If $X$ is beta distributed, how can you show that $1-X$ is also beta distributed? Can you just plug in $1-X$ into the Beta Density Function?
 A: For $0\le x\le 1$,
$$
\Pr(X \le x) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \int_0^x u^{\alpha-1} (1-u)^{\beta-1}\,du,
$$
and
$$
\Pr(1-X\le x) = \Pr(X \ge 1-x) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \int_{1-x}^1 u^{\alpha-1} (1-u)^{\beta-1}\,du. \tag 1
$$
Let $v=1-u$ so that $u = 1-v$ and $dv=-du$.
As $u$ goes from $1-x$ to $1$, then $v$ goes from $x$ to $0$.
Then $(1)$ becomes
$$
\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \int_x^0 (1-v)^{\alpha-1} v^{\beta-1}\, (-dv) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \int_0^x (1-v)^{\alpha-1} v^{\beta-1}\, dv.
$$
We've just interchanged the roles of $\alpha$ and $\beta$.
A: You can do that, because $x \rightarrow -x$ is linear transformation. For $a\neq0,b$ we have $g(x)=\frac{1}{|a|}\cdot f(ax)$, where $g(x)$ is probability density function of $aX$. Having density of $-X$ you can easily find density of $-X+1$.
A: Another way is to work with the PDF, since the beta distribution is continuous with no discrete atoms.  The form of the PDF is
$$
f_X(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha, \beta)}
$$
and it should be clear by inspection (from symmetry) that the corresponding PDF for $Y = 1-X$ can be written
$$
f_Y(y) = \frac{y^{\beta-1}(1-y)^{\alpha-1}}{B(\beta, \alpha)}
$$
