What is the Moment Generating Function for the Distribution f(x) = 2x?

[cse63146]

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) \int^1_0 e^{tx}2x dx = \frac{2xe^{tx}}{t} - \int^1_0 e^{tx}2 dx <br /> = \frac{2}{t}(xe^{tx} - e^t + 1)

b)

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

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}

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

 

[Billy Bob]

(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.

 

[cse63146]

Sorry, I'm not sure what you meant by " x should have been integrated out". Does that mean that I did the integral wrong?

 

[Billy Bob]

Sorry, I'm not sure what you meant by " x should have been integrated out".

I meant \int_a^b F(x,t)\,dx depends on t only, not x. Your (a) has an x in it so there is an (easy to fix) error.

 

[cse63146]

Is the 'x' you're referring to - \frac{2xe^{tx}}{t}. If so, I'm not sure how to get rid of it.

Sorry if I'm being difficult

 

[Billy Bob]

Sorry if I'm being difficult

You're not being difficult.

For integration by parts, you tried

\int_a^b u\,dv=uv-\int_a^b v\,du

but the correct formula

\int_a^b u\,dv=uv\bigr|_a^b-\int_a^b v\,du

 

[cse63146]

In uv|^1_0, the |^1_0 applies to both u and v, or just v?

 

[Billy Bob]

Both u and v

 

[cse63146]

Once I use uv|^1_0, I get this:

\frac{2e^t}{t} - \frac{2}{t}(e^t -1) = \frac{2}{t}.

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

- \frac{2}{nt^n}

but I can't evaluate it at 0 since the denominator = 0. Did I make a mistake somewhere?

 

[Billy Bob]

Double check your integration to see if it should be \frac{2e^t}{t} - \frac{2}{t^2}(e^t -1)=\frac{2te^t-2e^t+2}{t^2}

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.

 

[cse63146]

Can I leave it like this (cause I don't know how the to find the nth derivative of a quotient)

\frac{d^n \frac{2te^t-2e^t+2}{t^2}}{dt}

 

[Billy Bob]

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.