How to convert the parameters in a binomial distribution to those in a beta distribution? I know that the beta distribution is the generalized continuous case of the discrete binomial distribution. Let's say I have a binomial distribution, $B(N,p)$. I would like to know the corresponding $\alpha$ and $\beta$ such that the binomial CDF of any count, $k$, in $\{0,1,2,...,N\}$ equals to the beta CDF at $x = k/N$.
In terms of equations, I would like the functions $f(x)$ and $g(x)$ such that:


*

*$\alpha = f(N,p)$ and $\beta = g(N,p)$, and

*CDF of $B(k,n,p) = $ CDF of $Beta(k/N, \alpha, \beta)$ where


*

*$N$ in $\mathbb{N}$,

*$p$ in $(0,1)$, and

*$k$ in $\{0,1,2,...,N\}$
I have been in search for a while but I find no direct answer. I don't have much maths background and I am hoping for an easy answer in terms of $f(x)$ and $g(x)$.
 A: I do not know that the beta distribution is the generalized continuous case of the discrete binomial distribution, and I am curious why you think that. $p^x(1-p)^{n-x}$ is not really the same as $x^{\alpha-1}(1-x)^{\beta-1}$. What I do know is that the beta distribution provides a conjugate family for the parameter $p$ of the binomial distribution, but that is a rather different statement
You have a problem in that the cumulative distribution functions are not going to match like that.  For example, for $X$ binomial you have $P(X\le 0)=\frac{1}{(1-p)^n}$ while for $Y$ beta-distributed you have $P\left(Y \le \frac0n\right) =0$
So if you want an approximate match between the two distributions, an alternative could be to match the means and variances.  That could suggest using something like $\alpha=(n-1)p, \beta=(n-1)(1-p)$ or in the other direction $n=\alpha+\beta+1, p= \frac{\alpha}{\alpha+\beta}$ 
For example with $n=4, p=\frac13$ this might suggest $\alpha=1, \beta=2$, and the following chart is what the two cumulative distribution functions would look like; the relationship is not bad, but you would not use one to calculate the other

A: I am brand new to the site.
Here is Henry's plot using Ian's formula.  CDF of Binomial(k,n,p) = 1 - CDF of Beta(p, k+1, n-k).
In Excel, I did BINOM.DIST(i,4,0.333,TRUE) for i = 0 to 4.  Then, I did 1-BETA.DIST(0.333,1+i,4-i,TRUE) for i = 0 to 3.99.

This is because the jth order statistic $X_{(j)}$ of n i.i.d. Unif(0,1) r.v.s is distributed with pdf Beta(j, n-j+1). So, the cdf is the cumulative Beta distribution.
$P(X_{(j)} \le x) = \int_{0}^{x}Beta(\theta, j, n-j+1)d\theta$
In general, the cdf of the order statistic is given by
$P(X_{(j)} \le x) = \sum_{k=j}^{n} {n\choose k} F(x)^k(1-F(x))^{n-k}$
If X ~ Unif(0,1), then $F(x) = x$, so
$P(X_{(j)} \le x) = \sum_{k=j}^{n} {n\choose k} x^k(1-x)^{n-k}$
Change to sum from k = 0 to j-1 and subtract from 1.
$P(X_{(j)} \le x) = 1 -\sum_{k=0}^{j-1} {n\choose k} x^k(1-x)^{n-k}$
Let $i = j-1$,
$P(X_{(i+1)} \le x) = 1 -\sum_{k=0}^{i} {n\choose k} x^k(1-x)^{n-k}$
substitue $i + 1$ for $j$ in the Beta cdf integral above.
$P(X_{(i+1)} \le x)= \int_{0}^{x}Beta(\theta, i+1, n-i)d\theta$
Set the two right-hand sides equal, and we have
$ 1 - \sum_{k=0}^{i} {n\choose k} x^k(1-x)^{n-k}) = \int_{0}^{x}Beta(\theta, i+1, n-i)d\theta$
or
cdf of Binomial(i, n, x) =  1- cdf of Beta(x, i+1, n-i).
This is what I said above with x replacing p and i replacing k.
A: The simplest relationship between beta and binomial distributions is
cdf_binomial(N,i,p) = sf_beta(p,i+1,N-i) (where sf_beta, the survival function is 1-cdf_beta)

or equivalently, cdf_binomial(N,i,p) = cdf_beta(1-p,N-i,i+1)
where i >= 0 and i < N
