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.

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!

Quote:

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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:

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

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

$\displaystyle dx=\sigma dy$

Quote:

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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

Thanks, again very helpful!