Assume that I have $n$ variables that are each $X_i \sim \text{Beta}(\alpha, 1)$ distributed (with the same $\alpha$, i.i.d.).

Is there anything known about the distribution of the sum $Y=\sum_i X_i$? I've tried to understand the works of Pham-Gia, but couldn't entirely grasp them. He seems to use some modified generalized Beta $B^*$.

An approximation is also fine with me, as long as I can compute some quantiles. I'm mostly interested in quantiles.


2 Answers 2


Your $Y$ will have mean $\frac{n\alpha}{\alpha+1}$ and variance $\frac{n\alpha}{(\alpha+1)^2(\alpha+2)}$

If you want a reasonable approximation other than a normal distribution with that mean and variance, you could say that $\frac{1}{n}Y$ has mean $\frac{\alpha}{\alpha+1}$ and variance $\frac{\alpha}{n(\alpha+1)^2(\alpha+2)}$ and then find a Beta distribution with those moments that would give you $\alpha'=\alpha\frac{\alpha n+2n-1}{\alpha+1}$ and $\beta'=\frac{\alpha n+2n-1}{\alpha+1}$.

The density function for this approximation to $\frac{1}{n}Y$ would then be $\frac{\Gamma(\alpha'+\beta')}{\Gamma(\alpha')\Gamma(\beta')}x^{\alpha'-1}(1-x)^{\beta'-1}$ on $[0,1]$ and the density function for the corresponding approximation to $Y$ would be $\frac{\Gamma(\alpha'+\beta')}{n\Gamma(\alpha')\Gamma(\beta')}\left(\frac{x}{n}\right)^{\alpha'-1}\left(1-\frac{x}{n}\right)^{\beta'-1}$ on $[0,n]$.

As an example, consider $\alpha=3$ and $n=2$. You get a mean of $\frac32$ and variance of $\frac3{40}$, and for the approximation you get $\alpha'= \frac{27}{4},\beta'= \frac{9}{4}$. In the following plot, the black curve gives the actual density, the red curve the stretched-Beta approximation and the blue curve the normal approximation. The red curve is not too far away in this case.

enter image description here

Similarly the cumulative distributions look like the next chart and you can see that the quantiles of the red stretched-Beta approximation are much closer than the quantiles of the blue normal approximation here.

enter image description here


I will assume $X_i$ independent in this post. To get a feel for the problem, recall the special case of uniform distribution, corresponding to $\alpha=1$, i.e. $\mathrm{Beta}(1,1) \stackrel{d}{=} U(0,1)$.

The sum of $n$ iid uniform distribution was studied by J.O. Irwin and P. Hall, and the result is known as Irwin-Hall distribution, aka uniform sum distribution.

Already for $n=3$ the distribution density of the sum of three standard uniform variables approximates normal quite well:

enter image description here

The same approximation will work well for larger values of $n$ in your case as well. To write it out we need to compute mean and variance of the sum: $$ \mu_n = \mathbb{E}(\sum_{k=1}^n X_i) = n \frac{\alpha}{\alpha+1} \qquad \sigma_n^2 = \mathbb{Var}(\sum_{k=1}^n X_i) = \sum_{k=1}^n \mathbb{Var}(X_i) = \frac{n \alpha}{\alpha+2} \frac{1}{(\alpha+1)^2} $$

Thus the quantile function approximation is: $$ Q_n(q) \approx n \frac{\alpha}{\alpha+1} + \frac{1}{\alpha+1} \sqrt{ \frac{n \alpha}{\alpha+2} } Q_{\mathcal{N}(0,1)}(q) $$

For $n=2$ CDF can be worked out exactly, and can be inverted using numerical algorithms:

enter image description here

Added: The normal approximation can be truncated to $(0,n)$ interval to improve accuracy: $$ Q_{Y_n}(q) = \mu_n + \sigma_n Q_{N(0,1)}( (1-q) \Phi(-\mu_n/\sigma_n) + q \Phi((n-\mu_n)/\sigma_n) ) $$

  • $\begingroup$ "The same approximation will work well for larger values of n in your case as well" But probably not so well, is $\alpha$ is very different from 1. I would go this route, though, is $n$ is large. $\endgroup$
    – leonbloy
    Commented Nov 25, 2011 at 16:42
  • $\begingroup$ I need $\alpha$ to be different from 1, and $n$ to be larger than $2$. I'm particularly interested in $\alpha = .5$ (as this occurs in my situation for Euclidean distance, a prime example of an $L_p$ norm - $1/p$ is actually my equation for $\alpha$), and I'm interested in the development of this distribution when I increase $n$. Of course I do expect it to approximate a Gaussian distribution for large values of $n$; at the same time I know it will never be exactly Gaussian, as negative values are not possible here. $\endgroup$ Commented Nov 25, 2011 at 16:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .