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Consider independent trials, such that for each trial

a type 1 outcome will occur with probability $p_1= \frac{1}{2} + \frac{\theta}{4}$

a type 2 outcome will occur with probability $p_2= \frac{1}{2}−\frac{\theta}{2}$

a type 3 outcome will occur with probability $p_3= \frac{\theta}{4}$

where $\theta\in(0,1)$

(a) Letting $N_i$ be the number of type $i$ outcomes that occur in $n$ trails give an unbiased frequency substitution estimator of $\theta$ based on just $N_1$.

(b) Give the variance of the unbiased estimator requested in part (a)

(c) Given that the observed counts for the three possible outcomes that occur in $20$ trials are $n_1= 11$, $n_2= 5$, and $n_3= 4$, give the mle of $\theta$.

My Attempt:

(a) We have

$$p_1-\frac{1}{2}=\frac{\theta}{4} \Rightarrow \hat{\theta}=4p_1-2=4\left(\frac{N_1}{n}\right)-2$$

(b) We have

$$\begin{align*} \mathsf{Var}(\hat{\theta}) &=\mathsf{Var}(4\hat{p_1}-2)\\\\ &=\mathsf{Var}\left(4\left(\frac{N_1}{n}\right)-2\right)\\\\ &=\left(\frac{4}{n}\right)^2 \mathsf{Var}(N_1) \\\\ &=\left(\frac{16}{n^2}\right) np_1(1-p_1)\\\\ &=\frac{16}{n}\left(\frac{\theta+2}{4}\right)\left(\frac{2-\theta}{4}\right)\\\\ &=\frac{4-\theta^2}{n} \end{align*}$$

(c) We have

$$\begin{align*} L(\vec{\theta}\mid\vec{N}) &=\frac{(N_1+N_2+N_3)!}{N_1!N_2!N_3!}\left(\frac{1}{2}+\frac{\theta}{4}\right)^{N_1}\left(\frac{1}{2}-\frac{\theta}{2}\right)^{N_2}\left(\frac{\theta}{4}\right)^{N_3}\\\\ &=c\cdot\left(\frac{1}{2}+\frac{\theta}{4}\right)^{N_1}\left(\frac{1}{2}-\frac{\theta}{2}\right)^{N_2}\left(\frac{\theta}{4}\right)^{N_3}\\\\ \end{align*}$$

Upon taking the natural log of both sides, we get

$$\mathscr{L}(\theta\mid\vec{N})=\text{log}c+N_1 \text{log }\left(\frac{1}{2}+\frac{\theta}{4}\right)+N_2\text{log}\left(\frac{1}{2}-\frac{\theta}{2}\right)+N_3\text{log}\left(\frac{\theta}{4}\right)$$

Hence

$$0=\frac{d}{d\theta}\mathscr{L}(\theta\mid\vec{N})=\frac{N_1}{\theta+2}-\frac{N_2}{1-\theta}+\frac{N_3}{\theta}$$

Solving for $\theta$ is nontrivial so I used software to obtain

$$\hat{\theta}_{MLE}=\frac{-N_1+2N_2+N_3\pm \sqrt{9N_3^2+12N_2N_3+6N_3N_1+4N_2^2-4N_2N_1+N_1^2}}{-2n}$$

From here, I get that $\hat{\theta}_{MLE}=\{0.562, -0.712\}$ but since we are given that $\theta\in(0,1)$ it must be that $\hat{\theta}_{MLE}=0.562$. Is it okay to say this or would the mle not exist since it is not unique?

Are these valid solutions?

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  • $\begingroup$ Capital $N$ and small $n$ are switched in question a and c. $\endgroup$ – user158565 Oct 12 '18 at 18:28
  • $\begingroup$ part of them were fixed. use 0.562 as a solution. Then 11/2.562 + 5/0.438 + 4/0.562 > 20 $\ne$ 0. My calculation is wrong?. You forget the log after N_2 and N_3 in log likelihood. $\endgroup$ – user158565 Oct 12 '18 at 18:47
  • $\begingroup$ Whoops, I think I did the calculations with the logs present though. $\endgroup$ – Remy Oct 12 '18 at 18:50
  • $\begingroup$ There's a minus sign before $\frac{5}{0.438}$ $\endgroup$ – Remy Oct 12 '18 at 18:51
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    $\begingroup$ Sorry. Then you can say that $\hat \theta_{MLE}=0.562.$ not unique is common, Generally you can find the other solutions beyond your range. $\endgroup$ – user158565 Oct 12 '18 at 18:58

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