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The random variable X takes on the values 0, 1, or 2 according to the following probability distribution:

$P(X = 0) = P(X = 1) = p^2, \ and P(X = 2) = 1 - 2p^2, for \ p \in C.$

(1) Determine whether this family of probability distributions is complete when C is the open interval $(0, \frac{1}{\sqrt{2}})$.

(2) Suppose that C = $\lbrace \frac{1}{3}, \frac{2}{3} \rbrace$, that is, the only two possible values of p are $\frac{1}{3}$ and $\frac{2}{3}$. For a single X observation, find the maximum likelihood estimator, $\hat{p}_{MLE}$, for p.

For (1), I set $0 = E_p[g(x)] = g(0)p^2 + g(1)p^2 + g(2)(1-2p^2)$

$= g(0)p^2 + g(1)p^2 - 2g(2)p^2 + g(2) = p^2 \Bigl(g(0) + g(1) - 2g(2)\Bigr) + g(2) $. Thus $E_p[g(x)] = 0$ if $g(2) = 0$ and $g(1) = -g(0)$. Therefore it is not complete.

For (2), I set up a table and got $\hat{\theta}_{x=0,1}=\frac{2}{3}$ and $\hat{\theta}_{x=2}=\frac{1}{3}$.

Not sure if I am simplifying this too much. Any assistance is greatly appreciated.

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up vote 0 down vote accepted

I just quickly did the problem, and your answers match what I got, so you made the right assumptions and are, therefore, correct.

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Thanks! That makes me feel a lot better. – user45185 Feb 11 '13 at 7:25

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