Given a random variable X which is non-negative and integer valued and its corresponding probability generating function: $$G_X(s) = e^{s-1}$$
Find: $$P(X<2)$$
As per my calculations I get the answer: $$e^{0-1} + e^{1-1} = e^{-1} + e^{0} = e^{-1} + 1$$
Is this correct or am I missing something?
From elementary theory on Taylor series we know that the probability mass function of $X$ can be obtained by differentiating its probability generating function. That is, for each nonnegative integer $k$ we have $$ p_k := \mathbb P(X=k) = \lim_{s\downarrow0}\frac{G_X^{(k)}(s)}{k!}. $$ Since $\frac{\mathsf d}{\mathsf ds} G_X(s) = G_X(s)$, it follows that $$ p_k = \frac{e^{-1}}{k!}, $$ and hence $$ \mathbb P(X<2) = p_0 + p_1 = e^{-1}(1+1) = 2e^{-1}. $$