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  1. #1
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    Question Random variables

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

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

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

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

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

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

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

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

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

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

    Yes.
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    Question

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

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

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

    Further related question:

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

    How do I find what function $\displaystyle g$ is and then use the given formula? The solution given is $\displaystyle e^{-z/2}-e^{-z} \text{ ; }z>0$.
    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: $\displaystyle g(u, v) = e^{-v}$. 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:

    $\displaystyle g(u) = \int_{v = u/2}^{v = u} e^{-v} \, dv$ etc.
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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: $\displaystyle g(u, v) = e^{-v}$.
    Would you show how did you get joint pdf (I guess it stems from J)?
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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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