Math Problem Statement

Let X1,· · · , Xn be a random sample from a Poisson distribution with parameter λ, i.e., f (x) = λxe λ x! , x = 0, 1, . . . Find the MLE of the parameter λ.

Solution

The Maximum Likelihood Estimator (MLE) of the parameter λ\lambda in a Poisson distribution can be found as follows:

Step 1: Write the likelihood function

Given that X1,X2,,XnX_1, X_2, \dots, X_n are a random sample from a Poisson distribution with parameter λ\lambda, the probability mass function for each observation XiX_i is:

f(xi)=λxieλxi!,xi=0,1,2,f(x_i) = \frac{\lambda^{x_i} e^{-\lambda}}{x_i!}, \quad x_i = 0, 1, 2, \dots

The likelihood function L(λ)L(\lambda) for the entire sample is the product of the individual likelihoods:

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

L(λ)=λi=1nxienλi=1nxi!L(\lambda) = \frac{\lambda^{\sum_{i=1}^n x_i} e^{-n\lambda}}{\prod_{i=1}^n x_i!}

Step 2: Write the log-likelihood function

To simplify, we take the natural logarithm of the likelihood function, known as the log-likelihood function (λ)\ell(\lambda):

(λ)=logL(λ)=log(λi=1nxienλi=1nxi!)\ell(\lambda) = \log L(\lambda) = \log \left( \frac{\lambda^{\sum_{i=1}^n x_i} e^{-n\lambda}}{\prod_{i=1}^n x_i!} \right)

(λ)=(i=1nxi)logλnλi=1nlog(xi!)\ell(\lambda) = \left( \sum_{i=1}^n x_i \right) \log \lambda - n\lambda - \sum_{i=1}^n \log (x_i!)

Since the last term i=1nlog(xi!)\sum_{i=1}^n \log (x_i!) does not depend on λ\lambda, it can be ignored for maximizing the log-likelihood.

(λ)=(i=1nxi)logλnλ\ell(\lambda) = \left( \sum_{i=1}^n x_i \right) \log \lambda - n\lambda

Step 3: Differentiate the log-likelihood function

To find the MLE, we differentiate the log-likelihood function with respect to λ\lambda and set it equal to zero:

d(λ)dλ=i=1nxiλn=0\frac{d\ell(\lambda)}{d\lambda} = \frac{\sum_{i=1}^n x_i}{\lambda} - n = 0

Step 4: Solve for λ\lambda

Solving the equation for λ\lambda:

i=1nxiλ=n\frac{\sum_{i=1}^n x_i}{\lambda} = n

λ=1ni=1nxi\lambda = \frac{1}{n} \sum_{i=1}^n x_i

Thus, the MLE for λ\lambda is the sample mean:

λ^=Xˉ=1ni=1nXi\hat{\lambda} = \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i

Conclusion:

The Maximum Likelihood Estimator (MLE) for the parameter λ\lambda in the Poisson distribution is the sample mean λ^=Xˉ\hat{\lambda} = \bar{X}.

Do you have any further questions or need more details on this?

5 Related Questions:

  1. How does the Poisson distribution differ from the binomial distribution?
  2. What are the properties of a Poisson process?
  3. Can you explain the role of the likelihood function in statistical estimation?
  4. How is the MLE method used for other distributions, such as Gaussian?
  5. What is the relationship between the sample mean and the MLE in general?

Tip:

When computing the MLE, always remember to check whether the log-likelihood function has any critical points by differentiating it and verifying solutions for possible maxima.

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Math Problem Analysis

Mathematical Concepts

Maximum Likelihood Estimation
Poisson Distribution
Statistical Estimation

Formulas

Likelihood function: L(λ) = ∏(λ^xi e^−λ / xi!)
Log-likelihood function: ℓ(λ) = Σ(xi) log(λ) − nλ
MLE for λ: λ̂ = (1/n) Σ(Xi)

Theorems

Maximum Likelihood Estimation

Suitable Grade Level

Undergraduate Statistics