# Moment generating function of a continuous variable

## 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:

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

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vela
Staff Emeritus
Homework Helper
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.

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

vela
Staff Emeritus
Homework Helper
What's the limit of e-kx, k>0, as x goes to infinity?

-inf?

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

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

Oh. typo above.

vela
Staff Emeritus
Homework Helper
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?

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

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

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

vela
Staff Emeritus
Homework Helper
Yup, you got it!

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
Staff Emeritus
Homework Helper
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
Staff Emeritus
Homework Helper
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.

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
Staff Emeritus