# How exactly are the beta and gamma distributions related?

According to Wikipedia, the Beta distribution is related to the gamma distribution by the following relation:

$$\lim_{n\to\infty}n B(k, n) = \Gamma(k, 1)$$

Can you point me to a derivation of this fact? Can it be generalized? For example, is there a similar relation that results in something other than a constant 1 for the Gamma second parameter? What if we have

$$\lim_{n\to\infty,m\to\infty,n=mb}n B(k, m)$$

That is, the two variables go to infinity while maintaining a constant ratio b.

The reason I'm asking is because I'm trying to figure out how to simplify a hieraerchical bayesian model involving the beta distribution.

(This is my first post; sorry for the math notation, the MathJaX syntax was too daunting, but I'll try to learn)

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It's not MathJaX, it's LaTex, and there's a very very helpful post here: meta.math.stackexchange.com/questions/5020/… :) –  Daniel Sep 3 '12 at 22:32
Your "fact" is wrong. For any integer $k \ge 2$, $$\lim_{n \to \infty} n B(n,k) = \lim_{n \to \infty} \dfrac{\Gamma(k)}{(n+1)(n+2)\ldots(n+k-1)} = 0$$ –  Robert Israel Sep 3 '12 at 23:04
Perhaps you meant $$\lim_{n \to \infty} n^k B(n,k) = \Gamma(k)$$ –  Robert Israel Sep 3 '12 at 23:06
For fixed $b > 0$, $$B(n,bn) \sim \sqrt{2\pi/n} b^{bn-1/2}(1+b)^{1/2-n-bn} \ \text{as}\ n \to \infty$$ –  Robert Israel Sep 3 '12 at 23:12
Robert, I think you are talking about the Beta and Gamma functions, whereas my question concerns the Beta and Gamma distributions. –  Sten Linnarsson Sep 6 '12 at 7:49

Let $X_n$ denote a random variable with beta distribution $\mathrm B(k,n)$ and $Y_n=nX_n$. Then, for every $s\geqslant0$, $\mathrm E(Y_n^s)=n^s\mathrm E(X_n^s)$ and one knows $\mathrm E(X_n^s)$, hence $$\mathrm E(Y_n^s)=n^s\frac{\mathrm B(k+s,n)}{\mathrm B(k,n)}=n^s\frac{\Gamma(k+s)\Gamma(k+n)}{\Gamma(k+s+n)\Gamma(k)}\longrightarrow\frac{\Gamma(k+s)}{\Gamma(k)}.$$ This is $\mathrm E(Z^s)$ for any random variable $Z$ with gamma distribution $\Gamma(k)$ hence $Y_n\to Z$ in distribution.

Let $\widetilde X_n$ denote a random variable with beta distribution $\mathrm B(k,n/b)$, and $\widetilde Y_n=n\widetilde X_n$. Then, $\widetilde X_n$ is distributed like $X_{n/b}$ hence $\widetilde Y_n$ is distributed like $bY_{n/b}$ and $\widetilde Y_n\to bZ$ in distribution.

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Thanks! I'm actually shocked and amazed at the speed and quality of answers on the stackexchange sites. It's like having an extra brain with just slightly higher latency! –  Sten Linnarsson Sep 6 '12 at 8:07
This concerns the relationship between the Gamma and Beta distributions as opposed to the Gamma and Beta functions. Let $X \sim \mbox{Gamma}(\alpha, 1)$ and $Y \sim \mbox{Gamma}(\beta, 1)$ where the paramaterization is such that $\alpha$ is the shape parameter. Then
$$\frac{X}{X + Y} \sim \mbox{Beta}(\alpha, \beta).$$
To prove this, write the joint pdf $f_{X, Y} (x, y) = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} x^{\alpha - 1} y^{\beta - 1} e^{-(x + y)}$ (on $\mathbb R^2_+$) and make the transformation $U = \frac{X}{X + Y}$ and $V = X + Y$. The Jacobian of the transformation $X = VU, Y = V(1 - U)$ is equal to $V$ so the joint distribution of $U$ and $V$ has pdf $$\frac{v}{\Gamma(\alpha)\Gamma(\beta)} (vu)^{\alpha - 1} (v (1 - u))^{\beta - 1} e^{-v} = \frac{1}{\Gamma(\alpha)\Gamma(\beta)}v^{\alpha + \beta - 1} e^{-v} u^{\alpha - 1} (1 - u)^{\beta - 1}$$ (on $\mathbb R_+ \times [0, 1]$) and hence $U$ and $V$ are independent (because the pdf factors over $u$ and $v$) with $V \sim \mbox{Gamma}(\alpha + \beta, 1)$ and $U \sim \mbox{Beta}(\alpha, \beta)$ which is apparent from the terms $v^{\alpha + \beta - 1} e^{-v}$ and $u^{\alpha - 1}(1 - u)^{\beta - 1}$ respectively.