrs and equate to zero $$\frac{\partial}{\partial\mu}\mathcal{F}=0\implies \mu=\sum_{i=1}^n\frac{X_i}{n}$$ $$\frac{\partial}{\partial\sigma}\mathcal{F}=0\implies \sigma^2=\sum_{i=1}^n\frac{(X_i-\mu)^2}{n}$$

So, the estimated distribution, given $\bar X=\sum_{i=1}^n\frac{X_i}{n}$, is

$$X\sim \mathcal{N}\left(\bar X,\sqrt{\sum_{i=1}^n\frac{(X_i-\bar X)^2}{n}}\right)$$ Thus it follows that $$P(X>c)=1-\phi\left(\frac{c-\bar X}{\sqrt{\sum_{i=1}^n\frac{(X_i-\bar X)^2}{n}}}\right)$$

  • $\begingroup$ You're right, it's very unclear, but that's word-for-word from the book. But I think it's clear that they are intending for the value $c$ to appear somewhere, but I'm really not sure. $\endgroup$ – ereHsaWyhsipS Feb 14 '18 at 3:20
  • $\begingroup$ @ereHsaWyhsipS c is just a dummy variable. I edited the answer, as to use the wording of their solution. Let me know if you understand now. $\endgroup$ – Žiga Sajovic Feb 14 '18 at 3:40
  • $\begingroup$ This explain the procedure of finding the MLE of a normal distributions very well. The MLE and MoM estimator can coincide in some case, like this one. In the second line of the quoted solution from OP, I think it is more important to understand the functional invariance property of MLE. $\endgroup$ – BGM Feb 14 '18 at 9:43

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