# maximum likelihood estimate and the UMVUE

I have been working on this question and I am little confused about how to solve it.

To evaluate the prevalence of periodontal diseases in a population, suppose that $$x_i$$, $$i=1,\ldots,n$$ are the outcomes of $$n$$ iid observations of $$X$$ regarding the status of participants recruited in a clinical trial. $$X_i= 0$$ (if the patient does not have the periodontal disease) or $$1$$ (if the patient has the disease) in the population with the unkown survival rate $$p= P(X=0).$$

a) If it is known that $$0.3\leq P \leq 0.8,$$ find the MLE of $$P$$

b) if $$0 and $$n \geq 4$$, derive the UMVUE of $$g(p)= p^4 +3p^2-p^3 + 0.7$$

With question a, I do not understand how the range of $$p$$ would affect the MLE. Since the MLE depends on our X values. My initial approach was to find the MLE of bernoulli (p ) which is the mean but I do not know how the range of $$P$$ as given affect my answer.

For question b, I was thinking of using the rao blackwell theorem but I sincerely do not know how to set it up.

I have an exams in a few days and an understanding of this is very important.Looking forward to a response soon.

Thank you

• The restricted range of $p$ would naturally affect the MLE. Would you say sample mean is the most likely value of $p$ based on the data when say, $p<0.3$? May 30, 2019 at 22:03
• So do you mean I have to state that the mean must fall within 0.3 to 0.8? May 31, 2019 at 0:15
• Yes, and account for the cases $p<0.3$ and $p>0.8$ separately. May 31, 2019 at 3:45
• Please see our policy on self-study questions. You might want to add this tag also. May 31, 2019 at 19:20

I'm going to guess that where you wrote "If it is known that $$0.3\leq P \leq 0.8,$$ find the MLE of $$P$$", what you meant was "If it is known that $$0.3\leq p \leq 0.8,$$ find the MLE of $$p$$". Be careful about this sort of thing.
Let $$t=x_1+\cdots+x_n.$$
The likelihood function is $$L(p) = p^t (1-p)^{n-t} \text{ for } 0.3\le p \le 0.8.$$ Then we have $$\ell(p) = \log L(p) = t\log p + (n-t)\log(1-p)$$ and thus \begin{align} \ell\,'(p) = \frac t p - \frac{n-t}{1-p} = \frac{t - np}{p(1-p)} \quad\begin{cases} >0 & \text{if } 0 \le p < t/n, \\[5pt] = 0 & \text{if } p = t/n, \\[5pt] < 0 & \text{if } t/n Therefore \begin{align} & L \text{ increases on } [0,\, t/n], \\ & L \text{ decreases on } [t/n,\, 1]. \end{align} So: \begin{align} & \text{If } t/n \le 0.3 \text{ then } L \text{ decreases on } [0.3,\,0.8] \\ & \qquad \text{so the MLE is } 0.3. \\[10pt] & \text{If } t/n \ge 0.8 \text{ then } L \text{ increases on } [0.3,\,0.8] \\ & \qquad \text{so the MLE is } 0.8. \\[10pt] & \text{If } 0.3 < t/n < 0.8 \text{ then } L \text{ increases on } [0.3,\,t/n] \\ & \qquad \text{and decreases on } [t/n,\,0.8], \text{ so the MLE is } t/n. \end{align}
To find the UMVUE, note that you need to include a proof that it exists, since in some cases it doesn't. Note also that $$X_1X_2X_3X_4$$ is an unbiased estimator of $$p^4,$$ which you can condition on a complete sufficient statistic.