# Homework Help: Moment generating function of a continuous variable

1. Mar 23, 2010

### exitwound

1. The problem statement, all variables and given/known data

Find the moment generating of:

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

2. Relevant equations

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

3. 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:

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

2. Mar 23, 2010

### vela

Staff Emeritus
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.

3. Mar 23, 2010

### exitwound

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

4. Mar 23, 2010

### vela

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

5. Mar 23, 2010

### exitwound

-inf?

6. Mar 23, 2010

### vela

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

7. Mar 23, 2010

### exitwound

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

Oh. typo above.

8. Mar 23, 2010

### vela

Staff Emeritus
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?

9. Mar 23, 2010

### exitwound

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

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

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

10. Mar 23, 2010

### vela

Staff Emeritus
Yup, you got it!

11. Mar 23, 2010

### 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?

12. Mar 23, 2010

### vela

Staff Emeritus
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.

13. Mar 23, 2010

### vela

Staff Emeritus
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.

14. Mar 23, 2010

### 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 $\sigma ^2$

15. Mar 23, 2010

### vela

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

16. Mar 23, 2010

### 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.