Probability Question

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Homework Statement



Suppose that X has pdf

f(x) = k[itex]x^{2}[/itex] for -1[itex]\leq x \leq 1[/itex], 0 otherwise

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

Homework Equations





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.
 

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  • #2
Ray Vickson
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Homework Statement



Suppose that X has pdf

f(x) = k[itex]x^{2}[/itex] for -1[itex]\leq x \leq 1[/itex], 0 otherwise

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

Homework Equations





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.

Show your work. Your answers to (b) and (c) are wrong.
 
  • #3
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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.
 
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