# Combinations of random variables

• Apr 13th 2009, 08:08 AM
Sashikala
Combinations of random variables
Can anyone help me on the following Pls?

The random variable X has p.d.f given by
f(x) = x/2 0£ x £ 2
0 otherwise
(a) Find the c.d.f of X, FX(x).
The random variable Y = X2 and takes values over the range
0 £ y £ 4.
(b) Show that P(Y < y) = P(X £ Öy).
(c) Hence show that the c.d.f of Y, FY(y) is given by

FY(y) = 0, y < 0
y/4, 0 £ y £ 4
1, y > 4
• Apr 13th 2009, 06:31 PM
mr fantastic
Quote:

Originally Posted by Sashikala
Can anyone help me on the following Pls?

The random variable X has p.d.f given by
f(x) = x/2 0£ x £ 2
0 otherwise
(a) Find the c.d.f of X, FX(x).
The random variable Y = X2 and takes values over the range
0 £ y £ 4.
(b) Show that P(Y < y) = P(X £ Öy).
(c) Hence show that the c.d.f of Y, FY(y) is given by

FY(y) = 0, y < 0
y/4, 0 £ y £ 4
1, y > 4

Set up and do the required integrations. What have you tried? Where do you get stuck?
• Apr 13th 2009, 07:41 PM
Sashikala
for the part(a) I got
F(x)=0 , x< 0
square of x/4, 0£ x £ 2
1, x>2

Do we have to prove (b) with the
help of probability distribution table?
How to prove it using the p.d.f though
both are equal?
How to get at c.d.f of Y?
• Apr 13th 2009, 08:25 PM
mr fantastic
Quote:

Originally Posted by Sashikala
for the part(a) I got
F(x)=0 , x< 0
square of x/4, 0£ x £ 2
1, x>2

Do we have to prove (b) with the
help of probability distribution table?
How to prove it using the p.d.f though
both are equal?
How to get at c.d.f of Y?

$F_Y(y) = \Pr(Y < y) = \Pr(X^2 < y) = \Pr(-\sqrt{y} < X < \sqrt{y})$ $= \Pr(X < \sqrt{y})$

since $\Pr(X < 0) = 0$

$= F_X(\sqrt{y}) = \frac{y}{4}$.

$f_Y(y) = \frac{dF_Y}{dy}$.

The details are left for you to fill in.
• Apr 13th 2009, 10:01 PM
Sashikala
Combinations of random variables
Yes. I understood.
Thanks very much