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One observation is taken from a discrete distribution with a parameter $\theta$. There are 3 possible values of $\theta$: 1, 2, and 3. The PMF is given below.

enter image description here

What is the MLE of $\theta$?

I suspect that $X = 3$ has something to do with the answer, because it has non-zero probabilities for all 3 values of $\theta$.

I also know that the MLE cannot be terribly complicated, because there is only one observation to work with.

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  • $\begingroup$ This looks like the right answer! Thank you! Would you be so kind to write it as an answer, so that I can vote it as a solution? $\endgroup$ – MSE Oct 26 '18 at 18:18
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Let $\hat \theta $ be the maximum likelihood estimate of $\theta$.

$$\hat \theta = \begin{cases} 1, & \text{for } X = 0 \text{ or } 1\\ 2, & \text{for } X = 3\\ 3, & \text{for } X=4\\ 2 \text{ with prob 0.5 and 3 with prob 0.5}&\text{for } X=2 \end{cases} $$

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  • $\begingroup$ This may just be convention, but I'm not used to seeing a stochastic maximum likelihood estimate. $\endgroup$ – Frank Oct 26 '18 at 19:46
  • $\begingroup$ @a_statistician If $X=2$, then is it correct to say that $\hat{\theta} = 2.5$? I say this because the average value of the 2 values of $\theta$ for that particular value of $X$ is 2.5. $\endgroup$ – MSE Oct 28 '18 at 15:50
  • $\begingroup$ $\hat \theta = 2.5$ is not the answer for this problem. If $X=2$, find a fair coin, specify head to 2, throw coin, if head $\hat \theta = 2$, otherwise $\hat \theta = 3$ $\endgroup$ – user158565 Oct 28 '18 at 15:56
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It depends on the sample value x you have drawn. MLE maximizes the likelihood of observing the data with regard to the parameter. There is no MLE without data.

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    $\begingroup$ This is really more of a comment than an answer. $\endgroup$ – jbowman Oct 26 '18 at 17:35
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    $\begingroup$ In this question, the MLE is considered to be a function defined on the sample space, with image in the parameter space. The question provides enough information to define an MLE. $\endgroup$ – whuber Oct 26 '18 at 18:48
  • $\begingroup$ can I convert this to a comment? $\endgroup$ – bellmaneqn Oct 26 '18 at 19:21

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