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

[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.

 

[dirk_mec1]

Let's start simple:

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

 

[Fredrik]

Use ## instead of $ for inline LaTeX.

 

[pasmith]

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.

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 $$\sum_{x=r}^\infty \binom{x-1}{r-1} p^r (1-p)^{x-r} <br /> = (1 - q)^r \sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m.$$ 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?

 

[Ray Vickson]

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.

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!

 

[MrKushtrim]

Hi, thanks for the reply .

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 $$\sum_{x=r}^\infty \binom{x-1}{r-1} p^r (1-p)^{x-r} <br /> = (1 - q)^r \sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m.$$ 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?

It must equal 1. Though I have no idea how to prove it.

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!

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

 

[pasmith]

It must equal 1. Though I have no idea how to prove it.

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 $q$?

 

[MrKushtrim]

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 $q$?

$\frac{1}{(1 -q)^r }$
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 ?

 

[pasmith]

$\frac{1}{(1 -q)^r }$
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 ?

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 $$\frac{d^k}{dx^k} \left( \sum_{n=0}^\infty a_n x^n \right) <br /> = \sum_{n=0}^\infty a_{n+k} \frac{(n + k)!}{n!} x^n$$ 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.