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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?

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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 $m = 1-2^{-1/\beta}$.

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