Probability Question

1. Mar 12, 2013

GreenPrint

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

Suppose that X has pdf

f(x) = k$x^{2}$ for -1$\leq x \leq 1$, 0 otherwise

(a) What is k?
(b) What is E$x^{n}$ where n$\geq 0$ is an odd integer?
(c) What is E$x^{n}$ where n$\geq 0$ is an even integer?

2. Relevant equations

3. The attempt at a solution

For (a) I get 1.5
For (b) I get (-3/8)
for (c) I get (9/8)

C is my concern because the expected value for when n is even is greater than one. Is this a problem? My initial thought that it was because x takes on values greater than 1 and have a probability of occurring. I have checked my work and saw nothing wrong. So I just wanted to make sure that my answer to c was impossible before posting my math. It's very simple math and I don't see what I'm doing wrong. Thanks for any help that you can provide.

2. Mar 12, 2013

Ray Vickson

3. Mar 12, 2013

GreenPrint

Using the definition
$$E[x] = ∫^{\infty}_{-\infty} x f_{x}(x)\, dx$$
So in the case of raising to a power
$$E[x^{n}]=∫^{\infty}_{-\infty} x^{n} f_{x}(x)\, dx$$
We are given that
$$f_{x}(x) = \begin{cases} kx^{2} & \text{for }-1 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$$

In part A, I solved for k and got $k=\frac{3}{2}$.

Now for part B
$$E[x^{n}] = \frac{3}{2} \int^{1}_{-1} x^{n} x^{2}\,dx = \frac{3}{2} \int^{1}_{-1} x^{n+2}\,dx = \frac{3}{2} \left.\frac{x^{n+3}}{n+3}\right|^{1}_{-1} = \frac{3}{2} \left.\frac{x^{n} x^{3}}{n+3}\right|^{1}_{-1} = \frac{3}{2}\left(\frac{1^{n}1^{3}}{n+3}-\frac{(-1)^{n}(-1)^{3}}{n+3}\right) = \frac{3}{2}\left(\frac{1^{n}}{n+3}-\frac{(-1)^{n}(-1)^{3}}{n+3}\right).$$ I think the correct answer is then $\frac{3}{3+n}$ when n even, 0 when n odd.

I think the first time I did this problem I plugged in 1 and -1 for both n and x and is the reason why I got the answers I did.

Last edited by a moderator: Mar 13, 2013