I'm trying to understand a derivation of the density of a Chi-Square distribution with one degree of freedom that my professor gave in class. Here are the steps:

\begin{align} p(y) &= \int_{-\infty}^{+\infty} p(x,y)dx\\\ &= \int_{-\infty}^{+\infty}p(y|x)p(x)dx\\ &=\int_{-\infty}^{+\infty}\delta(y-x^2)\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx \end{align}

Now we use the substitution $u=x^2$ to get:

\begin{align} p(y) &= 2\int_{0}^{+\infty} \delta(y-u)\frac{1}{\sqrt{2\pi}}\frac{e^{\frac{-u}{2}}}{2\sqrt{u}}du\\ &= \frac{1}{\sqrt{2\pi}}\frac{e^{\frac{-y}{2}}}{\sqrt{y}} \end{align}

The substitution puzzles me for a few reasons. Firstly, it removes integrating over the negative values, therefore how can we evaluate the dirac delta? Secondly, where do we get that extra factor of 2 from which fortuitously cancels the one in the Jacobian? I've been looking at this a while, help anyone?


Note $\delta(y-(-x)^2) \frac{1}{\sqrt{2 \pi}} e^{-(-x)^2/2} = \delta(y-x^2) \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$, so $\int_{-\infty}^\infty \delta(y-x^2) \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx$ can be written as $\int_0^\infty \delta(y-(-x)^2) \frac{1}{\sqrt{2 \pi}} e^{-(-x)^2/2} dx + \int_0^\infty \delta(y-x^2) \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx$ where the first term corresponds to negative $x$ and the second to positive $x$ in the original integral. Thus, noting the equivalence mentioned, $\int_{-\infty}^\infty \delta(y-x^2) \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx= 2 \int_0^\infty \delta(y-x^2) \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx$. Then use the sifting property of the Dirac delta.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.