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!