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For finding when the median of a beta distribution is $\frac{1}{2}$, this answer says:

If a $\mathrm{Beta}(a,b)$ distribution has $a>b$ then $\mathbb P(X \le \frac12) \lt \frac12$ and its median is above $\frac12$, while if $a<b$ then then $\mathbb P(X \le \frac12) \gt \frac12$ and its median is below $\frac12$.

This can be proved by comparing the densities at $x$ and $1-x$ and then integrating over the half intervals. When $a=b$ you get $\mathbb P(X \le \frac12) = \frac12$ and a median of $\frac12$ by symmetry.

I don't understand the second part (where the property mentioned) is proved. What is meant by "comparing the densities", and why is that important? Is the integration actually to be done), or is there a trick to see that it is less than, or more than $\frac{1}{2}$

I understand that symmetry might be useful in this situation, but I don't where to start applying it.

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Let $a < b$ and $X \sim \text{Beta}(a, b)$. Then $$P(X \le 1/2) = c\int_0^{1/2} x^{a-1}(1-x)^{b-1} \, dx \overset{u=1-x}{=} c \int_{1/2}^1 (1-u)^{a-1} u^{b-1} \, du$$ and $$P(X > 1/2) = c\int_{1/2}^1 u^{a-1}(1-u)^{b-1} \, dx.$$

To show $P(X \le 1/2) > P(X > 1/2)$, it suffices to show $u^{b-1}(1-u)^{a-1} \ge u^{a-1}(1-u)^{b-1}$ when $u > 1/2$. This can be achieved by some rearranging.

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  • $\begingroup$ Is the inequality/probability reversed somewhere? Because when I check the wikipedia page, the P(X<0.5)>P(X>0.5) for say a=1<b=3 (green curve). And the same is the case with the inequality. I am getting the reverse. en.wikipedia.org/wiki/File:Beta_distribution_cdf.svg $\endgroup$ Commented Mar 21 at 8:40
  • $\begingroup$ @Starlight Apologies, updated my answer. $\endgroup$ Commented Mar 22 at 4:13
  • $\begingroup$ Thanks. I was starting to doubt my knowledge in math for some time (until I realized anyone could make a typo....) $\endgroup$ Commented Mar 22 at 5:38

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