p(x) = [ ((Theta)^x) * (e^(-Theta)) ] / x! X = 0,1,2,...

You may assume that E(X)= Theta and V(X) = Theta

A random sample were examined....

Show that the maximum likelihood estimator (MLE) of (Theta) is the sample mean.

Any help, or a point in the right direction would be great. Any worked solutions are also welcome!

Thankyou!