# Continuous random variable

1. Oct 19, 2015

### toothpaste666

1. The problem statement, all variables and given/known data
Suppose the distance X between a point target and a shot aimed at the point in a coin-operated target game is a continuous random variable with pdf

f(x) = { k(1−x^2), −1≤x≤1
0, otherwise.

(a) Find the value of k.

(b) Find the cdf of X.

(c) Compute P (−.5 < X ≤ .5).

(d) Find the expected distance between a point target and a shot aimed.

3. The attempt at a solution
a) $k\int_{-1}^1(1-x^2)dx$

$= k[\int_{-1}^1dx-\int_{-1}^1x^2dx]$

$= k[x\Big|_{-1}^1-\frac{1}{3}x^3\Big|_{-1}^1]$

= k(2-2/3) = 1

k(4/3) = 1

k = 3/4

b) $\frac{3}{4} \int_{-1}^X(1-x^2)dx$

c) $\frac{3}{4}[x\Big|_{-.5}^{.5}-\frac{1}{3}x^3\Big|_{-.5}^{.5}]$

= (3/4)(1-(1/3)[2(1/3)(1/8)])

= (3/4)(1-1/36)
= .7292

d) $\frac{3}{4}\int_{-1}^1x(1-x^2)dx$

$=\frac{3}{4}\int_{-1}^1(x-x^3)dx$

$= \frac{3}{4}[\int_{-1}^1xdx-\int_{-1}^1x^3dx]$

$= \frac{3}{4}[\frac{1}{2}x^2\Big|_{-1}^1-\frac{1}{4}x^4\Big|_{-1}^1]$

= 0

am I doing this right?

2. Oct 19, 2015

### SteamKing

Staff Emeritus
For b), instead of expressing the cdf as an integral, you should actually carry out the integration and express the cdf as a function of X.

3. Oct 19, 2015

### Ray Vickson

Not for (d), no. You computed $E X$, but what is wanted is $E |X|$. Also, I get a different answer for (c).

4. Oct 19, 2015

### toothpaste666

for part b)
$F(X) = \frac{3}{4}[x\Big|_{-1}^X-\frac{1}{3}x^3\Big|_{-1}^X]$

$F(X) = \frac{3}{4} [(X+1) - \frac{1}{3} (X^3 +1)]$

$F(X) = \frac{3}{4}[X + 1 - \frac{X^3}{3} - \frac{1}{3}]$

$F(X) = \frac{3}{4}[X - \frac{X^3}{3} + \frac{2}{3}]$

$F(X) = \frac{3X}{4} - \frac{X^3}{4} + \frac{1}{2}$

part c) I made a arithmetic error. it comes down to (3/4)(11/12) = .6875

part d)
I am not quite sure how to do this. plugging in |x| whereever there is an x gives me the same answer. should I switch the limits of integration from 0 to 2 instead of -1 to 1?

5. Oct 19, 2015

### Ray Vickson

No. Without doing any calculations you can see why $EX = 0$: it is because $x f(x)$ is an odd function on $[-1,1]$ so integrates to zero automatically. That is NOT the case for $|x| f(x)$, because this not now an odd function on $[-1,1]$. I won't say any more.

6. Oct 19, 2015

### toothpaste666

d) $\frac{3}{4}\int_{-1}^1|x|(1-x^2)dx$

$=\frac{3}{4}\int_{-1}^1(|x|-|x|x^2)dx$

$= \frac{3}{4}[\int_{-1}^1|x|dx-\int_{-1}^1|x|x^2dx]$

$= \frac{3}{4}[-\int_{-1}^0xdx + \int_0^1xdx-\int_{-1}^0x^3dx + \int_0^1x^3dx]$

$= \frac{3}{4}[-\frac{1}{2}x^2\Big|_{-1}^0 +\frac{1}{2}x^2\Big|_0^1-\frac{1}{4}x^4\Big|_{-1}^0 +\frac{1}{4}x^4\Big|_0^1]$

= (3/4) [(1/2) + (1/2) + (1/4) + (1/4)]
= (3/4)(3/2) = 9/8 = 1.125

7. Oct 19, 2015

### Ray Vickson

This cannot possibly be right: $|X| \leq 1$ for all non-zero probability values, so $E|X| \leq 1$ (and, in fact, $E|X| < 1$ strictly).

8. Oct 19, 2015

### toothpaste666

in that case I am lost =[

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted