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.

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

### Step 1: Write the likelihood function

Given that \( X_1, X_2, \dots, X_n \) are a random sample from a Poisson distribution with parameter \( \lambda \), the probability mass function for each observation \( X_i \) is:

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

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

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

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

### Step 2: Write the log-likelihood function

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

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

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

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

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

### Step 3: Differentiate the log-likelihood function

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

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

### Step 4: Solve for \( \lambda \)

Solving the equation for \( \lambda \):

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

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

Thus, the MLE for \( \lambda \) is the sample mean:

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

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

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.

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

Learn how to find the Maximum Likelihood Estimator (MLE) for the parameter λ in a Poisson distribution. This step-by-step guide covers the likelihood and log-likelihood functions, derivation of the MLE formula, and sample mean relationship. Suitable for undergraduate statistics students.

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