I am studying probability and statistics at my university but haven't had a solid math course in awhile(mostly forget algebra dealing with factorials)thus I am stuck with the following proof. According to my book there is a recurrence relation algorithm detailing the following:
$P(X=k+1) = c_k * P(X=k)$ with $c_k =[ \frac{(n-k)}{(k+1)} ] * \frac{p}{1-p}$ with X being a binomial random variable.
I have the following proof down thus far:
P(X=k+1)
$=(\binom{N}{K+1}) * p^{k+1} * (1-p)^{n-k-1}=$ As per Binomial definition
$= [ \frac{n!}{(k+1)!(n-k-1)!} ] * \frac{p}{1-p} * p^k * (1-p)^{n-k}$
$= [ \frac{n!}{(k+1)!(n-k-1)!} ] * \frac{p}{1-p} * P(X=k)$
And I am left to prove that $\frac{n!}{(k+1)!(n-k-1)!} = \frac{(n-k) }{ (k+1)}$
Any help with steps would be gratefully appreciated or if there is an issue with my proof thus far.
Thank You
We have $$\Pr(X=k+1)=\binom{n}{k+1}p^{k+1}(1-p)^{n-k-1}=\frac{n!}{(k+1)!(n-k-1)!}p^{k+1}(1-p)^{n-k-1}\tag{1}$$ and $$\Pr(X=k)=\binom{n}{k+1}p^{k}(1-p)^{n-k}=\frac{n!}{k!(n-k)!}p^{k}(1-p)^{n-k}.\tag{2}$$ Divide (1) by (2). In unsimplified form we get the following mess: $$\frac{n!k!(n-k)!p^{k+1}(1-p)^{n-k-1} }{n!(k+1)!(n-k-1)!p^k(1-p)^{n-k} }.$$ Note that the $n!$ in the top and bottom cancel. Note also that we have a $k!$ in the top and a $(k+1)!$ in the bottom. Since $(k+1)!=(k+1)k!$, we have $\frac{k!}{(k+1)!}=\frac{1}{k+1}$. A similar partial cancellation occurs with $\frac{(n-k)!}{(n-k-1)!}$, since $(n-k)!=(n-k)(n-k-1)!$.We leave the rest of the simplification to you.