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Homework Help: Two problems with random variables transformations

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data

    [1] A random variable X is distributed as fX(x) = 1/9*(1+x)^2 1{-1<= x<= 2}.

    a) Find the density function of Y = -X^2 + X + 2.
    b) Find the cummulative distribution function of Y = X1{-1<=X<=1} + 1{X>=1}

    [2] Find the function that transforms a variable X with fX(x) = e^(-x) 1{x>0} into a variable Y with fY(y) = 31{0<=y<=1/4} + 1/31{1/4<y<1}.

    3. The attempt at a solution

    The problem with exercise 1 is that the functions aren't injective in the domain of X. My first guess with a) was to use equivalent events:

    FY(y) = P(Y<=y) = P(-X^2 + X + 2 <= y)
    FY(y) = P(-X^2 + X + 2 - y<=0) = P([tex]X \le \frac{1}{2} - \frac{{\sqrt {9 - 4y} }}{2}[/tex]) + P([tex]X \ge \frac{1}{2} + \frac{{\sqrt {9 - 4y} }}{2}[/tex])

    Then, taking into account that X is absolutely continuous:

    FY(y) = P([tex]X \le \frac{1}{2} - \frac{{\sqrt {9 - 4y} }}{2}[/tex]) + (1 - P([tex]X \le \frac{1}{2} + \frac{{\sqrt {9 - 4y} }}{2}[/tex]))

    FY(y) = FX([tex]\frac{1}{2} - \frac{{\sqrt {9 - 4y} }}{2}[/tex]) + (1 - FX([tex]\frac{1}{2} + \frac{{\sqrt {9 - 4y} }}{2}[/tex])).

    Then I could derive term to term and find fY(y). Is this OK? I'm mainly worried about the unicity of all this, due to g(x) not being injective.

    For point b), on the other hand, I don't know how to interprete it. Should I consider this a problem of conditional probabilities? I mean, FY(y) = FX|-1<=X<=1(x) + ????


    For exercise 2, I need a hint. Best thing I could think was that, since fY(y) = fX(g-1(y))/(g'(g-1(y))), that in both cases g(x) must some sort of logarithmic function.

    Any recomendations? Thanks.
     
  2. jcsd
  3. Apr 15, 2010 #2

    lanedance

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    first thought, have you tired completing the square, may help make it an easier transformation, though will have a look thourgh the rest
     
  4. Apr 17, 2010 #3
    Anybody else?
     
  5. Apr 18, 2010 #4

    lanedance

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    notation is alittle difficult to read what are the bold 1s?
    This looks reasonable to me as you are splitting the function into 1:1 parts,

    Another equaivalent way, not necessarily easier but maybe clearer, is as follows

    First note range of Y is [0,1] and the function is 1:1 on the intervals [-1,1/2) & (1/2,2]

    so treat the contribution from each section seperately, first X<1/2, solve for X
    [tex] Y = -(X-\frac{1}{2})^2 + \frac{9}{4} [/tex]
    [tex] X = \sqrt{\frac{9}{4} -Y}+\frac{1}{2})[/tex]
    then find the derivative
    [tex] |\frac{dx}{dy}| = ...[/tex]

    the contribution to the density function of Y, from the region X<1/2
    [tex]f_{Y, (X<1/2)}(y) = f_X(x(y)) |\frac{dx}{dy}| [/tex]

    do the same for X>1/2 and sum the contributions to get the total distribution for Y
    [tex]f_{Y}(y) = f_{Y, (X<1/2)}(y)+f_{Y, (X>1/2)}(y) [/tex]
     
    Last edited: Apr 18, 2010
  6. Apr 18, 2010 #5

    lanedance

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    For b) i would read it as Y = X for -1<X<1 and one otherwise. What is the probabilty of X>1?
    this the probabilty Y should have for being 1 (check its not one anywhere else on the X domain).

    In effect the distribution for Y will be a combination of continuous & discrete - do you know about dirac delta functions?

    that said its pretty hard to read similar for 2, can you explain the notation?
     
  7. Apr 18, 2010 #6
    It's [tex]\int\limits_1^2 {{f_X}(x)dx}[/tex]

    It can't, because Y is X for -1<=X<1.

    I don't think it's necessary, since they ask me for the cummulative distribution function, not the density function.

    OK, here it goes again:

    [2] Find the function that transforms a variable X with [tex]{f_X}(x) = {e^{ - x}}*1\left\{ {x > 0} \right\}[/tex] into a variable Y with [tex]{f_Y}(y) = 3*1\left\{ {0 < y < 1/4} \right\} + 1/3*1\left\{ {1/4 < y < 1} \right\}[/tex], where 1{X [tex]\in[/tex]A} is the indicator function.

    Thanks.
     
  8. Apr 18, 2010 #7

    lanedance

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    ok so for 1b) as it cumulative distribrution function can you write it in terms of a step function?
     
  9. Apr 18, 2010 #8

    lanedance

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    for 2) i haven't tried yet but would start from
    [tex] f_Y(y) = f_X(x)|\frac{dx}{dy}|[/tex]

    also examine the cumulative probability at the change in [itex] f_Y(x) [/itex] at the point y = 1/4.

    as another idea you could look at how to make [itex] f_Z(z) = 1 [/itex], on [0,1] as a simpler first pass
     
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