Binomial Distribution satisfies Marcoff Chain

[ppedro]

1. The problem statement

Consider the Binomial Distribution in the form

$P_{N}(m)=\frac{N!}{(\frac{N+m}{2})!(\frac{N-m}{2})!}p^{\frac{N+m}{2}}q^{\frac{N-m}{2}}$

where $p+q=1$, $m$ is the independent variable and $N$ is a parameter.

Show that it satisfies the marcoff chain

$P_{N+1}\left(m\right)=pP_{N}\left(m-1\right)+qP_{N}\left(m+1\right)$

2. The attempt at a solution

I'm trying my solution starting from this:

$pP_{N}\left(m-1\right)+qP_{N}\left(m+1\right)$


$=p\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}p^{\frac{N+m-1}{2}}q^{\frac{N-m+1}{2}}+q\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m-1}{2}}$


$=\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}+\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}$


$=p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}\left(\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}+\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}\right)$


I can't go any further. If you can help I would appreciate.

 

[TSny]

(Edit) See what happens if you multiply by$\frac{N+1}{N+1}$.

 

[TSny]

My previous comment was based on a particular way that I went about it which got to the result. But, on review, I see that multiplying by (N+1)/(N+1) isn't necessary.

I would factor out the $N!$ in your expression.

The important thing is to get the two denominators in your expression to match the denominator in $P_{N+1}(m)$. For example, what could you multiply the numerator and denominator of $\frac{1}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}$ by to get the denominator in $P_{N+1}(m)$?

[By the way, welcome to PF, ppedro!]

 

[ppedro]

Hey TSny! Thanks for your reply. I see what you're suggesting but I'm not being able to compute it. The factorials are not helping me simplify the expression.

 

[TSny]

$P_{N+1}(m)$ has a denominator of $(\frac{N+m+1}{2})!(\frac{N-m+1}{2})!$.

Compare that to your first denominator $(\frac{N+m-1}{2})!(\frac{N-m+1}{2})!$.

To get your denominator to match the denominator in $P_{N+1}(m)$, you've got to somehow transform $(\frac{N+m-1}{2})!$ into $(\frac{N+m+1}{2})!$.

What can you multiply $(\frac{N+m-1}{2})!$ by to produce $(\frac{N+m+1}{2})!$?

 

[ppedro]

$\frac{(N+m+1)!}{(N+m-1)!}=\frac{(N+m+1)(N+m)(N+m-1)!}{(N+m-1)!}=(N+m+1)(N+m)$

 

[TSny]

No. Note that $\frac{N+m+1}{2} = \frac{N+m-1}{2} + 1$.

 

[ppedro]

Ok, I see your point. Thanks!