Properties of Poisson Distribution Estimators

StatisticsMaximum Likelihood EstimationHard

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2. Let $X_1, \dots, X_n$ be a sample from a $\text{Poisson}(\lambda)$ distribution (so they are independent $\text{Poisson}(\lambda)$ distributed random variables). (a) Give the formula for $\mathbb{P}(X_1 = x_1, \dots, X_n = x_n)$ (with $x_i$ nonnegative integers). (b) Show that $\overline{X} = \frac{1}{n} \sum_{i=1}^n X_i$ is the maximum likelihood estimator of $\lambda$. (c) Show that the Fisher information $I(\lambda)$ is equal to $1/\lambda$. (d) Use the Cramér-Rao bound to show that $\overline{X}$ has minimum variance among all unbiased estimators of $\lambda$. (e) Give a consistent estimator of $I(\lambda)$.

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Step 1

Hi Ashish, let's solve this problem on statistical inference for a Poisson distribution together.

Problem 2: Poisson Distribution

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Step 2

For part (a), we need the joint probability mass function for a sample of independent Poisson variables.

$$P(X_1=x_1, \dots, X_n=x_n) = \prod_{i=1}^{n} \frac{e^{-\lambda} \lambda^{x_i}}{x_i!}$$
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Step 3

We can simplify this product. The terms e to the negative lambda repeat n times, and the lambdas multiply, so we add their exponents.

$$P(X_1=x_1, \dots, X_n=x_n) = e^{-n\lambda} \frac{\lambda^{\sum x_i}}{\prod x_i!}$$
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Step 4

This expression is the likelihood function for the sample, which we will use to find the maximum likelihood estimator.

$$L(\lambda) = e^{-n\lambda} \lambda^{\sum x_i} \left( \prod x_i! \right)^{-1}$$
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Step 5

Now for part (b), we find the maximum likelihood estimator for lambda. We start by taking the natural log of the likelihood function.

$$l(\lambda) = \log(L(\lambda)) = -n\lambda + \left( \sum x_i \right) \log(\lambda) - \log \left( \prod x_i! \right)$$
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Step 6

Next, we take the derivative with respect to lambda and set it to zero.

$$\frac{d}{d\lambda} l(\lambda) = -n + \frac{1}{\lambda} \sum x_i = 0$$
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Step 7

Solving for lambda, we get the sample mean, which is the sum of x sub i divided by n.

$$\lambda = \frac{1}{n} \sum x_i = \overline{X}$$
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Step 8

Moving to part (c), we compute the Fisher information. Recall that the Fisher information for a single observation is the negative expectation of the second derivative of the log-likelihood.

$$l_i(\lambda) = -\lambda + x_i \log(\lambda) - \log(x_i!)$$
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Step 9

Taking the first derivative gives us negative one plus x sub i over lambda.

$$\frac{d}{d\lambda} l_i(\lambda) = -1 + \frac{x_i}{\lambda}$$

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About This Question

Subject
Statistics
Topic
Maximum Likelihood Estimation
Difficulty
Hard
Question Type
Open Ended

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