The PDF of the Normal Distribution

[AvgStudent]

Hi all, I'm looking at the derivation of the probability density function of the normal distribution. So far, I've come up with the following double integral but am unsure how to continue.
The objective is to find the value of A such that:
[math]\int_{0}^{\infty} \int_{0}^{\infty} e^{\frac{-k(x^2+y^2)}{2}} \,dy\,dx=\frac{1}{4A^2}[/math]
How do I integrate such integral?

 

[blamocur]

Hi all, I'm looking at the derivation of the probability density function of the normal distribution. So far, I've come up with the following double integral but am unsure how to continue.
The objective is to find the value of A such that:
[math]\int_{0}^{\infty} \int_{0}^{\infty} e^{\frac{-k(x^2+y^2)}{2}} \,dy\,dx=\frac{1}{4A^2}[/math]
How do I integrate such integral?

You integrate it by switching to polar coordinates:
[math]\int_0^\infty\int_0^\infty e^{\frac{-k(x^2+y^2)}{2}} \,dy\,dx = \int_0^{\frac{\pi}{2}}\int_0^\infty e^{\frac{-kr^2}{2}} r dr d\theta[/math][math]= \int_0^{\frac{\pi}{2}} d\theta \int_0^\infty e^{-k\left(\frac{r^2}{2}\right)} d\left(\frac{r^2}{2}\right)[/math][math]= \frac{\pi}{2} \int_0^\infty e^{-{ku}} du = \frac{\pi}{2k}[/math]
Note: if your integrals had [imath](-\infty,\infty)[/imath] limits instead of [imath](0,\infty)[/imath] then the upper limit for [imath]\theta[/imath] would be [imath]2\pi[/imath] and the final answer [imath]\frac{2\pi}{k}[/imath].

 

[AvgStudent]

You integrate it by switching to polar coordinates:
[math]\int_0^\infty\int_0^\infty e^{\frac{-k(x^2+y^2)}{2}} \,dy\,dx = \int_0^{\frac{\pi}{2}}\int_0^\infty e^{\frac{-kr^2}{2}} r dr d\theta[/math][math]= \int_0^{\frac{\pi}{2}} d\theta 2 \int_0^\infty e^{\frac{-kr^2}{2}} d(r^2)[/math][math]= \frac{\pi}{2} \int_0^\infty e^{-\frac{ku}{2}} du = \frac{\pi}{k}[/math]
Note: if your integrals had [imath](-\infty,\infty)[/imath] limits instead of [imath](0,\infty)[/imath] then the upper limit for [imath]\theta[/imath] would be [imath]2\pi[/imath] and the final answer [imath]\frac{4\pi}{k}[/imath].

Wow, I didn't think to use polar coordinates, but [imath]x^2+y^2[/imath] should've been my first clue.
From your second line to the third line, make sure I get it. You let u=r^2, then du/dr=2r. After the substitution, it becomes 2r*r.
Another question on your note about the limits of integration. Why does [imath](0,\infty)[/imath] is [imath]\pi /2[/imath], but [imath](-\infty,\infty)[/imath] is [imath]2\pi[/imath]? I would think that if [imath](-\infty,\infty)[/imath] is [imath]2\pi[/imath], then [imath](0,\infty)[/imath] would be half of that which is [imath]\pi[/imath], instead of [imath]\pi/2[/imath].

As for the value of A. Since [imath]\frac{\pi}{2k}=\frac{1}{4A^2} \Rightarrow A=\sqrt{\frac{k}{2\pi}}[/imath]. Right?

 

[blamocur]

I've edited my post to use [imath]u=\frac{r^2}{2}[/imath], hopefully getting it right this time.
As for changing the integration limits, we would be switching from one (positive) quadrant of [imath]\mathbb R^2[/imath] to all of [imath]\mathbb R^2[/imath], which explains the factor of 4.

P.S. I've realized belatedly that I've broken the rules by providing the whole solution instead of hints, but at least you saw an error in the first edition of my post, and I'll leave to you the verification of the formula for A. I've numerically checked my latest edition, and it seems correct.

 

[AvgStudent]

I've edited my post to use [imath]u=\frac{r^2}{2}[/imath], hopefully getting it right this time.
As for changing the integration limits, we would be switching from one (positive) quadrant of [imath]\mathbb R^2[/imath] to all of [imath]\mathbb R^2[/imath], which explains the factor of 4.

P.S. I've realized belatedly that I've broken the rules by providing the whole solution instead of hints, but at least you saw an error in the first edition of my post, and I'll leave to you the verification of the formula for A. I've numerically checked my latest edition, and it seems correct.

Got it. Thanks!