# Uniform random variable

• Mar 10th 2010, 12:51 AM
ChaosticMoon
Uniform random variable
Hi I have a question about uniform random variable

If X is uniformly distributerd over (a,b) what random variable, having a linear relation with X, is uniformly distributed over (0,1)?

I knew the answer should be very simple, but I don't know how should I proceed. The fact I do know is that the probability density function of the random variable must be
f(x) = { 1 0<x<1
0 otherwise }

But how should I manipulate so that the new random variable would fall into this interval while having a linear relation with X?

One attempt I had was U=(X-a)/(b-a) so that U is in the right interval, but I think U doesn't even have a linear relation to X. So yeah, I'm completely clueless. ]
Thanks.
• Mar 10th 2010, 10:22 AM
Moo
Quote:

Originally Posted by ChaosticMoon
One attempt I had was U=(X-a)/(b-a) so that U is in the right interval, but I think U doesn't even have a linear relation to X. So yeah, I'm completely clueless. ]
Thanks.

But that's a linear relation to X : $\frac{X-a}{b-a}=\frac{1}{b-a}\cdot X-\frac{a}{b-a}$
which is in the form mx+n

and if you check with the cumulative density function, you'll see that you're not wrong.
• Mar 10th 2010, 10:26 AM
CaptainBlack
Quote:

Originally Posted by ChaosticMoon
Hi I have a question about uniform random variable

If X is uniformly distributerd over (a,b) what random variable, having a linear relation with X, is uniformly distributed over (0,1)?

I knew the answer should be very simple, but I don't know how should I proceed. The fact I do know is that the probability density function of the random variable must be
f(x) = { 1 0<x<1
0 otherwise }

But how should I manipulate so that the new random variable would fall into this interval while having a linear relation with X?

One attempt I had was U=(X-a)/(b-a) so that U is in the right interval, but I think U doesn't even have a linear relation to X. So yeah, I'm completely clueless. ]
Thanks.

Consider:

$U=\frac{X-a}{b-a}$

Why is this what you seek, in particular what does it mean to say something has a linear association with X?

CB