0
$\begingroup$

In class, my professor computed the density of a gamma random variable by taking the derivative of its CDF, but he skipped many steps. I am trying to go through the derivation carefully but cannot reproduce his final result.

Let $k$ be the shape and $\mu$ be the scale. Then the CDF for a gamma random variable $T$ is

$$ F(t) = 1 - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i}}{i!} $$

Using the product rule, I get

$$ \begin{align} f(t) &= \frac{\partial}{\partial t} F(t) \\ &= \frac{\partial}{\partial t} \Big( 1 - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i}}{i!} \Big) \\ &= - \sum_{i=0}^{k-1} \frac{\partial}{\partial t} \Big( \frac{e^{-\mu t} (\mu t)^{i}}{i!} \Big) \\ &= - \sum_{i=0}^{k-1} \frac{1}{i!} \frac{\partial}{\partial t} \Big( e^{-\mu t} (\mu t)^{i} \Big) \end{align} $$

where

$$ \frac{\partial}{\partial t} \Big( e^{-\mu t} (\mu t)^{i} \Big) = e^{-\mu t} (-\mu)(\mu t)^i + e^{-\mu t} i (\mu t)^{i-1} $$

Putting everything together, we get

$$ f(t) = \sum_{i=0}^{k-1} \frac{\mu e^{-\mu t} (\mu t)^i}{i!} - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i-1}}{(i-1)!} $$

But here I am stuck. I know that

$$ f(t) = \frac{\mu e^{-\mu t} (\mu t)^{k-1}}{(k-1)!} $$

but am not sure how to get there.

$\endgroup$
  • 1
    $\begingroup$ You have some omissions in your product and chain rules line: $$ \frac{\partial}{\partial t} (\dots)= \mathrm{e}^{- \mu t} (-\mu) (\mu \underline{t})^i + \mathrm{e}^{-\mu t} \underline{i} \mu (\mu t)^{i-1} \text{.} $$ $\endgroup$ – Eric Towers Mar 14 at 15:21
  • $\begingroup$ Thanks. The first one was a typo but the second was a mistake and critical for factoring. $\endgroup$ – gwg Mar 14 at 21:32
1
$\begingroup$

a) note that the second sum starts from $i=1$;
that's because when you derivate $\frac{(\mu t)^i}{i!}$ you get $\frac{i(\mu t)^{i-1}}{i!}$ which is $0$ for $i=0$, and only for $1 \le i$ becomes $\frac{(\mu t)^{i-1}}{(i-1)!}$
b) factor out $e^{-\mu t}$
c) you are left with a telescoping sum

$\endgroup$
  • $\begingroup$ Thanks, this worked. Just to clarify, I factored out $\mu e^{- \mu t}$. If a term in a series is undefined, is it mathematically acceptable to just remove that term by updating the index (your first point above). $\endgroup$ – gwg Mar 14 at 21:34
  • 1
    $\begingroup$ Glad it helped. But concerning the index of second sum, no, $0$ is not removed because $(i-1)!$ is otherwise not defined (which is not allowed, and just indicates there is a fault somewhere), it is the result of the derivation : I added a note in the answer. $\endgroup$ – G Cab Mar 14 at 23:39
1
$\begingroup$

You might also want to manually check what's going on with the first one or two terms in the sum for $F$ as you take the derivative, since introducing "$(-1)!$" is not happening. (That is, you have divided by zero in your shown work.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.