Moment generating function and moments of standard normal distribution

[spitz]

Homework Statement

Find the MGF and all the moments for $X\sim N(0,1)$

2. The attempt at a solution

For the MGF, I have:

$$M_X(s)=\displaystyle\int_{-\infty}^{\infty}e^{sx}\frac{e^{x^2/2}}{\sqrt{2\pi}}\,dx = \ldots=e^{s^2/2}$$

Next I found that:
$$M'_X(0)=E[X]=0$$
$$M''_X(0)=E[X^2]=1$$
$$E[X^3]=0$$
$$E[X^4]=3$$
$$\ldots$$
$$E[X^{ODD}]=\{0\}$$
$$E[X^{EVEN}]=\{1,3,15,105,945,\ldots\}$$
Is it enough to write:
$$E[X^k]=M_X^{(k)}(0)=\frac{d^k}{ds^k}e^{s^2/2}$$

Am I totally off track here? How would I prove this?

 

[LCKurtz]

Is it enough to write:
$$E[X^k]=M_X^{(k)}(0)=\frac{d^k}{ds^k}e^{s^2/2}$$

Am I totally off track here? How would I prove this?

It looks OK to me. How to prove it...it is a general property of Moment Generating Functions and the reason they are called that. If$$
M(s) = E(e^{sx}) = \int_{-\infty}^{\infty}e^{sx}f(x)\, dx$$and you differentiate with respect to s you get:$$
M'(s) = \int_{-\infty}^{\infty}xe^{sx}f(x)\, dx$$If you evaluate that at $s=0$ you get$$
M'(0)=\int_{-\infty}^{\infty}xf(x)\, dx = E(X)$$Each time you differentiate with respect to $s$ you get another $x$ out in front giving you the next moment.

 

[Ray Vickson]

Homework Statement

Find the MGF and all the moments for $X\sim N(0,1)$

2. The attempt at a solution

For the MGF, I have:

$$M_X(s)=\displaystyle\int_{-\infty}^{\infty}e^{sx}\frac{e^{x^2/2}}{\sqrt{2\pi}}\,dx = \ldots=e^{s^2/2}$$

Next I found that:
$$M'_X(0)=E[X]=0$$
$$M''_X(0)=E[X^2]=1$$
$$E[X^3]=0$$
$$E[X^4]=3$$
$$\ldots$$
$$E[X^{ODD}]=\{0\}$$
$$E[X^{EVEN}]=\{1,3,15,105,945,\ldots\}$$
Is it enough to write:
$$E[X^k]=M_X^{(k)}(0)=\frac{d^k}{ds^k}e^{s^2/2}$$

Am I totally off track here? How would I prove this?

Your MGF formula is seriously wrong: you need $e^{-x^2/2}$ in the integrand, not your $e^{x^2/2}.$ However, your result
$$M_X(s) = e^{s^2/2}$$ is correct, as are your subsequent results.

As for how to prove it: just use standard theorems about differentiation under the integral sign. The normal density goes to 0 quickly enough for large |x| that you will not have any problems meeting the hypotheses of the required theorems.

RGV