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Math Help - Prove that Y1 and Y2 are independent

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    Prove that Y1 and Y2 are independent

    Let X1 and X2 be a random sample of size 2 from a distribution with probability density function f(x) = exp(-x) , 0 <x < infinity. and f(x)= 0 elsewhere.

    Now, let Y1= X1+ X2.
    Y2 = X1/(X1+X2).

    Prove that Y2 and Y2 are independent.
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  2. #2
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    Quote Originally Posted by cryptic26 View Post
    Let X1 and X2 be a random sample of size 2 from a distribution with probability density function f(x) = exp(-x) , 0 <x < infinity. and f(x)= 0 elsewhere.

    Now, let Y1= X1+ X2.
    Y2 = X1/(X1+X2).

    Prove that Y2 and Y2 are independent.
    The joint pdf of X_1 and X_2 is f(x_1, x_2) = e^{-x_1} e^{-x_2}.

    X_1 = Y_1 Y_2 and X_2 = Y_1 (1 - Y_2). Therefore |J| = Y_1.

    Therefore the joint pdf of Y_1 and Y_2 is g(y_1, y_2) = f(x_1(y_1, y_2), x_2(y_1, y_2)) \, J = \, ....

    Now show that g(y_1, y_2) can be written as a product of the form h_1(y_1) \cdot h_2(y_2).


    Aside: If Y_1 and Y_2 are independent then Cov(Y_1, Y_2) = 0. The converse is NOT true.

    Cov(Y_1, Y_2) = E(Y_1 Y_2) - E(Y_1) E(Y_2) = E(X_1) - [E(x_1) + E(x_2)] E\left(\frac{X_1}{X_1 + X_2} \right)  = 1 - (1 + 1) E\left(\frac{X_1}{X_1 + X_2} \right) = 1 - (2) \left(\frac{1}{2} \right) = 0, as expected.

    Note:  I = \int_0^{+\infty} \int_0^{+\infty} \frac{x_1}{x_1 + x_2} e^{-x_1} e^{-x_2} \, dx_1 \, dx_2 = \int_0^{+\infty} \int_0^{+\infty} e^{-x_1} e^{-x_2} \, dx_1 \, dx_2 - I.

    Therefore E\left(\frac{X_1}{X_1 + X_2} \right) = I = \frac{1}{2}.
    Last edited by mr fantastic; February 9th 2009 at 02:54 AM.
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    Quote Originally Posted by cryptic26 View Post
    Let X1 and X2 be a random sample of size 2 from a distribution with probability density function f(x) = exp(-x) , 0 <x < infinity. and f(x)= 0 elsewhere.

    Now, let Y1= X1+ X2.
    Y2 = X1/(X1+X2).

    Prove that Y2 and Y2 are independent.
    Anyone wishing to contribute to this thread can pm me. In light of another thread being completely vandalised by edit-deletes, I'm closing this thread.
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