Let $a, b, \alpha, β$ be real numbers such that $a <b, α> 0$ and $β> 0$. If the random variable $X$ is distributed beta with parameters $α$ and $β$, find the density function of the random variable $Y: = (b - a) X + a $.
This is my solution: I know that the density function is $$ f_X(x)=\frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1} $$ with $B(\alpha,\beta)=\displaystyle\int_0^{1}x^{\alpha-1}(1-x)^{\beta-1} dx$.
And by a theorem (i don't know the name of the theorem) $$ f_Y(y)= \left\{ \begin{array}{lcc} f_X(h^{-1}(x))\left|\frac{\partial(h^{-1}(x))}{\partial y}\right| & if & y=h(x) \\ \\ 0 & if & y\neq h(x) \end{array} \right. $$ So as $h(x)=(b-a)x+a$, we have $h^{-1}(y)=\frac{y-a}{b-a}=x$. Then $\frac{\partial(h^{-1}(x))}{\partial y}=\frac{\partial}{\partial y}\left(\frac{y-a}{b-a}\right)=\frac{1}{b-a}$.
Finally i have $$f_Y(y)= \left\{ \begin{array}{lcc} \frac{1}{B(\alpha,\beta)}\left(\frac{y-a}{b-a}\right)^{\alpha-1}\left(1-\frac{y-a}{b-a}\right)^{\beta-1} \left(\frac{1}{b-a}\right)& if & y=h(x) \\ \\ 0 & if & y\neq h(x) \end{array} \right. $$ is okey my solution? thanks for the help
I would say that since $X\in(0, 1)$ and $Y=(b-a)X+a$, then the image of $(0,1)$ is $(a,b)$. These are the values that the random variable $Y$ can take. So instead of saying $y=h(x)$ you could put the actual bounds of Y like this:
$$g(Y)=\begin{cases}\frac{(Y-a)^{\alpha-1}(b-Y)^{\beta-1}}{B(\alpha,\beta)(b-a)^{\alpha+\beta-1}}&\text {if } a<Y<b\\ 0&\text{otherwise}\end{cases}$$
Otherwise it looks fine.