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I have a question where I'm given a discrete random variable $X$ with possible outcomes and distributions, respectively:

$X = 0, 1, 2$

$P(X) = 1 - 2\theta, 1 - \theta, 3\theta - 1$.

There is one observation, and it is that of $X = 1$. The question goes on to say that $\frac{1}{3} \leq \theta < \frac{1}{2}$, and asks to find the MLE given this info.

Answer: The answer is $\theta = 0$. This makes sense intuitively, but specifically asks to calculate the MLE, with taking the derivative of the log likelihood function. But,

$$\frac{\partial\log L(X\mid \theta)}{\partial\theta} = \frac{\partial}{\partial \theta} \log(1-\theta) = -\frac{1}{1-\theta} = 0,$$

which is undefined at $0$. What am I missing here?

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If the likelihood is $1-\theta$ and $1/3 \le \theta < 1/2$, then the choice of $\theta$ in this range that maximizes the likelihood is $\theta=1/3$. No need for derivatives.

(Note that the conditions $1/3 \le \theta < 1/2$ are necessary for the problem to make sense. If $\theta < 1/3$, then $P(X=2) < 0$; if $\theta > 1/2$ then $P(X=0)<0$.)

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  • $\begingroup$ This is what I thought... $\theta = \frac{1}{3}$ not $0$? The answer to problem says that $\theta = 0$. I figured there was something I was missing, but maybe a typo in the question then. $\endgroup$
    – jj8989
    Jun 26, 2017 at 5:58

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