# Proof within Univariate Normal

• Aug 20th 2009, 10:30 PM
ynotidas
Proof within Univariate Normal
Show that
E(X)=μ<br><br>thats all the information i've been given.<br>I dont know where to start.<br>
• Aug 20th 2009, 10:45 PM
Chris L T521
Quote:

Originally Posted by ynotidas
Show that
E(X)=μ<br><br>thats all the information i've been given.<br>I dont know where to start.<br>

First, note that the pdf of $\displaystyle \mathcal{N}\left(\mu,\sigma^2\right)$ is $\displaystyle f\!\left(x\right)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$

Now, $\displaystyle E\left(x\right)=\int_{-\infty}^{\infty}xf\!\left(x\right)\,dx=\frac{1}{\s qrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx$

Making the substitution $\displaystyle u=\frac{x-\mu}{\sigma}\implies \sigma\,du=\,dx$.

Thus, we have $\displaystyle \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx\xrightarrow{u=\left(x-\mu\right)/\sigma}{}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\sigma u+\mu\right)e^{-\frac{1}{2}u^2}\,du$ $\displaystyle =\frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du+\frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du$

For the first integral:

$\displaystyle \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du$

Let $\displaystyle z=\tfrac{1}{2}u^2\implies \,dz=u\,du$

Thus, $\displaystyle \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2} {}\frac{\sigma}{\sqrt{2\pi}}\int_{\infty}^{\infty} e^{-z}\,dz=\frac{\sigma}{\sqrt{2\pi}}\cdot0=0$.

Now for the second integral:

$\displaystyle \frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du$

Recall that the Gaussian integral $\displaystyle \int_{0}^{\infty}e^{-ax^2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\implies\ int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$.

So it follows that

$\displaystyle \frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du=\frac{\mu}{\sqrt{2\pi}}\cdot\s qrt{\frac{\pi}{1/2}}=\frac{\mu}{\sqrt{2\pi}}\cdot\sqrt{2\pi}=\mu$.

Therefore, $\displaystyle E\left(x\right)=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx=\mu$.

Does this make sense?
• Aug 20th 2009, 10:51 PM
ynotidas
Thanks Chris,

I'm going to have to look over it again and again, and analyse everystep to see if i understand all of it, i dont want to waste your time by asking a stupid question ehheh...

THANKS AGAIN! :D
• Aug 20th 2009, 10:54 PM
ynotidas
BUT, in saying that,

In the beginning, why did you integrate the E[x]??

also do you know any websites that can help me in understanding more on this? thanx!!
• Aug 20th 2009, 10:57 PM
CaptainBlack
Quote:

Originally Posted by ynotidas
BUT, in saying that,

In the beginning, why did you integrate the E[x]??

also do you know any websites that can help me in understanding more on this? thanx!!

$\displaystyle E(x)$ was not integrated, it is defined to be the integral of $\displaystyle x\times f(x)$ where $\displaystyle f(x)$ is the pdf of $\displaystyle x$, and that is what was done.

CB
• Aug 20th 2009, 11:02 PM
Chris L T521
Quote:

Originally Posted by ynotidas
BUT, in saying that,

In the beginning, why did you integrate the E[x]??

also do you know any websites that can help me in understanding more on this? thanx!!

I didn't integrate the expected value. That's how the expected value is defined for continuous distributions (like the normal):

$\displaystyle E\left(X\right)=\int_{-\infty}^{\infty}xf\!\left(x\right)\,dx$, where $\displaystyle f\!\left(x\right)$ is the pdf of the distribution.

Here's a site that evaluated the integral the way I did: The Normal Distribution
• Aug 20th 2009, 11:03 PM
matheagle
Note that
Thus, $\displaystyle \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2} {}\frac{\sigma}{\sqrt{2\pi}}\int_{\infty}^{\infty} e^{-z}\,dz=\frac{\sigma}{\sqrt{2\pi}}\cdot0=0$.
is a lot of work.
The integrand is an odd function so the integral must be zero.
I just commented on that in my review of that calc book.
Noticing that a function is even or odd can save some work.

As for
$\displaystyle \frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du$

I first show that the normal is a valid density with that polar coordinate trick (multiplying it by itself)
and then we have

$\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du=1$
since this is a valid density of a N(0,1).
• Aug 20th 2009, 11:06 PM
ynotidas
OH okkkkkkkkkkkkkk i'll keep reading down..

thanx everyone!
• Aug 20th 2009, 11:59 PM
ynotidas
How do you get the integration of e^-z from -infinity to +infinity to be equal to zero?

• Aug 21st 2009, 12:09 AM
matheagle
Just integrate and let the bounds go to infinity, calc 2.
• Aug 21st 2009, 11:18 AM
Chris L T521
Quote:

Originally Posted by ynotidas
How do you get the integration of e^-z from -infinity to +infinity to be equal to zero?

After the substitution $\displaystyle z=\tfrac{1}{2}u^2$, the limits of integration became positive (and the same)!

So it follows from integration properties that $\displaystyle \int_a^a f\!\left(x\right)\,dx=0$.

That's why $\displaystyle \int_{\infty}^{\infty}e^{-z}\,dz=0$