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Given marginal pdfs of X and Y, find pdf of Z=X-Y

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    The probability density function of the random variables X and Y are given by:

    [tex] f_1(x)= \begin{cases} 2 & -\frac{1}{4}\le x\le \frac{1}{4} \\ 0 & \text{elsewhere} \end{cases} [/tex]
    and
    [tex] f_2(y) \begin{cases} \frac{1}{2} & 0\le y \le 2 \\ 0 & \text{elsewhere} \end{cases} [/tex]

    respectively.

    a) Find the probability density function of the random variable Z=X-Y .
    b) What is the probability that Z will assume a value greater than zero?


    2. Relevant equations

    Not sure yet.

    3. The attempt at a solution

    There isn't an example like this in my book. I'm not sure how to go from marginals to the new variable thing, which I couldn't solve in an ordinary manner anyway! Sad sad sad. Am I supposed to make the marginals into a regular f(x,y), or is there some direct way to get to the Z?
     
  2. jcsd
  3. Nov 27, 2011 #2

    D H

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    I assume your book tells you how to compute the distribution of a sum of random variables such as W=X+Y.

    One way to look at this is to invent a new random variable U=-Y. (Use Z=X-Y=X+(-Y)=X+U.) What does the distribution of this variable U look like? of X+U?
     
  4. Nov 27, 2011 #3

    Ray Vickson

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    Unless you are given more information you cannot do the question:you need to know something about the joint distribution of the pair (X,Y). In particular, are X and Y independent? If they *are* independent, just let Y1 = -Y and look at X+Y1. The distribution of Y1 is easy to get, and surely the distribution of X+Y1 must be obtainable from material in your textbook or notes.

    RGV
     
  5. Nov 27, 2011 #4
    What I typed is all I have.
     
  6. Nov 27, 2011 #5

    D H

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    So assume they are independent. As both Ray and I noted, your text or notes must have something to say about the sum of two independent random variables.
     
  7. Nov 27, 2011 #6
    Hmm, it looks like if they are independent then [itex] f(x,y)=f_1(x)f_2(y) [/itex]
    From there, it's like any other random variable problem. Thanks for the suggestion. :)
     
  8. Nov 27, 2011 #7

    I like Serena

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    Hi Arcana! :smile:

    For adding or subtracting independent distributions, we have the convolution rule for distributions.

    Suppose X and Y are independent probability distributions with probability density functions fX(x) and fY(y), and cumulative probability function FX(x) and FY(y).

    If U=X+Y, then
    [tex]P(U \le u)
    = P(X + Y \le u)
    = \int_{-\infty}^{\infty} f_X(x) P(x+Y \le u) \textrm{ d}x
    = \int_{-\infty}^{\infty} f_X(x) P(Y \le u - x) \textrm{ d}x
    [/tex]
    so
    [tex]P(U \le u)
    = \int_{-\infty}^{\infty} f_X(x) F_Y(u-x) \textrm{ d}x
    [/tex]

    And if you want to know the probability density of U, we have:
    [tex]f_U(u)= {d \over du}F_U(u) = {d \over du}P(U \le u)[/tex]
     
  9. Nov 27, 2011 #8
    great, thanks!
     
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