# Prove that the function given with f(x) is a PDF

1. May 28, 2014

### MrKushtrim

Hi,
Could someone help me prove that the function given with
$f(x) = \binom{x-1}{r-1} p^{r}(1-p)^{x-r}$ is a probability density function, where $x= r, r+1,..., \infty$ and $0<p<1$

I thought to solve it somehow by using the binomial theorem, but since it's the upper part that's changing on the binomial coefficients, this proved futile.

Last edited: May 28, 2014
2. May 28, 2014

### dirk_mec1

Let's start simple:

Can you prove that the value is always larger than zero?

3. May 28, 2014

### Fredrik

Staff Emeritus
Use $instead of  for inline LaTeX. 4. May 28, 2014 ### pasmith You need to show that $$\sum_{x=r}^\infty \binom{x-1}{r-1} p^r (1-p)^{x-r} = 1.$$ Setting $q = 1- p$ and $m = x - r$ gives $$\sum_{x=r}^\infty \binom{x-1}{r-1} p^r (1-p)^{x-r} = (1 - q)^r \sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m.$$ What does the series on the right have to equal if the left hand side is to equal 1, and can you prove that it does? 5. May 28, 2014 ### Ray Vickson For discrete random variables, I prefer to use symbols like$j$and$k$instead of$x$. So, in this notation: what properties must be possessed by $$f_k = {k-1 \choose r-1} p^r (1-p)^{k-r}, \, k = r, r+1, r+2, \ldots$$ in order that it be a legitimate probability mass function of a discrete random variable? Note that$f_k$is NOT a probability density function; it is a probability mass function. There is a world of difference! 6. May 28, 2014 ### MrKushtrim Hi, thanks for the reply . It must equal 1. Though I have no idea how to prove it. Thanks, I had the terminology mixed up. To be a probability mass function ,$f_k$has to be positive, and the sum over all$ k $has to be 1. It's easy to see that$f_k$is positive, but I cannot seem to prove the second property 7. May 29, 2014 ### pasmith Remember, you need to show that $$(1 -q)^r \sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m = 1$$ for every $0 < q < 1$. This suggests that the series $$\sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m$$ must be the binomial expansion of ... what function of $q$? 8. May 29, 2014 ### MrKushtrim$ \frac{1}{(1 -q)^r } $But how would I prove that ? --- Edit: I checked a wikipedia article about binomial series, and it turns out that the series $$\sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m$$ is of this form , where$β= r-1$and$z=q##

Thanks for the help :)
Though, is there any other way to prove it, without directly using the formula ?

Last edited: May 29, 2014
9. May 29, 2014

### pasmith

Ultimately the only way to sum power series is to relate them to power series of known functions.

There are other ways of showing that $\sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m = (1 - q)^{-r}$. For example, using the formula $$\frac{d^k}{dx^k} \left( \sum_{n=0}^\infty a_n x^n \right) = \sum_{n=0}^\infty a_{n+k} \frac{(n + k)!}{n!} x^n$$ we see that $$\sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m = \frac{1}{(r - 1)!} \frac{d^{r-1}}{dq^{r-1}} \left( \sum_{m=0}^\infty q^m \right).$$ $\sum_{m=0}^\infty q^m$ is a geometric series and is easily summed.