Let X∼Poisson(k−1)X∼Poisson(k−1) with Var(X)=1Var(X)=1, then find the value of the constant kk
We know that if "X" follows a Poisson distribution of parameter "\\lambda" , i.e. "X\\sim \\textrm{Poisson}(\\lambda)" , then the variance of "X" is "\\textrm{Var}(X)=\\lambda".
If we apply it to our case we have:
Thus, the value of the constant "k" is "k=2" .
Comments
Leave a comment