Question

Let X_1,X_2,...X_n be a random sample from a Poisson distribution with parameter λ.Find an estimator of λ using
(i) the method of moments;
(ii) the method of maximum likelihood.
Also,compare the estimators obtained in parts i) and ii).

Answer 1

Answer on Question #85923 – Math – Statistics and Probability

Question

Let $X_1, X_2, \ldots, X_n$ be a random sample from a Poisson distribution with parameter $\lambda$. Find an estimator of $\lambda$ using

(i) the method of moments;

(ii) the method of maximum likelihood.

Also, compare the estimators obtained in parts i) and ii).

Solution

(i) MME

We know that $E(X) = \lambda$, from which we have a moment estimator of $\lambda$ as

$$\hat{\lambda} = \frac{1}{n} \sum_{i=1}^{n} X_i$$

Also, because we have $Var(X) = \lambda$, equating the second moments, we can see that $\lambda = E(X^2) - (E(X))^2$ so that

$$\hat{\lambda} = \frac{1}{n} \sum_{i=1}^{n} X_i^2 - \left(\frac{1}{n} \sum_{i=1}^{n} X_i\right)^2$$

Both are moment estimators of $\lambda$. Thus, the moment estimators may not be unique. We generally choose $\overline{X}$ as an estimator of $\lambda$, for its simplicity.

(ii) MLE

We have the probability mass function

$$p(x) = \frac{\lambda^x e^{-\lambda}}{x!}, \quad x = 0, 1, 2, \ldots, \quad \lambda > 0$$

Hence, the likelihood function is

$$L(\lambda) = \prod_{i=1}^{n} \frac{\lambda^{x_i} e^{-\lambda}}{x_i!} = \frac{\lambda^{\sum_{i=1}^{n} x_i} e^{-n\lambda}}{\prod_{i=1}^{n} x_i!}$$

Taking the natural logarithm, we have

$$\ln L(\lambda) = \sum_{i=1}^{n} x_i \ln \lambda - n\lambda - \sum_{i=1}^{n} \ln(x_i!)$$

Differentiate both sides with respect to $\lambda$

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

Find the value of $\lambda$ which maximizes $L(\lambda)$

$$\frac{d}{d\lambda}(L(\lambda)) = 0 \Rightarrow \frac{1}{\lambda} \sum_{i=1}^{n} x_i - n = 0$$

That is

$$\lambda = \frac {1}{n} \sum_ {i = 1} ^ {n} x _ {i} = \overline {{x}}$$

The second derivative

$$\frac {d ^ {2}}{d \lambda^ {2}} (L (\lambda)) = - \frac {1}{\lambda^ {2}} \sum_ {i = 1} ^ {n} x _ {i} < 0 \mathrm {f o r a l l} \lambda$$

Therefore, $MLE$ of $\lambda$ is

$$\hat {\lambda} = \overline {{X}},$$

which is the same as Method of Moments estimator.

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