Moment generating function of a continuous variable

[exitwound]

Homework Statement

Find the moment generating of:

f(x)=.15e^{-.15x}

Homework Equations

M_x(t)= \int_{-\infty}^{\infty}{e^{tx}f(x)dx}

The Attempt at a Solution

I get down to the point (if I've done my calculus correctly) and gotten:

\frac{.15e^{(t-.15)x}}{t-.15} \Bigr| _{0}^{\infty}

and I don't know how to evaulate this. Am I right this far? And if I am, how does this equate to the answer in the back of the book:

\frac{.15}{.15-t}

 

[vela]

Remember that x is your variable of integration, not t. Assume t<0.15, otherwise the integral won't converge, and just plug in the limits like you would normally do.

 

[exitwound]

That's what I'm having trouble with. How do I evaluate it at infinity?

 

[vela]

What's the limit of e^-kx, k>0, as x goes to infinity?

 

[exitwound]

-inf?

 

[vela]

Instead of guessing, try plotting the function or plugging numbers into your calculator to see if you can see what the limit is.

 

[exitwound]

I'm not guessing. I plotted it out, and as x goes to -inf, e goes to inf.

Oh. typo above.

 

[vela]

I think you're screwing up the signs. You have a function of the form e^-kx where k>0 (this is why you need t<0.15). What does it do as x goes to positive infinity, which is the upper limit of the integral?

 

[exitwound]

Oh. I see. My head wouldn't release the reading error.

\frac{-.15e^{(t-.15)\infty}}{t-.15} - \frac{-.15e^{(t-.15)0}}{t-.15}

0-\frac{.15}{t-.15} = \frac{-.15}{t-.15}=\frac{.15}{.15-t}

Do I have it yet? I'm way too tired to think clearly.

 

[vela]

Yup, you got it!

 

[exitwound]

Okay. I have to take the second derivative of that to find the Variance of the pdf.

First derivative = \frac{.15}{(.15-t)^2}

Second derivative = \frac{.3}{(.15-t)^3}

I've even confirmed the second derviative with Wolfram Alpha. However, when I plug 0 into it (to find the second moment, or the variance), I get 88.88. But the book says 44.44. Any idea what I'm doing wrong?

 

[vela]

I think the book is wrong. I multiplied f(x) by x² and integrated to calculate E(x²) and got the same answer you did.

 

[vela]

Oh, wait. Were you looking for E(x²) or the variance? If it's the latter, you need to subtract off the square of the mean.

 

[exitwound]

Well, V(X) = E(x²)-[E(X)]² but it also says that if I take the 2nd derivative of the moment generating function, I get V(X) or \sigma ^2

 

[vela]

That's not correct. The moment generating function only allows you to calculate E(xⁿ) by taking the n-th derivative.

 

[exitwound]

So if E(x²) is the second derivative, I have to subtract off [E(X)]^2, or 44.44, which makes it 44.44 just like the book says.

I think I understand that now. Thanks.