# Expectation value of continuous random variable

1. Nov 7, 2005

Hi.. i am doing this question for Probability Theory, to find E[x] of a continuous random variable

E[x] = the integral from (0 to infinity) of 2x^2 * e^(-x^2) dx

So I used integration by parts...

u = x^2
du = 2xdx

dv = e^(-x^2) <--- ahh... how do you integrate that. (it dosn't look like it could be)

Anybody have any ideas?

2. Nov 7, 2005

looks like X~Normal random variable..
why does your EX have x^2?

to integrate exp(-x^2), you need to use a trick. multiply integral by itself to form a double integral... result should have exp(-(x^2 + y^2)) as part of integrand. recall x^2+y^2=r^2 and dx dy = r dr d0

but i don't think probability theory should be an exercise in integration. use symmetry to arrive at EX should be. ie. is the density function even or odd? what do you know about integration over symmetric intervals?

edit: integration by parts should be
u = 2x
dv = x*exp(-x^2) dx <-- you should know how to integrate this, use change of variable

Last edited: Nov 7, 2005
3. Nov 8, 2005

### benorin

sqrt(pi)/2

$$\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}$$

Pf:
Put $$I=\int_{0}^{\infty}e^{-x^{2}}dx$$ so that $$I^2=\int_{x=0}^{\infty}e^{-x^{2}}dx\int_{y=0}^{\infty}e^{-y^{2}}dy=\int_{y=0}^{\infty}\int_{x=0}^{\infty}e^{-(x^{2}+y^{2})}dxdy$$.

Now Transform to polar coordinates, and note that the first quadrant (e.g. QI) is one quarter of an infinite plane in rectangluar coordinates, so too is it one quarter of an infinite circle in polar coordinates (you can prove it with using squeeze theorem if your so inclined); you get

$$I^2=\int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}^{\infty}e^{-r^{2}}rdrd\theta=\frac{1}{2}\int_{\theta=0}^{\frac{\pi}{2}}d\theta\int_{u=0}^{\infty}e^{-u}du=\frac{\pi}{4}\lim_{M \rightarrow \infty}(1-e^{-M})=\frac{\pi}{4}$$

Therefore, $$I=\frac{\sqrt{\pi}}{2}$$.

4. Nov 8, 2005

what about if its just e^-(x^2) if its not a definit integral.

I mean.

Simply what would be the integral of e^-(x^2) ?

5. Nov 8, 2005

### VietDao29

Others have shown you that:
$$\int \limits_0 ^ \infty e ^ {-x ^ 2}dx = \frac{\sqrt{\pi}}{2}$$
Now you can use integration by parts:
You can choose u = x, dv = 2xe-x ^ 2 instead of choosing u = 2x2 and dv = e-x ^ 2 dx (the 2nd poster has pointed that out!).
So u = x, dv = 2xe-x ^ 2
du = dx, v = ...
Can you go from here?

6. Nov 8, 2005

### benorin

Ulgy

Ugly. I should define ugly, ugly means incapable of being expressed by a finite number of elementary functions--you know, ugly. Use series: here, like this

$$e^{-x^{2}} = \sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k}}{k!}$$

so that

$$\int e^{-x^{2}}dx = \int \sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k}}{k!}dx= \sum_{k=0}^{\infty} (-1)^{k}\int \frac{x^{2k}}{k!}dx=\sum_{k=0}^{\infty} (-1)^{k}\int \frac{x^{2k}}{k!}dx=\sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k+1}}{(2k+1)k!} + C$$

where interchanging the order of summation and integration is justified by the uniform convergence of that power series for $e^{-x^{2}}$ (for every bounded interval for x).

Last edited: Nov 8, 2005
7. Nov 8, 2005