Finding the PDF using method of distribution of a beta distribution Given that X is a beta distribution with parameters alpha and beta find the PDF of $Y=1-X$.
$P(Y<=y)=P(x>1-y) = 1 - P(x<1-y)$
The next step is to use the CDF of a beta distribution to calculate $P(x<1-y)$.
I know that the cdf of a beta distribution is 
$F(x) = I_{x}(p,q) = \frac{\int_{0}^{x}{t^{p-1}(1-t)^{q-1}dt}}{B(p,q)}
\hspace{.2in} 0 \le x \le 1; p, q > 0$ 
where $B(\alpha,\beta) = \int_{0}^{1} {t^{\alpha-1}(1-t)^{\beta-1}dt}$ .
I'm stuck on how do I plug 1-Y into the CDF equation.
 A: I'm going to let $F_X$ be the CDF of $X$. Your steps are correct: notice that
$$P(Y \leq y) = P(1 - X \leq y) = P(X \geq 1 - y) = 1 - P(X \leq 1-y) = 1 - F_{X}(1-y)\text{.}$$
Do not bother computing the CDF of $X$ at this point. (It's way too tedious.) Instead, you should know that 
$$\dfrac{\text{d}}{\text{d}y}\left[P(Y \leq y)\right] = f_{Y}(y)$$
where $f_{Y}$ is the pdf of $Y$, so 
$$\dfrac{\text{d}}{\text{d}y}\left[P(Y \leq y)\right] = \dfrac{\text{d}}{\text{d}y}\left[ 1 - F_{X}(1-y) \right]= -F^{'}_{X}(1-y)\cdot (-1) =   f_{Y}(y)$$
where the chain rule for derivatives is used for the last step (derivative of the outside function with the original inside, times the derivative of the inside function; i.e., derivative of $F_{X}$ with $1-y$ in it multiplied by the derivative of $1-y$). 
Of course, $F_{X}^{'}$ is just the derivative of the CDF of $X$, or simply the PDF. So $$-F^{'}_{X}(1-y)\cdot (-1)  = f_{X}(1-y) = f_{Y}(y)\text{.}$$
Take your PDF for $X$ and just substitute $1-y$ in for $x$. Then that's your PDF for $Y$, for $0 \leq y \leq 1$.
A: By what you wrote, we have
$$F_Y(y)= 1-\frac{\int_{0}^{1-y}{t^{p-1}(1-t)^{q-1}dt}}{B(p,q)}.$$
Differentiate, using the Fundamental Theorem of Calculus. We get
$$\frac{{(1-y)^{p-1}y^{q-1}}}{B(p,q)}$$
(for $0\le y\le 1$, and $0$ elsewhere). 
