I need to generate random numbers from Beta distribution using random variables from Uniform distribution.
If I have two random variables $Y_1=U_1^{1/\alpha}$ and $Y_2=U_1^{1/\beta}$,
and If $Y_1+Y_2<= 1$ then $X=Y_1/(Y_1+Y_2)$ is from Beta distribution with parameters $\alpha\ and\ \beta$.

This is what I've done so far:
I've generated 10,000 random variables from a Uniform distribution.
I then used those rv to generate $Y_1 and\ Y_2 $ using $Y_1=U_1^{1/\alpha}$ and $Y_2=U_1^{1/\beta}$.
Then I generated $X$ using the formula $X=Y_1/(Y_1+Y_2)$.

Here is the R code I used:

# my plot
alpha <- 2
beta <- 4

u <- runif(10000,0,1)

y1 <- u^(1/alpha)
y2 <- u^(1/beta)

x <- data.frame(y1,y2 ) %>% 
  filter(y1+y2<=1) %>% # check that y1+y1 <=1
  mutate(x = y1/(y1+y2)) %>% # generate x random variable

# plot x

# plot from beta distribution
hist(rbeta(10000,shape = 2,shape2 =4)) 

But when I plot this, the histogram seems to be going in the opposite direction to when I just generate rv from Beta distribution. Here is my plot: enter image description here

And here is the plot if I generate rv from actual beta distribution:
enter image description here

Why does my curve look different to the one from Beta distribution?
This question seems very similar, but the solution uses qbeta() function. I think I need to use my $X=Y_1/(Y_1+Y_2)$

  • 3
    $\begingroup$ Hint: You need the $Y_i$ to be independent. $\endgroup$
    – whuber
    May 26, 2018 at 10:53
  • 5
    $\begingroup$ I believe it is a bit rude to downvote this question. The question is well explained and developed. The answer may be simple, but, it is a minor mistake anyone of us can do. $\endgroup$
    – Jon Nagra
    May 27, 2018 at 17:33
  • 3
    $\begingroup$ @whuber I think I got it. I generated only 1 sample from the Uniform distribution, where as I need to get generate 2 If I change my code to this, it looks like it's working: y1 <- runif(10000,0,1)^(1/alpha) y2 <- runif(10000,0,1)^(1/beta) $\endgroup$
    – jmich738
    May 28, 2018 at 9:40

1 Answer 1


I just had the same problem with the distribution creation, thanks for the latest reply to the original post. Please find below a viable solution to create one RV in Python:

def beta(a,b):
    rv1 = np.random.rand()**(1/a)
    rv2 = np.random.rand()**(1/b)

    while (rv1+rv2) > 1:
        rv1 = np.random.rand()**(1/a)
        rv2 = np.random.rand()**(1/b)

    return = rv1 / (rv1+rv2)
  • 1
    $\begingroup$ The code would crash on u1+u2 since the variables are not defined. $\endgroup$
    – Tim
    Jul 16, 2021 at 11:46
  • $\begingroup$ copy paste error, apologies. It is rv1 and rv2 of course, thanks for pointing it out and the immediate downvote. $\endgroup$
    – Guenter
    Jul 17, 2021 at 10:45
  • $\begingroup$ What is the point of the while loop?? As far as I can tell, all it does is slow down the algorithm by an arbitrarily large factor. $\endgroup$
    – whuber
    Jul 21, 2021 at 15:46
  • $\begingroup$ I haven't found a proper explanation for this, but when testing this without the condition in the loop, the results are not following the beta distribution. I tested the results versus the distribution given a grid of input variables. $\endgroup$
    – Guenter
    Jul 22, 2021 at 22:05
  • $\begingroup$ The while loop is performing the rejection sampling, so when (rv1+rv2)>1 it rejects the pair and tries again. $\endgroup$
    – Henry
    Apr 12 at 15:10

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