Find the variance of $Y$ where we are given the density $f_X(x)$

Question

Suppose $X$ has density $f_X(x) = \frac{1}{\beta}e^{-x/\beta}$ for $X\geq 0$. Find the variance of $Y = X^{\delta}$ for $\delta > 0$.

We need to find the pdf $f_Y(y)$ first wish we can find by applying the formula $$f_Y(y) = f_X(g^{-1}(y)\left|\frac{d}{dy}g^{-1}(y)\right|$$ We have $Y = X^{\delta}$ so $X = g^{-1}(y) = Y^{1/\delta}$ thus $$f_X(g^{-1}(y)) = \frac{1}{\beta} e^{-y^{1/\delta}/\beta} \ \ \ \text{and} \ \ \ \frac{d}{dy}(Y^{1/\delta}) = \frac{1}{\delta}Y^{1/\delta - 1}$$ Hence $$f_Y(y) = \frac{1}{\beta\delta}e^{-y^{1/\delta}/\beta}y^{1/\delta - 1}$$

I am not sure if this is right so far, any suggestions are greatly appreciated.

Answer 1

You only need $E[Y]$ and $E[Y^2]$. You don't actually have to compute its density.

$$E[Y] = \int_{0}^{\infty}\frac{1}{\beta}x^{\delta}e^{-\frac{x}{\beta}}\,dx$$ This is the gamma function. $E[Y] = \beta^{\delta}\Gamma(\delta+1)$. It is easy to see that $E[Y^2] = \beta^{2\delta}\Gamma(2\delta+1)$.