Let Y1, Y2, . . . , Yn be independent random variables,

Step 1 of 4

Given, \(Y_{1}, Y_{2}, \ldots, Y_{n}\) are independent random variables, where \(Y \sim \operatorname{beta}(\alpha, \beta)\).

Also, given, \(\alpha=\beta=2\)

The probability distribution function of \(Y\) is given by,

\(f(y)=\left\{\begin{array}{cc} \frac{1}{\beta(\alpha, \beta)} Y^{\alpha-1}\left((1-Y)^{\beta-1}\right) & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right.\)

That is,        \(f(y)=\left\{\begin{array}{cc} \frac{1}{\beta(2,2)} Y(1-Y) & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right.\)

In terms of gamma function, \(\beta(\alpha, \beta)=\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}\)

So,  \(\beta(2,2)=\frac{\Gamma(2) \Gamma(2)}{\Gamma(4)}=\frac{1 ! \times 1 !}{3 !}=\frac{1}{6}\)

Thus, \(f_{Y}(y)=6 y(1-y)\).