1 Attachment(s)

Expecatation of Normal Distribution

Hi everybody,

I attached the first part to derive the expectation E[X] of the Normal distribution.

I understand that **(x-µ)/****σ** is replaced by **y**. But why is there an additional **σ** at the very end of the second part before **dy**. Why is this sigma added? A bit confusing for me..

Thanks for your help.

1 Attachment(s)

Re: Expecatation of Normal Distribution

Also, I have a problem to understand the integration by parts when solving for E(X^2).

Again I attached the solution. In don't understand how the first term in the second row is integrated:

I would use f(x)=y^2 ; f'(x)=2y; g'(x)=exp(-0.5(y^2) ; g(x) = (2*pi)^0.5

However I would get 0 with this approach for that term

...In the solution, I dont get what was used for f(x), g'(x) etc...

Again, thanks for any help. This forum is great!

Re: Expecatation of Normal Distribution

Quote:

Originally Posted by

**Marmy** Hi everybody,

I attached the first part to derive the expectation E[X] of the Normal distribution.

I understand that **(x-µ)/****σ** is replaced by **y**. But why is there an additional **σ** at the very end of the second part before **dy**. Why is this sigma added? A bit confusing for me..

Thanks for your help.

If:

$\displaystyle y=\frac{x-\mu}{\sigma}$

$\displaystyle dy=\frac{1}{\sigma}dx$

$\displaystyle dx=\sigma dy$

Re: Expecatation of Normal Distribution

Quote:

Originally Posted by

**Marmy** Also, I have a problem to understand the integration by parts when solving for E(X^2).

Again I attached the solution. In don't understand how the first term in the second row is integrated:

I would use f(x)=y^2 ; f'(x)=2y; g'(x)=exp(-0.5(y^2) ; g(x) = (2*pi)^0.5

However I would get 0 with this approach for that term

...In the solution, I dont get what was used for f(x), g'(x) etc...

Again, thanks for any help. This forum is great!

Split:

$\displaystyle -y^2e^{-\frac{y^2}{2}}= y \left[-y e^{-\frac{y^2}{2}} \right]$

then the term in the square brackets is the derivative of $\displaystyle e^{-\frac{y^2}{2}}$

CB

Re: Expecatation of Normal Distribution

Thanks, again very helpful!