Show that

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

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- August 20th 2009, 11:30 PMynotidasProof 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> - August 20th 2009, 11:45 PMChris L T521
- August 20th 2009, 11:51 PMynotidas
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 - August 20th 2009, 11:54 PMynotidas
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!! - August 20th 2009, 11:57 PMCaptainBlack
- August 21st 2009, 12:02 AMChris L T521
I didn't integrate the expected value. That's how the expected value is defined for continuous distributions (like the normal):

, where is the pdf of the distribution.

Here's a site that evaluated the integral the way I did: The Normal Distribution - August 21st 2009, 12:03 AMmatheagle
Note that

Thus, .

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

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

and then we have

since this is a valid density of a N(0,1). - August 21st 2009, 12:06 AMynotidas
OH okkkkkkkkkkkkkk i'll keep reading down..

thanx everyone! - August 21st 2009, 12:59 AMynotidas
How do you get the integration of e^-z from -infinity to +infinity to be equal to zero?

- August 21st 2009, 01:09 AMmatheagle
Just integrate and let the bounds go to infinity, calc 2.

- August 21st 2009, 12:18 PMChris L T521