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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?

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  • $\begingroup$ 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. $\endgroup$
    – Fabien
    Jan 22, 2020 at 14:16

2 Answers 2

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$\mathbb{E}(Y^2) = \mathbb{E}(2 \cdot X) = 2 \cdot \mathbb{E}(\chi^2_{n}) = 2 \cdot n$

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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 $\{ \Gamma(\alpha) = \int\limits_{0}^{+ \infty} x^{\alpha - 1} \cdot e^{-x} \ dx \}$

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    $\begingroup$ You can just add this to your first answer. That is custom here. Very rarely (only if there is a special reason) more than one answer from the same author appears. $\endgroup$
    – drhab
    Jan 22, 2020 at 14:22

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