Find the maximum likelihood estimator for the parameter λ of the Poisson 
distribution on the basis of a sample of size n . Also, find its variance.

To find the **Maximum Likelihood Estimator (MLE)** for the parameter \(\lambda\) of a Poisson distribution and its variance, we'll proceed step-by-step.

### 1. The Poisson Distribution

A random variable \(X\) is said to have a Poisson distribution with parameter \(\lambda > 0\) if its probability mass function (PMF) is given by:

\[
P(X = x; \lambda) = \frac{\lambda^x e^{-\lambda}}{x!}, \quad x = 0, 1, 2, \ldots
\]

### 2. Likelihood Function

Given a sample of size \(n\), say \(X_1, X_2, \ldots, X_n\), that are independent and identically distributed (i.i.d.) from a Poisson distribution with parameter \(\lambda\), the likelihood function \(L(\lambda)\) is the joint probability of observing the sample dataTo find the **Maximum Likelihood Estimator (MLE)** for the parameter \(\lambda\) of a Poisson distribution and its variance, we'll proceed step-by-step.

### 1. The Poisson Distribution

A random variable \(X\) is said to have a Poisson distribution with parameter \(\lambda > 0\) if its probability mass function (PMF) is given by:

\[
P(X = x; \lambda) = \frac{\lambda^x e^{-\lambda}}{x!}, \quad x = 0, 1, 2, \ldots
\]

### 2. Likelihood Function

Given a sample of size \(n\), say \(X_1, X_2, \ldots, X_n\), that are independent and identically distributed (i.i.d.) from a Poisson distribution with parameter \(\lambda\), the likelihood function \(L(\lambda)\) is the joint probability of observing the sample 

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

The log-likelihood function, \(\ell(\lambda)\), is easier to work with:

\[
\ell(\lambda) = \log L(\lambda) = \left( \sum_{i=1}^n x_i \right) \log \lambda - n\lambda - \sum_{i=1}^n \log(x_i!)
\]

### 3. Maximum Likelihood Estimator (MLE)

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

\[
\frac{d\ell(\lambda)}{d\lambda} = \frac{\sum_{i=1}^n x_i}{\lambda} - n = 0
\]

Solving for \(\lambda\), we get:

\[
\hat{\lambda} = \frac{\sum_{i=1}^n X_i}{n} = \bar{X}
\]

Thus, the **MLE of \(\lambda\) is \(\hat{\lambda} = \bar{X}\), the sample mean.**

### 4. Variance of the MLE

For a Poisson-distributed random variable \(X_i \sim \text{Poisson}(\lambda)\), the mean and variance are both equal to \(\lambda\):

\[
E[X_i] = \lambda, \quad \text{Var}(X_i) = \lambda
\]

The MLE, \(\hat{\lambda} = \bar{X}\), is the average of \(n\) i.i.d. Poisson random variables. Therefore, the variance of \(\hat{\lambda}\) is:

\[
\text{Var}(\hat{\lambda}) = \text{Var}\left( \frac{1}{n} \sum_{i=1}^n X_i \right)

Learn how to find the Maximum Likelihood Estimator (MLE) for the parameter λ of a Poisson distribution based on a sample of size n. Also, understand the variance of the MLE. This detailed solution covers the Poisson distribution PMF, likelihood function, log-likelihood function, and variance calculation.

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