The MGF is exp[lambda*(e^t - 1)].
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The exponential distribution and the Poisson distribution.
The moment generating function is M(t) = Expected value of e^(xt) = SUM[e^(xt)f(x)] and for the Poisson distribution with mean a inf = SUM[e^(xt).a^x.e^(-a)/x!] x=0 inf = e^(-a).SUM[(ae^t)^x/x!] x=0 = e^(-a).e^(ae^t) = e^[a(e^t -1)]
Yes.
Why belong exponential family for poisson distribution
Divide the total number of incidents by the total time. The result, representing the average number of incidents per unit of time, is the mean as well as the variance of the Poisson distribution.