As a discrete probability distribution, the Poisson distribution has a single parameter \(\lambda\), called rate or expected value. In fact \(\lambda\) is not only its population mean, but also its variance.
The probability mass function \(f\) of a Poisson distribution is as follows:\[f(k; \lambda) = \frac{\lambda^k e^{-\lambda}}{k!}\]for a positive \(\lambda\).
Remembering that the Maclaurin series of the exponential function is \[e^x = \sum_{k = 0}^\infty \frac{x^k}{k!},\]it is clear that \(\sum_{k = 0}^\infty f(k, \lambda) = 1.\)
More generally, all of the cumulants of a Poisson distribution are equal to its parameter \(\lambda\). But you may wonder if only the Poisson distribution has the constant, positive cumulants?
The answer is yes. To see the fact, note that, if the probability distribution of a discrete variable \(X\) has constant, positive cumulants \(\lambda\), then it has the same (finite) moments as a Poisson distribution does. Then the moment-generating function\[M_X(t) = e^{\lambda (e^t - 1)}\] is shared too. It uniquely determines the distribution.