# A Moments of normal distribution

1. Jan 6, 2017

### senobim

I have calculated characteristic function of normal distribution $$f_{X}(k)=e^{(ika-\frac{\sigma ^{2}k^{2}}{2})}$$ and now I would like to find the moments, so I know that you could expand characteristic function by Taylor series

$$f_{X}(k)=exp(1+\frac{1}{1!}(ika - \frac{\sigma^2k^2}{2})+\frac{1}{2!}(ika - \frac{\sigma^2k^2}{2})^2+\frac{1}{3!}(ika - \frac{\sigma^2k^2}{2})^3+...)$$

$$f_{X}(k)=exp(1+\frac{(ik)}{1!}\left \langle X^1 \right \rangle+\frac{(ik)^2}{2!}\left \langle X^2 \right \rangle+\frac{(ik)^3}{3!}\left \langle X^3 \right \rangle+...)$$

and the moments will be
$$\left \langle X^n \right \rangle$$

Now the problem is that I completely forgot how to evaluate Taylor series.
Could you help me to calculate for example second moment? I know what the answer should be, but I couldn't get it right.

Last edited: Jan 6, 2017
2. Jan 6, 2017

### jambaugh

Take the n-th derivative of the taylor series and evaluate at zero.
BTW your exponential function should not appear within the taylor expansion... this expansion IS the exponential funcition.

3. Jan 6, 2017

### senobim

I am trying to calculate first derivate of term <x1>

$$\frac{d}{dk}(ika-\frac{\sigma^2k^{2} }{2})=(ia - \sigma ^{2}k)$$

now I am evaluating it at 0

$$f_{X}(0)= ia$$

And what will happen with a i term?

4. Jan 6, 2017

### mathman

$f'_X(0)=iE(X)$, not E(X).

5. Jan 7, 2017

Thank you!