# PDF Problem

• October 4th 2008, 04:05 PM
Xan21
PDF Problem
Heres a problem im stuck on.

"Random variable X has the probability density function f(x) = 1/3 for -1 < x < 2 and zero elsewhere. Find the probability density function of http://www.mathhelpforum.com/math-he...43dcf1e1-1.gif

Would it be
http://www.mathhelpforum.com/math-he...8cdef263-1.gif for 0<y<1 , 0 elsewhere
• October 4th 2008, 10:35 PM
mr fantastic
Quote:

Originally Posted by Xan21
Heres a problem im stuck on.

"Random variable X has the probability density function f(x) = 1/3 for -1 < x < 2 and zero elsewhere. Find the probability density function of http://www.mathhelpforum.com/math-he...43dcf1e1-1.gif

Would it be
http://www.mathhelpforum.com/math-he...8cdef263-1.gif for 0<y<1 , 0 elsewhere

Not quite. According to your pdf, Pr(Y > 1) = 0 ...... Is that correct ....?

If -1 < x < 2 it's clear that 0 < x^2 < 4 => 0 < y < 4. Therefore f(y) = 0 for y > 4 or y < 0.

So you have to find the pdf for Y when $1 \leq y < 4$. For such values of y, the cumulative density function is given by $F(y) = \Pr(Y < y) = \Pr(X^2 < y) = \Pr(-1 < X < \sqrt{y})$.
• October 9th 2008, 08:02 AM
embria
Sorry i don't understand this
Hi,
I have a question why for 0<y<4http://www.mathhelpforum.com/math-he...c4339481-1.gif.

Why the Pr(X^2 < y)= Pr(-1<X<y^1/2) and not = Pr(-(y^1/2)<X<y^1/2) ?

Thanks.
• October 9th 2008, 03:28 PM
mr fantastic
Quote:

Originally Posted by embria
Hi,
I have a question why for 0<y<4http://www.mathhelpforum.com/math-he...c4339481-1.gif.

Why the Pr(X^2 < y)= Pr(-1<X<y^1/2) and not = Pr(-(y^1/2)<X<y^1/2) ?

Thanks.

Yes it is $\Pr( - \sqrt{y} \leq X \leq \sqrt{y})$. But note:

1. $\Pr( - \sqrt{y} \leq X \leq \sqrt{y}) = \Pr( - \sqrt{y} \leq X \leq -1) + \Pr( -1 \leq X \leq \sqrt{y})$.

2. It's given that -1 < x < 2. So the pdf for X is zero if x < -1. So $\Pr( - \sqrt{y} \leq X \leq -1) = 0$ when $1 \leq y < 4$.

3. So $\Pr( -1 \leq X \leq \sqrt{y})$ is the only calculation required when $1 \leq y < 4$.