Appendix B. Proofs for ChapterÂ 9

Theorem 1

Let $upper X tilde upper P o i s left-parenthesis lamda right-parenthesis$ and $upper Y tilde upper P o i s left-parenthesis mu right-parenthesis$ . If $Z = X + Y$ and $Ï = Î» + Î¼$ , then $upper Z tilde upper P o i s left-parenthesis tau right-parenthesis$ .

Proof

The definition of the Poisson process is:

P (X = k) = \frac{{Î»}^{k} e^{- Î»}}{k!}

Consider $P (X + Y = k)$ . This could happen if $upper X equals 0$ and $Y = k$ , or $upper X equals 1$ and $Y = k - 1$ , etc. So:

\begin{matrix} P (X + Y = k) & = P (X = 0, Y = k) + P (X = 1, Y = k - 1) + ... + P (X = k, Y = 0) \\ = \frac{{Î»}^{0} e^{- Î»}}{0!} \frac{{Î¼}^{k} e^{- Î¼}}{k!} + \frac{{Î»}^{1} e^{- Î»}}{1!} \frac{{Î¼}^{k - 1} e^{- Î¼}}{(k - 1)!} + ... + \frac{{Î»}^{k} e^{- Î»}}{k!} \frac{{Î¼}^{0} e^{- Î¼}}{0!} \\ = {â}_{i = 0}^{k} \frac{{Î»}^{i} e^{- Î»}}{i!} \frac{{Î¼}^{k - i} e^{- Î¼}}{(k - i)!} \end{matrix}

We can pull out $e^{- Î»}$ and $e^{- Î¼}$ :

P (X + Y = k) = e^{- Î»} e^{- Î¼} {â}_{i = 0}^{k} \frac{1}{i! (k - i)!} {Î»}^{i} {Î¼}^{k - i}

But $\frac{1}{i! (k - i)!} = (\binom{k}{i}) \frac{1}{k!}$ , and we can pull the $StartFraction 1 Over k factorial EndFraction$ out of the sum, too, so:

P (X + Y = k) = \frac{e^{- Î»} e^{- Î¼}}{k!} {â}_{i = 0}^{k} (\binom{k}{i}) {Î»}^{i} {Î¼}^{k - i}

Now, since exponents are additive:

\begin{matrix} e^{- Î»} e^{- Î¼} & = e^{- Î» - Î¼} \\ = e^{- (Î» + Î¼)} \\ = e^{- Ï} \end{matrix}

Finally, the binomial theorem states:

{â}_{k = 0}^{n} (\binom{n}{k}) a^{k} b^{n - k} = {(a + b)}^{n}

So:

\begin{matrix} P (X + Y = k) & = \frac{e^{- Î»} e^{- Î¼}}{k!} {â}_{i = 0}^{k} (\binom{k}{i}) {Î»}^{i} {Î¼}^{k - i} \\ = \frac{e^{- (Î» + Î¼)}}{k!} {(Î» + Î¼)}^{k} \\ = \frac{{Ï}^{k} e^{- Ï}}{k!} = Pois (Ï) \end{matrix}

Theorem 2

If $upper X tilde upper E x p o n left-parenthesis lamda right-parenthesis$ and $upper Y tilde upper E x p o n left-parenthesis mu right-parenthesis$ then $P (X < Y) = \frac{Î»}{Î¼ + Î»}$ .