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Uniform Random Variables

  1. May 2, 2014 #1
    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.

    Next, I attempted the same thing for X^2 by trying to follow a similar thread: https://www.physicsforums.com/showthread.php?t=398718

    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. jcsd
  3. May 2, 2014 #2

    Ray Vickson

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    You haven't really missed anything. Your treatment may be longer than necessary, but it is OK.
     
  4. May 3, 2014 #3
    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.
     
  5. May 3, 2014 #4

    Ray Vickson

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    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.
     
  6. May 3, 2014 #5
    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?
     
  7. May 3, 2014 #6

    Ray Vickson

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