I've seen that the median of the Beta Distribution cannot be defined by a closed form analytic expression but in most sites I see people give that, when the parameters $\alpha$ and $\beta$ are the same, the median is $1/2$. Moreover, I've seen other cases such as $\text{median} = 1 - 2^{-1/\beta}$ whenever $\alpha = 1$ and $\beta > 0$, but I am not being able to get these results by working out this:

$$\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \Gamma(\beta)} \int_0^{m(\alpha,\beta)} x^{\alpha-1} (1-x)^{\beta-1} \, dx = \frac12.$$

Can someone help me to work out the expression to get a closed formula for these cases?


The case $\alpha = \beta$ is easy since the PDF is symmetric about the point $x=1/2$.

For $\alpha = 1$ and $\beta > 0$, note that $\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} = \beta$ so $$\int_0^m f(x)\,dx = \beta\int_0^m (1-x)^{\beta-1} \, dx = [-(1-x)^\beta ]_{x=0}^m = 1 - (1-m)^\beta.$$ Setting this equal to $1/2$ yields $m = 1-2^{-1/\beta}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.