# Calculating pdf, expectation and variance of random variable

Suppose that X follows a chi-square distribution $$\chi_n^2$$ and that $$Y=\sqrt{2X}$$. Find the pdf of $$Y$$ and show that $$\mathbb{E}(Y) = \frac{\Gamma((n+1)/2)}{\Gamma(n/2)}$$ and $$\mathbb{E}(Y^2) = 2n$$.

I have calculated the pdf using the change of variable formula as $$f_Y(y) = y^{n-1} e^{-y^2/4}/(2^{n-1} \Gamma(1/2n))$$. However how would i calculate $$\mathbb{E}(Y)$$ and $$\mathbb{E}(Y^2)$$? Is there a trick to calculating the integral

$$\mathbb{E}(Y) = \int^{\infty}_{0} \frac{y e^{-y^2/4}}{2^{n-1}\Gamma(1/2n)} dy$$ as it looks messy?

My guess would be $$\mathbb{E}(Y^2)$$ would follow from a similar calculation?

• It seems you forgot a $n$ in exponent in the integral, but if you mean calcultaing $\int_0^\infty y^ne^{-y^2/4}$ I would try to link this to the moments of Gaussian variables. Jan 22, 2020 at 14:16

$$\mathbb{E}(Y^2) = \mathbb{E}(2 \cdot X) = 2 \cdot \mathbb{E}(\chi^2_{n}) = 2 \cdot n$$
Integral of type $$\int\limits_{0}^{+ \infty} x^p \cdot e^{-x^2} \ dx$$ is computed easily by making a substitution $$\{ x^2 = t \}$$ and making use of Euler Gamma function $$\{ \Gamma(\alpha) = \int\limits_{0}^{+ \infty} x^{\alpha - 1} \cdot e^{-x} \ dx \}$$