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Math Help - Random variables

  1. #1
    Member courteous's Avatar
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    Question Random variables

    \[ p_{X,Y}(x,y) = \left\{ \begin{array}{ll}<br />
e^{-x} & \mbox{; $x>y>0$};\\<br />
0 & \mbox{; $otherwise$}.\end{array} \right. \]

    Find p_X(x) and p_Y(y). Are they independent?

    Find also P(2<Y<3), P(Y<X) and P(2Y<X).
    I'm only having trouble with (formalizing) last two requirements: P(Y<X) and P(2Y<X).

    p_X(x)=...=xe^{-x}\text{ ; }x > 0
    p_Y(y)=...=e^{-y}\text{ ; }y > 0
    P(2<Y<3)=e^{-2}-e^{-3}

    Now the last two:
    P(Y<X) is obviously 1, but I don't know how to setup the corresponding definite integral(s).

    Similarly for P(2Y<X) which is reasonably enough \frac{1}{2} (I've peeked at the solutions, but still) ... as would P(nY<X) = \frac{1}{n}, wouldn't it?
    Last edited by courteous; August 28th 2010 at 03:02 AM. Reason: mr fantastic already answered, so I undid the edit
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  2. #2
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    Quote Originally Posted by courteous View Post
    I'm only having trouble with (formalizing) last two requirements: P(Y<X) and P(2Y<X).

    p_X(x)=...=xe^{-x}\text{ ; }x > 0
    p_Y(y)=...=e^{-y}\text{ ; }y > 0
    P(2<Y<3)=e^{-2}-e^{-3}

    Now the last two:
    P(Y<X) is obviously 1, but I don't know how to setup the corresponding definite integral(s).

    Similarly for P(2Y<X) which is reasonably enough \frac{1}{2} (I've peeked at the solutions, but still) ... as would P(nY<X) = \frac{1}{n}, wouldn't it?
    You should have been taught to draw the required region of integration and then set up the required double integral:

    \displaystyle \Pr(2Y < X) = \Pr(Y < X/2) = \int_{x = 0}^{+\infty} \int_{y = 0}^{y = x/2} f(x, y) \, dy \, dx.

    Yes.
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  3. #3
    Member courteous's Avatar
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    Question

    Thank you ... \frac{1}{2} indeed.

    Further related question:
    Let Z=X+Y.
    What is distribution of Z?
    I'm told:
    "If Y=g(X,Z) holds and g(x,z) is a "well behaved" function, monotonous for each x in z (in z?!),
    then p_Z(z)=\int_{-\infty}^{\infty}p_{X,Y}\big(x,g(x,z)\big)\left|\fr  ac{\partial g}{\partial z}(x,z)\right|dx."

    How do I find what function g is and then use the given formula? The solution given is e^{-z/2}-e^{-z} \text{ ; }z>0.
    Last edited by courteous; August 28th 2010 at 04:00 AM. Reason: well behaved
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    Quote Originally Posted by courteous View Post
    Thank you ... \frac{1}{2} indeed.

    Further related question:

    I'm told:
    "If Y=g(X,Z) holds and g(x,z) is a "well behaved" function, monotonous for each x in z (in z?!),
    then p_Z(z)=\int_{-\infty}^{\infty}p_{X,Y}\big(x,g(x,z)\big)\left|\fr  ac{\partial g}{\partial z}(x,z)\right|dx."

    How do I find what function g is and then use the given formula? The solution given is e^{-z/2}-e^{-z} \text{ ; }z>0.
    I prefer not to use Z. I'll use U instead.

    Use the change of variable theorem:

    Make the transformation U = X + Y, V = X. The inverse transformation is X = V, Y = U - V. Then |J| = 1.

    Get the joint pdf of U and V: g(u, v) = e^{-v}. Integrate wrt to v to get the marginal, which is the required pdf. Note that 0 < y < x => 0 < u - v < v => v < u < 2v and so the region of integration is between the lines v = u/2 and v = u where v > 0:

    g(u) = \int_{v = u/2}^{v = u} e^{-v} \, dv etc.
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  5. #5
    Member courteous's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Then |J| = 1.
    I've lost you here ... what is J, a Jacobian matrix?
    Quote Originally Posted by mr fantastic View Post
    Get the joint pdf of U and V: g(u, v) = e^{-v}.
    Would you show how did you get joint pdf (I guess it stems from J)?
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  6. #6
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    Quote Originally Posted by courteous View Post
    I've lost you here ... what is J, a Jacobian matrix?

    Would you show how did you get joint pdf (I guess it stems from J)?
    Yes.

    See for example Walpole, Meyers and Meyers: Probability and Statistics for Engineers and Scientists. It is also explained here: Transformations of Variables (scroll down).
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