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Moment Generating Function

  • Thread starter cse63146
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  • #1
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Homework Statement



Let f(x) = 2x 0<x<1

a) Determing the Moment Generating function M(t) of X
b) Use the MGT to determine all moments about the origin
c) Give the 3rd central moment called the skewness

Homework Equations





The Attempt at a Solution



a) [tex]\int^1_0 e^{tx}2x dx = \frac{2xe^{tx}}{t} - \int^1_0 e^{tx}2 dx
= \frac{2}{t}(xe^{tx} - e^t + 1)[/tex]

b)

[tex]E\left(X^n\right)=M^{(n)}(0)=\left.\frac{\mathrm {d}^n M_(t)}{\mathrm{d}t^n}\right|_{t=0}[/tex]

[tex]E\left(X^n\right)=M^{(n)}(0)=\left.\frac{\mathrm {d}^n \frac{2}{t}(xe^{tx} - e^t + 1)}{\mathrm{d}t^n}\right|_{t=0}[/tex]

Is that what I'm supposed to do for part b)?
 
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Answers and Replies

  • #2
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(a) can't be right because the x should have been integrated out. It's easy to correct what you did wrong.

(b) is the correct method, OR it might be easier to express the answer to (a) as a known power series, depending on what it really turns out to be.
 
  • #3
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Sorry, I'm not sure what you meant by " x should have been integrated out". Does that mean that I did the integral wrong?
 
  • #4
392
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Sorry, I'm not sure what you meant by " x should have been integrated out".
I meant [tex]\int_a^b F(x,t)\,dx[/tex] depends on t only, not x. Your (a) has an x in it so there is an (easy to fix) error.
 
  • #5
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Is the 'x' you're referring to - [tex]\frac{2xe^{tx}}{t}[/tex]. If so, I'm not sure how to get rid of it.

Sorry if I'm being difficult
 
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  • #6
392
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Sorry if I'm being difficult
You're not being difficult.

For integration by parts, you tried

[tex]\int_a^b u\,dv=uv-\int_a^b v\,du[/tex]

but the correct formula

[tex]\int_a^b u\,dv=uv\bigr|_a^b-\int_a^b v\,du[/tex]
 
  • #7
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In [tex]uv|^1_0,[/tex] the [tex]|^1_0[/tex] applies to both u and v, or just v?
 
  • #8
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Both u and v
 
  • #9
452
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Once I use [tex]uv|^1_0[/tex], I get this:

[tex]\frac{2e^t}{t} - \frac{2}{t}(e^t -1) = \frac{2}{t}[/tex].

So when I apply I take the nth derivative I get:

[tex] - \frac{2}{nt^n}[/tex]

but I can't evaluate it at 0 since the denominator = 0. Did I make a mistake somewhere?
 
  • #10
392
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Double check your integration to see if it should be [tex]\frac{2e^t}{t} - \frac{2}{t^2}(e^t -1)=\frac{2te^t-2e^t+2}{t^2}[/tex]

This is made continuous at t=0, which you can verify by l'Hopital or by substituting the Maclaurin series for e^t. In fact, I think it might be easier to use Maclaurin series for e^t to find the moments as well, but I admit I didn't try differentiating.
 
  • #11
452
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Can I leave it like this (cause I don't know how the to find the nth derivative of a quotient)

[tex]\frac{d^n \frac{2te^t-2e^t+2}{t^2}}{dt}[/tex]
 
  • #12
392
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I would prefer to substitute the power series for e^t, simplify, then get the derivatives at 0 from that, to see if there is a pattern to the nth moment.
 

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