Question

Show that $$ \int_{\mu}^{\infty} \frac{1}{\Gamma(k)} z^{k-1} e^{-z} d z=\sum_{x=0}^{k-1} \frac{\mu^{x} e^{-\mu}}{x !}, \quad k=1,2,3, \ldots $$ This demonstrates the relationship between the cdfs of the gamma and Poisson distributions. Hint: Either integrate by parts \(k-1\) times or obtain the "antiderivative" by showing that $$ \frac{d}{d z}\left[-e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1) !} z^{k-j-1}\right]=z^{k-1} e^{-z} $$

Short answer

By using the chain rule for differentiation and by evaluating the integral we have demonstrated the relationship between gamma and Poisson distribution cdfs by verifying the given equation.

Step-by-step solution

Simplify the right hand side and take the derivative

Let's start by taking the derivative of \(-e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1) !} z^{k-j-1}\). This will be equal to the integrand of the left hand side. Using the Chain Rule, this derivative is \(\frac{d}{d z}\left[-e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1) !} z^{k-j-1}\right] = z^{k-1}e^{-z}.\)

Compare the left and right sides

After taking the derivative in Step 1, we observe that it is equal to the integrand of the left hand side of the given equation. This implies that the integral of the left hand side is equal to \(-e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1) !} z^{k-j-1}\).

Show the integral and sum are equal

When evaluating the integral expressions from \(\mu\) to \(\infty\), you get \(\sum_{x=0}^{k-1} \frac{\mu^{x} e^{-\mu}}{x !}\). We already know that this is equal to the expression \(-e^{-\mu} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1) !} \mu^{k-j-1}\) when plugged in \(\mu\) instead of \(z\) in the equation from the previous step. Hence this verifies the given equation.