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

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In summary, your MGF formula is seriously wrong; you need e^{-x^2/2} in the integrand, not e^{x^2/2}. However, your result M_X(s) = e^{s^2/2} is correct, as are your subsequent results.
  • #1
spitz
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Homework Statement



Find the MGF and all the moments for [itex]X\sim N(0,1)[/itex]

2. The attempt at a solution

For the MGF, I have:

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

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

Am I totally off track here? How would I prove this?
 
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  • #2
spitz said:
Is it enough to write:
[tex]E[X^k]=M_X^{(k)}(0)=\frac{d^k}{ds^k}e^{s^2/2}[/tex]

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.
 
  • #3
spitz said:

Homework Statement



Find the MGF and all the moments for [itex]X\sim N(0,1)[/itex]

2. The attempt at a solution

For the MGF, I have:

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

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

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

Your MGF formula is seriously wrong: you need [itex] e^{-x^2/2}[/itex] in the integrand, not your [itex] e^{x^2/2}.[/itex] However, your result
[tex]M_X(s) = e^{s^2/2} [/tex] 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
 

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

1.

What is the moment generating function (MGF) of a normal distribution with mean 0 and standard deviation 1?

The MGF of a normal distribution with mean 0 and standard deviation 1 is given by the formula e^(t^2/2).

2.

How do you find the moments of a normal distribution with mean 0 and standard deviation 1?

The moments of a normal distribution with mean 0 and standard deviation 1 can be found by taking the derivatives of the MGF at t=0. The nth moment is equal to the nth derivative of the MGF at t=0.

3.

Why is the MGF useful in finding moments of a normal distribution?

The MGF allows us to find moments of a normal distribution by using a simple formula, instead of having to integrate over the entire distribution. This makes it a more efficient and convenient method for finding moments.

4.

What are the properties of the MGF of a normal distribution with mean 0 and standard deviation 1?

The MGF of a normal distribution with mean 0 and standard deviation 1 has the following properties:

  • It is always defined for all real values of t.
  • It is a continuous function.
  • It is non-negative for all values of t.
  • It is convex, meaning the graph of the MGF is always above its tangent lines.
  • It uniquely determines the distribution, meaning no two distributions can have the same MGF.

5.

Can the MGF of a normal distribution with mean 0 and standard deviation 1 be used to find moments of other distributions?

Yes, the MGF of a normal distribution with mean 0 and standard deviation 1 can be used to find moments of other distributions. This is because the MGF of a distribution is a unique function that characterizes that distribution, and can be used to find moments regardless of the specific distribution.

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