Assume that Y has a beta distribution with parameters ?

Step 1 of 4

a) We have to find the density function of U = 1-Y

Given that Y is a beta distribution with \(\alpha\) and \(\beta\)

The probability density function of beta distribution is

\(f(y)=\frac{y^{\alpha-1}(1-y)^{\beta-1}}{\Gamma(\alpha) \Gamma(\beta) / \Gamma(\alpha+\beta)}\)

Now the cumulative distribution function of beta distribution is

\(F(y)=\int_{0}^{y} \frac{y^{\alpha-1}(1-y)^{\beta-1}}{\Gamma(\alpha) \Gamma(\beta) / \Gamma(\alpha+\beta)} d y\)

Now 

\(\begin{aligned}
F_{U}(y) & =P(U \leq y) \\
& =P(1-Y \leq y) \\
& =P(Y \geq 1-y) \\
& =1-P(Y \leq 1-y) \\
& =1-\int_{0}^{1-y} \frac{y^{\alpha-1}(1-y)^{\beta-1}}{\Gamma(\alpha) \Gamma(\beta) / \Gamma(\alpha+\beta)} d y
\end{aligned}\)

The probability density function

\(\begin{aligned}
f(y) & =F^{\prime}(y) \\
& =\frac{d}{d y}\left(1-\int_{0}^{1-y} \frac{y^{\alpha-1}(1-y)^{\beta-1}}{\Gamma(\alpha) \Gamma(\beta) / \Gamma(\alpha+\beta)} d y\right) \\
& =-\frac{(1-y)^{\alpha-1} y^{\beta-1}}{\Gamma(\alpha) \Gamma(\beta) / \Gamma(\alpha+\beta)}
\end{aligned}\)

Hence \(f_{U}(y)=-\frac{(1-y)^{\alpha-1} y^{\beta-1}}{\Gamma(\alpha) \Gamma(\beta) / \Gamma(\alpha+\beta)}\)