# Uniform Random Variables

1. May 2, 2014

### frzncactus

1. The problem statement, all variables and given/known data

If X is a random variable uniformly distributed over (0,1), and a, b are constants, what can you say about the random variable aX + b? What about X^2?

2. Relevant equations

For uniformity of notation, let

f(x) = probability density function of x
F(a) = distribution function of x
g(x) = density function of Y = aX + b
G(a) = distr function of Y
h(x) = density function of Z = X^2
H(a) = distr function of Z

3. The attempt at a solution

I came across this problem as I was studying for finals, and wasn't quite sure about the answer. I've been stumped for the past two days, and created an account here out of hopelessness just so I could ask this question (so please help? it's not even for a grade, I promise!). I think the answer is that aX + b is uniform, and that X^2 is not. Here's my attempt at a derivation:

G(Y) = P{Y < y) = P{aX+b < y) = P{X < (y-b)/a} = F((y-b)/a)
Now I differentiate with respect to y:
g(y) = d/dy (G(y)) = d/dy (F((y-b)/a)) = 1/a * f((y-b)/a) by chain rule

Since the integral of the probability density function of anything is 1, I integrated g(y) from x=0 to x=1, i.e. y = -b/a to (1-b)/a after modifying the original (0,1) bounds. At this point, I got confused because I wasn't entirely sure if the bounds are supposed to be changed that way, and I'm not sure whether I should be integrating over x or y. Heck, I have no idea if I'm overthinking the problem in the first place.

Putting the result from that thread into my notation, I think that the distribution and density functions of X^2 would be:
H(x) = 1 - F(-sqrt x) + F(sqrt x) = 1 + F(sqrt x)
h(x) = [1/(2 sqrt x)] * (h(sqrt x))

I feel like there's something very simple that I don't understand yet. Please enlighten me!

2. May 2, 2014

### Ray Vickson

You haven't really missed anything. Your treatment may be longer than necessary, but it is OK.

3. May 3, 2014

### frzncactus

Thank you for the response! Was my conclusion that ax+b is uniform correct? I am uncertain about how it can fit the form for a uniform distribution.

4. May 3, 2014

### Ray Vickson

Of course Y = aX+b is uniformly distributed, and your derivation was correct: on a range of y, the cdf $P(Y \leq y)$ is linear in $y$, so the derivative is constant over the range.

5. May 3, 2014

### frzncactus

OHHH. To confirm my understanding, a cdf with a constant derivative (with respect to a uniform random variable) would therefore have a constant density, and thus would be uniform as well. Was that a correct statement?

6. May 3, 2014

### Ray Vickson

I think you know the answer to that! Just use the definitions, etc.