This is from measure theoretic probability class.

Let $Z$ have the Gaussian distribution with mean $0$ and variance $b$. Show that, then, $X=Z^2$ has the gamma distribution with shape index $a=1/2$ and scale parameter $c=1/2b$.

The instructor indicated using the following theorem:


Let $X$ be a random variable taking values in $(E,\mathscr{E})$. If $\mu$ is the distribution of $X$, that is, $\mu =\mathbb{P}\circ X^{-1}$, then,

$\mathbb{E}f\circ X=\mu f$

for every positive $\mathscr{E}$-measurable function $f$.

Concersely, if this holds for some measure $\mu$ and all positive $\mathscr{E}$-measurable functions, then $\mu$ is the distribution of $X$.

I guess this problem can be solved by computing $\mathbb{E}f\circ Z $, where $f(x)=x^2$,and applying the theorem to say $\mathbb{E}f\circ Z=\mu f$ and then identify $\mu$. But I really can't see how exactly, or is this a completely wrong direction?

  • $\begingroup$ You don't need measure theory here...take a look at my answer on this post math.stackexchange.com/questions/795138/… $\endgroup$ – afedder Jun 12 '14 at 17:59
  • $\begingroup$ Another plausible answer (first one) can be found here math.stackexchange.com/questions/792556/… $\endgroup$ – afedder Jun 12 '14 at 18:03
  • $\begingroup$ Note that $\frac{Z}{\sqrt{b}} \sim N(0,1)$. Thus, $\frac{Z^2}{b} \sim \chi_1^2$. Now, to see the density of $Z^2$, use the hint about random variables multiplied by constants in my answer to the other post given above. $\endgroup$ – afedder Jun 12 '14 at 18:08

Let $g$ be positive and measurable. Then we want to bring $$ {\rm E}[g(X)]={\rm E}[g(Z^2)]=\int_\mathbb{R} g(z^2) f_Z(z)\,\mathrm dz $$ onto the form $$ \int_\mathbb{R} g(x) h(x)\,\mathrm dx $$ for some non-negative, measurable function $h$ because then $h$ is the density of $X$. First note that $z\mapsto g(z^2) f_Z(z)$ is symmetric around $0$ and hence $$ {\rm E}[g(X)]=2\int_0^\infty g(z^2)f_Z(z)\,\mathrm dz. $$ Now, do a change of variables with $x=z^2$ to obtain $h$ (and thus the density of $X$).

  • $\begingroup$ Thank you. That was helpful! $\endgroup$ – stph Jun 12 '14 at 21:42

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