PDF of Normal Distribution (Part 2)

[AvgStudent]

I arrived at the following density function for the normal distribution from the previous thread. However, it is incomplete as the value of k is missing.
[math]p(x)=\sqrt{\frac{k}{2\pi}}e^{-\frac{kx^2}{2}}[/math]
Background: The expected value is defined as, [math]\mu=\int_{-\infty}^{\infty}xp(x) \,dx[/math]Since the function p(x) is odd, the mean is 0. However, the variance,[imath]\sigma^2[/imath], requires additional work, which is defined as:
[math]\sigma^2=\int_{-\infty}^{\infty}x^2p(x) \,dx=\int_{-\infty}^{\infty}x^2 \sqrt{\frac{k}{2\pi}}e^{-\frac{kx^2}{2}}\,dx=2\sqrt{\frac{k}{2\pi}}\int_{0}^{\infty}x^2 e^{-\frac{kx^2}{2}}\,dx[/math]
My question how do I integrate? Ultimately, finding the value of k.
I've tried integration by parts, but it didn't get me anywhere.
[imath]u=x^2[/imath] and [imath]v=e^{-\frac{kx^2}{2}}[/imath]

 

[blamocur]

Look at the derivative of [imath]xe^{-\frac{kx^2}{2}}[/imath].

 

[AvgStudent]

Look at the derivative of [imath]xe^{-\frac{kx^2}{2}}[/imath].

I meant [imath]dv=e^{-\frac{kx^2}{2}}[/imath], not v in the original post.
Anyways, the derivative of [imath]xe^{-\frac{kx^2}{2}} = \mathrm{e}^{-\frac{kx^2}{2}}-kx^2\mathrm{e}^{-\frac{kx^2}{2}} [/imath]
Not sure how this helps?

 

[blamocur]

Integration by parts?

 

[AvgStudent]

Integration by parts?

Ok, I see
[imath]u=x[/imath] and [imath]dv=xe^{-\frac{kx^2}{2}}[/imath]

 

[AvgStudent]

After the integration by parts this is what I have:

[math]\sigma^2 =2\sqrt{\frac{k}{2\pi}} \left[\lim_{M\to\infty} -\frac{x}{k}e^{\frac{-kx^2}{2}}\biggr\rvert_{0}^{M} +\int_{0}^{\infty} \frac{e^{\frac{-kx^2}{2}}}{k} \,dx \right][/math]
1) I think the limit is 0. Can you confirm?
2) How to evaluate the integral?

 

[blamocur]

1) I also think the limit is 0.
2) Use the result from your previous thread.

 

[AvgStudent]

[math]\int_{0}^{\infty} \frac{e^{\frac{-kx^2}{2}}}{k} \,dx=\frac{1}{k}\sqrt{\frac{2\pi}{k}}?[/math]

 

[blamocur]

Looks right to me. How did you get it?

 

[AvgStudent]

Let [imath]u=\sqrt{\frac{k}{2}}x \Rightarrow dx=\sqrt{\frac{2}{k}}du[/imath]
[math]\int_{0}^{\infty} \frac{e^{\frac{-kx^2}{2}}}{k} \,dx=\frac{1}{k}\sqrt{\frac{2}{k}}\int_{0}^{\infty} e^{-u^2} \,du=\frac{1}{k}\sqrt{\frac{2}{k}}\sqrt{\pi}= \frac{1}{k}\sqrt{\frac{2\pi}{k}}[/math]I'm still missing a 1/2 somewhere, only because I know what k supposed to be.

 

[blamocur]

You assume that the value of [imath]\int_0^\infty e^{-u^2}du[/imath] is a given, but I meant that you can actually figure it out from the 2D case.

 

[AvgStudent]

I think I know what you mean now:
From previously:
[math]\int_{0}^{\infty} \int_{0}^{\infty} e^{\frac{-k(x^2+y^2)}{2}} \,dy\,dx= \left(\int_{0}^{\infty}e^{\frac{-kx^2}{2}} \,dx \right) \left(\int_{0}^{\infty}e^{\frac{-ky^2}{2}} \,dy \right)=\frac{1}{4A^2}= \left(\frac{1}{2A}\right)\left(\frac{1}{2A}\right)[/math]Then:
[math]\int_{0}^{\infty}e^{\frac{-kx^2}{2}} \,dx =\frac{1}{2A}= \frac{1}{2}\sqrt{\frac{2\pi}{k}}[/math]Together:
[math]\sigma^2 =2\sqrt{\frac{k}{2\pi}} \left[\int_{0}^{\infty} \frac{e^{\frac{-kx^2}{2}}}{k} \,dx \right]= \frac{2}{k}\sqrt{\frac{k}{2\pi}}\left(\frac{1}{2}\sqrt{\frac{2\pi}{k}}\right)=\frac{1}{k} \Rightarrow k=\frac{1}{\sigma^2}[/math]

 

[AvgStudent]

Thanks a lot @blamocur! Feels good after torturing myself for the past few days

 

[AvgStudent]

For completeness, the probability density function of the normal distribution is defined as:
[math]p(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x}{\sigma})^2}[/math]The general equation for the normal distribution with mean [imath]\mu[/imath] and standard deviation [imath]\sigma[/imath] is created by a simple horizontal shift of the above distribution:
[math]p(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}[/math]