Solution
= Solution
Normalization of the <probability mass function> requires
$$
1=A\sum_{k=0}^\infty\frac{\lambda^k}{k!}=Ae^\lambda.
$$
Thus
$$
\boxed{A=e^{-\lambda}}.
$$
The resulting law is the <Poisson distribution> with parameter $\lambda$.
= Solution
Normalization of the <probability mass function> requires
$$
1=A\sum_{k=0}^\infty\frac{\lambda^k}{k!}=Ae^\lambda.
$$
Thus
$$
\boxed{A=e^{-\lambda}}.
$$
The resulting law is the <Poisson distribution> with parameter $\lambda$.