Finding MGF & Moments of X~N(0,1)

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SUMMARY

The moment generating function (MGF) for the standard normal distribution \(X \sim N(0,1)\) is correctly derived as \(M_X(s) = e^{s^2/2}\). The moments calculated include \(E[X] = 0\), \(E[X^2] = 1\), \(E[X^3] = 0\), and \(E[X^4] = 3\). The general property of MGFs states that the \(k\)-th moment can be found using \(E[X^k] = M_X^{(k)}(0) = \frac{d^k}{ds^k} e^{s^2/2}\). The discussion emphasizes the importance of using the correct integrand, which should include \(e^{-x^2/2}\) for the normal density function.

PREREQUISITES
  • Understanding of moment generating functions (MGFs)
  • Familiarity with the standard normal distribution \(N(0,1)\)
  • Knowledge of differentiation under the integral sign
  • Basic calculus, particularly derivatives of exponential functions
NEXT STEPS
  • Study the properties of moment generating functions in probability theory
  • Learn about the derivation of moments for various probability distributions
  • Explore the application of differentiation under the integral sign in calculus
  • Investigate the implications of even and odd moments in statistical analysis
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Students and professionals in statistics, probability theory, and data analysis, particularly those focusing on the properties of the normal distribution and moment generating functions.

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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?
 
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spitz said:
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.
 
spitz said:

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
 

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