1). Let \( x_{2}, x_{2}, \ldots x_{n} \) be an i.i.d. sample from a Poisson distribution with parameter \( \mu \) \[ P\left(X=\frac{x}{\mu}\right)=\frac{\mu^{x} e^{-\mu}}{x!} \] i.e. Find the MLE of the parameter \( \mu \)

To find the maximum likelihood estimate (MLE) of the parameter \( \mu \) for an i.i.d. sample from a Poisson distribution, we start with the likelihood function for the sample \( x_1, x_2, \ldots, x_n \) given by: \[ L(\mu) = \prod_{i=1}^{n} \frac{\mu^{x_i} e^{-\mu}}{x_i!} \] This can be simplified to: \[ L(\mu) = \frac{\mu^{\sum x_i} e^{-n\mu}}{\prod_{i=1}^{n} x_i!} \] To find the MLE, we take the logarithm of the likelihood function (the log-likelihood): \[ \ell(\mu) = \log L(\mu) = \sum_{i=1}^{n} \left( x_i \log \mu - \mu - \log(x_i!) \right) \] We can focus on the terms that involve \( \mu \): \[ \ell(\mu) = \left( \sum_{i=1}^{n} x_i \right) \log \mu - n\mu - \sum_{i=1}^{n} \log(x_i!) \] Next, we differentiate the log-likelihood with respect to \( \mu \) and set it equal to zero: \[ \frac{d\ell}{d\mu} = \frac{\sum_{i=1}^{n} x_i}{\mu} - n = 0 \] Solving for \( \mu \), we get: \[ \frac{\sum_{i=1}^{n} x_i}{\mu} = n \implies \mu = \frac{\sum_{i=1}^{n} x_i}{n} \] Thus, the MLE of \( \mu \) is: \[ \hat{\mu} = \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \] In summary, the MLE of the parameter \( \mu \) for a Poisson distribution based on the sample \( x_1, x_2, \ldots, x_n \) is simply the sample mean \( \bar{x} \).