The random variables X1,..,Xn are independent Poisson random variables with a common parameter Lambda . Find the maximum likelihood estimate of Lambda based on observed values x1,...,xn.

(x1 + x2 +...+ xn) / n

To find the maximum likelihood estimate of Lambda based on observed values x1,...,xn, we first need to write down the likelihood function.

The likelihood function, L(Lambda | x1,...,xn), represents the probability of observing the given values x1,...,xn for a given value of Lambda.

Since the random variables X1,...,Xn are independent Poisson random variables with a common parameter Lambda, the probability mass function of each random variable Xi is given by:

P(Xi = xi | Lambda) = (e^(-Lambda) * Lambda^xi) / xi!

The joint probability mass function of observing the given values x1,...,xn for the random variables X1,...,Xn, respectively, is the product of the individual probability mass functions:

L(Lambda | x1,...,xn) = P(X1 = x1 | Lambda) * P(X2 = x2 | Lambda) * ... * P(Xn = xn | Lambda)

Taking the natural logarithm of the likelihood function, we can simplify the calculations:

log L(Lambda | x1,...,xn) = log P(X1 = x1 | Lambda) + log P(X2 = x2 | Lambda) + ... + log P(Xn = xn | Lambda)

= log ((e^(-Lambda) * Lambda^x1) / x1!) + log ((e^(-Lambda) * Lambda^x2) / x2!) + ... + log ((e^(-Lambda) * Lambda^xn) / xn!)

= (log e^(-Lambda) + log Lambda^x1 - log x1!) + (log e^(-Lambda) + log Lambda^x2 - log x2!) + ... + (log e^(-Lambda) + log Lambda^xn - log xn!)

= (-Lambda + x1 * log Lambda - log x1!) + (-Lambda + x2 * log Lambda - log x2!) + ... + (-Lambda + xn * log Lambda - log xn!)

To find the maximum likelihood estimate of Lambda, we need to find the value of Lambda that maximizes the log likelihood function.

To do this, we take the derivative of the log likelihood function with respect to Lambda and set it equal to zero:

d/dLambda (log L(Lambda | x1,...,xn)) = (-1 + x1 / Lambda) + (-1 + x2 / Lambda) + ... + (-1 + xn / Lambda) = 0

Simplifying, we get:

-n + (x1 + x2 + ... + xn) / Lambda = 0

Rearranging the equation, we have:

Lambda = (x1 + x2 + ... + xn) / n

Therefore, the maximum likelihood estimate of Lambda based on the observed values x1,...,xn is:

Lambda_hat = (x1 + x2 + ... + xn) / n

To find the maximum likelihood estimate (MLE) of Lambda based on observed values x1,...,xn, we need to maximize the likelihood function L(Lambda | x1,...,xn).

The likelihood function L(Lambda | x1,...,xn) is the probability density function (PDF) of the observed values x1,...,xn, given the parameter Lambda. Since the random variables X1,...,Xn are independent Poisson random variables with a common parameter Lambda, the probability mass function (PMF) of each Xi is given by:

P(Xi = xi) = (e^(-Lambda) * Lambda^xi) / xi!

The joint probability mass function (PMF) of x1,...,xn, given Lambda, can be obtained by multiplying the individual PMFs because the Xi's are independent:

P(X1 = x1, ..., Xn = xn) = P(X1 = x1) * ... * P(Xn = xn)

Taking the log of the likelihood function makes it easier to work with, as it transforms the product of probabilities into a sum of logarithms:

log L(Lambda | x1,...,xn) = log P(X1 = x1) + ... + log P(Xn = xn)

Using the PMF of the Poisson distribution, we can substitute the probabilities into the log-likelihood function:

log L(Lambda | x1,...,xn) = (xi * log(Lambda) - Lambda - log(xi!)) + ... + (xn * log(Lambda) - Lambda - log(xn!))

Simplifying the expression:

log L(Lambda | x1,...,xn) = (x1 + ... + xn) * log(Lambda) - n * Lambda - log(x1! * ... * xn!)

To find the MLE of Lambda, we differentiate the log-likelihood function with respect to Lambda and set the derivative equal to zero:

d/dLambda (log L(Lambda | x1,...,xn)) = (x1 + ... + xn)/Lambda - n = 0

Solving for Lambda, we get:

Lambda = (x1 + ... + xn) / n

Thus, the maximum likelihood estimate of Lambda based on observed values x1,...,xn is (x1 + ... + xn) / n.