Salmon sometimes carry a parasite anisakis simplex which they pick up when feeding on krill at sea. The number of parasites on each fish might be assumed a random variable $X$ having a probability distribution $p(x)$ below:

$$p(x)= \frac{\theta^x e^{-\theta}}{x!}\,,\quad x= 0,1,2...$$

You may assume that $E(X) = \theta$ and $V(X) = \theta$.

A random sample of $n$ salmon were examined and the number of parasites per fish, $x_1,$ $x_2,$ $\ldots,$ $x_n$ counted.

Show that the Maximum Likelihood Estimator (MLE) of $\theta$ is the sample mean.

Show that the MLE is unbiased for $\theta$. Find an expression for the variance of the MLE in terms of the parameter $\theta$ and hence show that the MLE is a consistent estimator of $\theta$.

A random sample of $n = 20$ salmon taken from a particular Scottish river were examined and the number of parasites recorded as follows:

xi : 2  0  4  0  1  0  3  2  0  5  1  2  0  3  0  1  6  2  2  7

Using these data, estimate $\theta$ giving approximate 95% confidence limits for the true value.

Show that the probability of one or more parasites being found on a fish is given by

$$P(X \geq 1) = 1 - e^{-\theta}$$

A random sample of $n = 60$ fish were taken from a different river and examined. Suppose all that is known is that $43$ fish were found to have at least one parasite and $17$ salmon were found to have no parasites. Find an expression for the likelihood of this and hence obtain the MLE of $\theta$ given this information.

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    $\begingroup$ You might have more success if you narrow down what it is that you are asking. Is the MLE procedure challenging, or do you have questions about how to show that the MLE of $\theta$ is unbiased and consistent? Are you uncertain about confidence intervals? The FAQ provides guidance for asking good questions. As it stands, it looks like you're asking us to do your homework. $\endgroup$ – Sycorax says Reinstate Monica Mar 27 '13 at 17:24

For the first question, you find the MLE through the following steps:

(1) take the likelihood function of the PMF (this is a Poisson PMF in this example):

$$\cal{L}(\theta) = \frac{1}{\prod_{i=1}^nx!}\,\theta^{\sum_{i=1}^n x_i}\,e^{-n \theta}$$

(2) Since the function and the log of the function are maximized at the same place and since the log will make this easier to work, take the log of the likelihood.

$$\log \cal{L}(\theta) = \log \frac{1}{\prod_{i=1}^nx!}\, + {\sum_{i=1}^n x_i} \log \theta - n \theta$$

(3) take the derivative with respect to $\theta$

$$\frac{d}{d\theta}\log \cal{L}(\theta) = \frac{\sum_{i=1}^n x_i}{\theta} - n$$

(4) set the derivative equal to zero to see the value of $\theta=\hat\theta$ where it's maximized:

$$\frac{\sum_{i=1}^n x_i}{\hat\theta} - n = 0$$

$$\frac{\sum_{i=1}^n x_i}{\hat\theta} = n $$

(5) Solve for $\hat{\theta}$

$$\frac{\sum_{i=1}^n x_i}{\hat\theta} = n\, ;$$

$$\hat\theta = \frac{\sum_{i=1}^n x_i}{n}$$

You can see that theta is equal to the sample mean. In order to make sure that you've found a maximum, and not a minimum, you should take the second derivative of the function and make sure that it's negative.

$$\frac{d^2}{d\theta^2} = -\frac{\sum_{i=1}^n x_i}{\theta^2}$$

Since the second derivative is less than zero (both n the total observations and ${\sum_{i=1}^n x_i}$ are both positive), you've found a maximum.

Hope this helps for your first answer. this should get you started. maybe you should try the rest and where you get caught up ask for help. that tends to elicit better responses.

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    $\begingroup$ +1 (You will find the $\TeX$ works much better when you include entire formulas, rather than mixing and matching ordinary text with markup. E.g., "$L(\theta)$" (\$L(\theta)\$) looks much better than "L($\theta$)" (L(\$\theta\$)). Center the formulas with double escapes (\$\$).) $\endgroup$ – whuber Mar 28 '13 at 4:40

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