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Let $X$ be a non-negative random variable with $$ P_X(n)=P(X=n)=A\frac{k^n}{n!},\quad n\in\mathbb N,\;n\geq 0. $$ How do I find $A$ and $k$ if I know that $E[X]=a$?

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I edited your question, so that it's hopefully easier to read. Please make sure that the question is correctly stated. – Stefan Hansen Dec 11 '12 at 12:02
very very much , is correct – Samvel Malintsyan Dec 11 '12 at 12:08
up vote 2 down vote accepted

This is a Poisson distribution. The answer is $A=\exp(-a)$.

Notice that because $P_X$ is a probability mass function, then \begin{eqnarray*} 1 & = & \sum_{n = 0}^{\infty} P_X \left( n \right)\\ & = & A \sum_{n = 0}^{\infty} \frac{k^n}{n!}\\ & = & A \exp \left( k \right) \end{eqnarray*} Therefore, $A = \exp(- k)$. (here we used the series expansion of the exponential function.)

It is also known that $E[X]=k$. (the mean of the Poisson distribution is equal to its parameter.)

Finally $k=a$.

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