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TedG

  • 2 years ago

The random variable X has a truncated Poisson distribution with mean mu: the probability mass function is pX(x) =(e^(-mu)*mu^(x))/(1-e^(-mu)x!) ; x = 1; 2; 3; : : : : (i) Given that sum(i=0 to infinity) k^(i)/i!= e^k , show that the probability generating function for this distribution is (e^(mu*t) -1)/(e^(mu)-1) (ii) Hence find the moment generating function for this distribution and use it to derive the first two moments, E(X) and E(X^2).

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