# Random Variables and Distribution

• Nov 11th 2009, 05:36 PM
azdang
Random Variables and Distribution
Let X be a random variable with distribution function F that is continuous. Show that Y = F(X) is uniform.

I'm sort of confused by what this is even asking, but I tried to put together some information.

F = $\int_{-\infty}^x f(u)du$

I tried to look in my book for anything and also found something:

P(a < x < b) = $P(a \leq X \leq b) = P(a \leq X < b) = F(b) - F(a)$

I know that for Y~unif(A): $f(y) = \frac{1_A(y)}{|A|}$

I'm not sure if any of this is relevant (or perhaps even right), but I was trying to put anything together because I'm just not sure what to do. Any help would be appreciated. Thank you!
• Nov 11th 2009, 07:59 PM
theodds
X is a random variable, so the associated value of the cdf that you get from that is a random variable also. If you assume that F(X) is monotone increasing over the support (which is extraordinarily common), then this problem is very straightforward: for fixed 0 < y < 1
$
P(Y \le y) = P(F_X (X) \le y) = P(F_X ^ {-1} (F_X(X)) \le F_X ^{-1} (y) ) =$
$P(X \le F ^{-1} (y) ) = F_X (F ^ {-1} (y)) = y$

which is the CDF for a uniform (0, 1). In F(X) is not monotone increasing (i.e. it is flat in places), then this method won't work, since F(X) won't have an inverse. It turns out that the same proof works if you define $F_X ^ {-1} (y) = inf \{x: F_X (x) \ge y\}$, but I wouldn't be surprised if they expected you to ignore this case.
• Nov 11th 2009, 08:09 PM
azdang
Thanks! I actually ended up figuring out basically the same thing, so it's good to know I'm on the right track. Thank you again :)
• Nov 12th 2009, 12:10 PM
Moo
Quote:

but I wouldn't be surprised if they expected you to ignore this case.
That's why they said F is continuous.

And it's not surprising, it's quite common to talk about the pseudo-inverse of F, huh ?