0
$\begingroup$

Probability Question:

If $Y=X^2$ and $X\sim N(1,1)$, what is the PDF of $Y$?

I know that the CDF of $Y$ is $$P(Y \leq y) = P(X^2 \leq y) = P(X \leq \sqrt{y}) - P(x \leq -\sqrt{y})$$

Thus, the CDF of Y is equal to the standard normal CDF $\Phi(\sqrt{y}-1) - \Phi(-\sqrt{y}-1)$

I'm trying to find the PDF of Y by taking the derivative of that, but a solution that my professor posted states that the $$PDF = \frac1{ 2\sqrt{y}} \phi(\sqrt{y}) - \phi(-\sqrt{y}).$$

*($\phi$ is the pdf of X)

Where does the $\frac1{2\sqrt{y}}$ come from? Can anyone please explain the derivative of that?

Thanks!

  • 0
    Instead of going through cdf, it is better to use the direct formula for pdf of the transformed rv, see e.g. my answer here: http://math.stackexchange.com/questions/1531572/density-of-x2-when-x-has-uniform-1-2-distribution/. – 2017-01-07

1 Answers 1

2

It's the chain rule.

$$ \frac{d}{dy}\Phi(\sqrt{y}-1) = \frac{1}{2\sqrt{y}}\Phi'(\sqrt{y}-1)=\frac{1}{2\sqrt{y}}\phi(\sqrt{y}-1) $$

where $\phi$ is the normal pdf $$ \phi(x) =\frac{e^{-x^2/2}}{\sqrt{2\pi}} $$

By the same token, $$ \frac{d}{dy}\Phi(-\sqrt{y}-1) = -\frac{1}{2\sqrt{y}}\phi(-\sqrt{y}-1), $$ so the final expression is $$ \frac{d}{dy}\left(\Phi(\sqrt{y}-1)-\Phi(-\sqrt{y}-1)\right) = \frac{1}{2\sqrt{y}}\left(\phi(\sqrt{y}-1) + \phi(-\sqrt{y}-1)\right) $$

  • 0
    Thanks! I'm a little confused about the second part though. Since the CDF is Φ(sqrt(y)-1) - Φ(-sqrt(y)-1), what should we do about the Φ(-sqrt(y)-1) part? – 2017-01-07
  • 0
    what's the second part? Why the derivative of $\Phi$ is $\phi$? – 2017-01-07
  • 0
    Oh never mind I got it. I was a little confused about what to do for the chain rule for the Φ(-sqrt(y)-1) part, but I got it. Thanks!! – 2017-01-07