MHB Finding Moments of $X\sim N(0,1)$ using MGF

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The moment generating function (MGF) for the standard normal distribution \(X \sim N(0,1)\) is derived as \(M_X(s) = e^{s^2/2}\). The first two moments calculated are \(E[X] = 0\) and \(E[X^2] = 1\), with higher moments showing that odd moments are zero and even moments follow a specific pattern: \(E[X^{EVEN}] = \{1, 3, 15, 105, 945, \ldots\}\). The relationship \(E[X^k] = M_X^{(k)}(0)\) is confirmed, where \(M_X^{(k)}(0)\) represents the \(k\)-th derivative of the MGF evaluated at zero. The discussion emphasizes the need for a clear proof of this relationship and the calculation of moments using the MGF.
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I have to find the moment generating function and find all the moments of $X\sim N(0,1)$

For the MGF, I have:

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

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?
 
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Jason said:
I have to find the moment generating function and find all the moments of $X\sim N(0,1)$

For the MGF, I have:

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

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?

Th \(k\)-th moment is \(k!\) times coefficient of \(s^k\) in the MacLauren series expansion of the MGF.

CB
 
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