Paragraph

Using the f(x) generated in the previous theorem
\begin{align*} \mu & = E[X] \\ & = \sum_{x=0}^{\infty} x \cdot \frac{(\lambda T)^x}{x!} e^{-\lambda T}\\ & = \lambda T e^{-\lambda T} \sum_{x=1}^{\infty} \frac{(\lambda T)^{x-1}}{(x-1)!} \\ & = \lambda T e^{-\lambda T} \sum_{k=0}^{\infty} \frac{(\lambda T)^k}{k!} \\ & = \lambda T e^{-\lambda T} e^{\lambda T} \\ & = \lambda T \end{align*}
which confirms the use of \(\mu\) in the original probability formula.
in-context