Question

You are given the pdf \(\phi(x)=\dfrac{1}{4\sqrt{2\pi}}e^{-x^2/32}\) \(\-\infty 11 years ago

Answer 1

I'm not exactly sure about the details of deriving the CDF of the normal distribution given the PDF (as shown on the wiki page, http://en.wikipedia.org/wiki/Normal_distribution ; I'm not terribly familiar with the error function \(\text{erf}\)). You can verify, though, that \(\Phi'(x)=\phi(x)\) by the fundamental theorem of calculus.

For #2, you have
\[\begin{align*}
G(w)&=P(W\le w)\\
&=P(X^2\le w)\\
&=P(|X|\le \sqrt w)\\
&=P(-\sqrt w\le X\le \sqrt w)\\
&=P(X\le \sqrt w)-P(X\le -\sqrt w)\\
&=\Phi(\sqrt w)-\Phi(-\sqrt w)
\end{align*}\]
Again, this assumes you have the closed form of \(\Phi\), which is the main difficulty here (in my opinion).

#3 and #4 should be easy enough once you determine \(G(w)\).

Answer 2

Yes, the most difficult part is part 1 :(

Answer 3

Let's see what we can do. Here's the definition for the error function: http://mathworld.wolfram.com/Erf.html
\[\newcommand{\erf}{\text{erf}}
\erf(x)=\frac{2}{\sqrt \pi}\int_0^x e^{-t^2}dt\]
Here's what we want to find:
\[\begin{align*}
\Phi(x)&=\int_{-\infty}^x\frac{1}{4\sqrt{2\pi}}e^{-t^2/32}~dt\end{align*}\]
Set \(\dfrac{t}{4\sqrt 2}=s\), then
\[\begin{align*}
\Phi(x)&=\frac{1}{\sqrt\pi}\int_{-\infty}^{x/4\sqrt2} e^{-s^2}~ds\\\\
&=\frac{1}{2}\left(\underbrace{\frac{2}{\sqrt\pi}\int_{0}^{x/4\sqrt2} e^{-s^2}~ds}_{\text{def. of erf}}\right)+\underbrace{\frac{1}{\sqrt\pi}\int_{-\infty}^0 e^{-s^2}~ds}_{\text{this is a known result}}\\\\
&=\frac{1}{2}\erf\left(\frac{x}{4\sqrt2}\right)+\frac{1}{2\sqrt\pi}\int_{-\infty}^\infty e^{-s^2}~ds
\end{align*}\]

Answer 4

Wow, it is a mess. :)
Thanks for the link. :)

Answer 5

I know \(\int_{0}^{\infty}e^{-x^2}dx=\dfrac{\sqrt{\pi}}{2}\)

Answer 6

That is Gaussian integral.

Answer 7

Oops, I meant
\[\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}\]
But okay, that's good. Less work!

Answer 8

Alright, so now to find \(G(w)\) you have
\[\begin{align*}
G(w)&=\Phi(\sqrt w)-\Phi(-\sqrt w)\\\\
&=\left[\frac{1}{2}\erf\left(\frac{\sqrt w}{4\sqrt2}\right)+\frac{1}{2}\right]-\left[\frac{1}{2}\erf\left(\frac{-\sqrt w}{4\sqrt2}\right)+\frac{1}{2}\right]\\\\
&=\frac{1}{2}\erf\left(\frac{\sqrt w}{4\sqrt2}\right)+\frac{1}{2}\erf\left(\frac{\sqrt w}{4\sqrt2}\right)\\\\
&=\erf\left(\frac{\sqrt w}{4\sqrt2}\right)
\end{align*}\]
The second step is due to \(\erf(x)\) being an odd function. From here you can find \(g(w)\) by taking the derivative. Recall that
\[\begin{align*}G(w)&=\erf\left(\frac{\sqrt w}{4\sqrt2}\right)\\\\
&=\frac{2}{\sqrt\pi}\int_0^{\sqrt w/4\sqrt2} e^{-t^2}~dt\\\\
G'(w)&=\frac{2}{\sqrt\pi}e^{-(\sqrt w/4\sqrt2)^2}\times\frac{d}{dw}\left[\frac{\sqrt w}{4\sqrt2}\right]\\\\
g(w)&=\frac{1}{4\sqrt{2\pi}}\frac{e^{-w/32}}{\sqrt w}
\end{align*}\]

Answer 9

Wow!! It's a big help. :)

Answer 10

Showing that \(\displaystyle\int_0^\infty g(w)~dw=1\) is simple enough, I think. It'll involve the error function. Quick tip: substitute \(y=\sqrt w\).

Answer 11

Yes, Sir. I know how to do it. :)