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Math Help - Functionn of two random variables.

  1. #1
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    Functionn of two random variables.

    Could anyone help me with this question please?

    Given X and Y are continuous random variables with the following joint probability density function:

      <br /> <br />
 f(x,y) =<br />
 \begin{cases}<br />
 {\theta}^2  e^{-{\theta} y} &  0 < x < y \\<br />
 0 & \text{otherwise}<br />
 \end{cases}<br /> <br />

    i) Obtain the joint probability density function of X and Y − X.
    ii) Show that X and Y − X are independent random variables and write down their (marginal)
    probability density functions. Identify these distributions

    For part (i)

    I let U = X and Y = V - X,

    My Jacobian value was 1,
    and then I got  f(u,v) =  {\theta}^2  e^{-{\theta}  , (v-u)} & \text    ,u > 0, (v-u) > u Is this the correct joint p.d.f for X and Y-X?

    Also for the marginal of X, I got  f X(x) =   e^{-{\theta x}  but I'm not quite sure.
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  2. #2
    MHF Contributor matheagle's Avatar
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    don't you want v=y-x?
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  3. #3
    MHF Contributor matheagle's Avatar
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     f_X(x) =   e^{-{\theta x}  is not a valid density
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  4. #4
    MHF Contributor matheagle's Avatar
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    Let U = X and V = Y - X,

    then the Jacobian is 1

    and  f(u,v) =  {\theta}^2  e^{-{\theta}(v+u)} I(u>0)I(v>0)

    So  f_U(u) =   \theta e^{-\theta u} I(u>0)

    and

     f_V(v) =   \theta e^{-\theta v}I(v>0)
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  5. #5
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    Quote Originally Posted by matheagle View Post
    Let U = X and V = Y - X,

    then the Jacobian is 1

    and  f(u,v) =  {\theta}^2  e^{-{\theta}(v+u)} I(u>0)I(v>0)

    So  f_U(u) =   \theta e^{-\theta u} I(u>0)

    and

     f_V(v) =   \theta e^{-\theta v}I(v>0)
    Thanks, I thought I was close to it.
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