if the probability density of $x$ is given by

Question

If the pdf of $X$ is given by

$$f(x)=\begin{cases} 1+x & \text{for } -1< x \leq 0 \\ 1-x & \text{for } 0<x<1 \\ 0 & \text{else} \end{cases}$$

and $U=X$ and $V=X^2$

then show $U$ and $V$ are dependent.

How would I show they are dependent I know if two random variable $X$ and $Y$ are dependent.

Then

$$E(XY)=E(X)E(Y)$$

I know the covariance of $COV(U,V)=0-(0)(1/12/)$ So it is zero.

score 4 · Accepted Answer · answered Oct 17 '14 at 21:57

4

Hint: compute $$ P(|U| < 1/2) \\ P(V < 1/4) \\ P(|U| < 1/2, V < 1/4) $$

answered Oct 17 '14 at 21:57

mookid

I see I do these integral then I show that $E(UV)$ does not equal $E(U)$ $E(V)$ – Fernando Martinez Oct 18 '14 at 15:15
2

you show that $E(f(U)g(V)) \neq E(f(U))E(g(V))$, with $f(U) = 1_{|U| < 1/2}$ and $g(V) = 1_{V < 1/4}$. – mookid Oct 18 '14 at 15:17

1 Answers1