# 3rd continous random variable question

• Oct 1st 2008, 07:23 PM
ban26ana
3rd continous random variable question
Consider the continuous random variable X with probability density function as:

Find
a) the value of C
b) the moment generating function of X
c) the second moment about origin of X

I could do parts b and c, but I don't know how to find C unless it is in the f(x) function.
• Oct 1st 2008, 07:27 PM
icemanfan
Think of question A this way: what is the value of C that solves the equation $\displaystyle \int _0 ^C 2(1-x) \, dx = 1$?
• Oct 1st 2008, 07:31 PM
ban26ana
Nevermind, I screwed up a negative. I got C=1. Thank you for your explanation!
• Oct 1st 2008, 07:35 PM
icemanfan
We have:

$\displaystyle \int _0 ^C 2(1-x) \, dx = 1$

$\displaystyle \int _0 ^C 2 - 2x \, dx = 1$

$\displaystyle 2x - x^2 | _0 ^C = 1$

$\displaystyle (2C - C^2) - (2(0) - 0^2) = 1$

$\displaystyle 2C - C^2 = 1$

$\displaystyle C^2 - 2C + 1 = 0$

$\displaystyle (C - 1)(C - 1) = 0$

$\displaystyle C = 1$

Good job on getting that before I did.
• Oct 1st 2008, 07:36 PM
ban26ana
Thanks. I had screwed up a negative in my original calculations. I got it. You made it very easy.
• Oct 1st 2008, 07:51 PM
ban26ana
Okay, so for the moment generating function, I found the integral to be

Since the denominator is t squared, and we're evaluating from 0 to 1, does this mean that the mgf doesn't exist?
• Oct 1st 2008, 11:03 PM
mr fantastic
Quote:

Originally Posted by ban26ana
Okay, so for the moment generating function, I found the integral to be
$\displaystyle m(t) = E\left( e^{tX} \right) = \int_{-\infty}^{+\infty} e^{xt} f(x) \, dx = 2 \int_{0}^{1} e^{xt} (1 - x) \, dx = \frac{2(e^t - t - 1)}{t^2}$.
Confirm that $\displaystyle E(X) = \frac{dm}{dt}$ evaluated at t = 0.