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Thread: Multivariate Probability Distribution

  1. #1
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    Multivariate Probability Distribution

    Question: f(x, y) = 1, where x,y are between 0 and 1 OR = 0, elsewhere
    a) What is P(X - Y > 0.5)?
    b) What is P(XY < 0.5)?

    Honestly I have no idea how to go about this question. Can anybody point out how to get it started?
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  2. #2
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    Re: Multivariate Probability Distribution

    Quote Originally Posted by Greenmango View Post
    Question: f(x, y) = 1, where x,y are between 0 and 1 OR = 0, elsewhere
    a) What is P(X - Y > 0.5)?
    b) What is P(XY < 0.5)?

    Honestly I have no idea how to go about this question. Can anybody point out how to get it started?
    First, define $X,Y$. Once you know that, you can start working on the problem.
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    Re: Multivariate Probability Distribution

    What do you mean by "define X and Y"? Find their marginal probability distributions?
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    Re: Multivariate Probability Distribution

    Quote Originally Posted by Greenmango View Post
    What do you mean by "define X and Y"? Find their marginal probability distributions?
    Ah, I did not understand Multivariate Probability Distribution to be Marginal Probability Distribution. In general, $X,Y$ are random variables, not specifically related to $f$. Perhaps your book defines them canonically somehow to relate to $f$. Perhaps $X = \int_{-\infty}^x f(x,y)dy$ and $Y = \int_{-\infty}^y f(x,y)dx$?

    Or is $f$ the probability distribution, letting you know that $0\le x,y \le 1$ with probability 1 and outside of that range with probability 0?
    Last edited by SlipEternal; Aug 17th 2018 at 12:16 PM.
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    Re: Multivariate Probability Distribution

    The second option, f(x,y) is the joint probability distribution for X and Y.
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  6. #6
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    Re: Multivariate Probability Distribution

    $$P(X-Y>0.5) = \int_{-\infty}^\infty \int_{-\infty}^{x-0.5}f(x,y)dydx = \dfrac{1}{8}$$

    See what you can do for part B.
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  7. #7
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    Re: Multivariate Probability Distribution

    I am not really sure how to give a hint, so here is my solution:

    If $x<0.5$ or $y<0.5$, then $XY<0.5$. Also if $x>0.5$ and $y>0.5$, you can still have $xy<0.5$ if $y<\dfrac{1}{2x}$.

    Thus, you have:

    $$\begin{align*}P(XY<0.5) & = P(X<0.5\text{ or }Y<0.5)+P( X>0.5\text{ and }Y>0.5\text{ and } XY < 0.5 ) \\ & = \dfrac{3}{4}+\int_{0.5}^1\int_{0.5}^{1/2x}dydx \\ & = \dfrac{3}{4}+\dfrac{1}{2}\left(\ln 2 - \dfrac{1}{2}\right)\end{align*}$$

    So, you get

    $$P(XY<0.5) = \dfrac{3+\ln\left(\dfrac{4}{e}\right)}{4}$$
    Last edited by SlipEternal; Aug 17th 2018 at 12:59 PM.
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  8. #8
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    Re: Multivariate Probability Distribution

    Quote Originally Posted by SlipEternal View Post
    $$P(X-Y>0.5) = \int_{-\infty}^\infty \int_{-\infty}^{x-0.5}f(x,y)dydx = \dfrac{1}{8}$$


    See what you can do for part B.

    \begin{align*}P(X-Y>0.5) &= \int_{-\infty}^\infty \int_{-\infty}^{x-0.5}1dydx \\
    &= \int_{0}^1 \int_{0}^{x-0.5}1dydx \\
    &= \int_{0}^1 (x - \frac{1}{2}-0)dx \\
    &= (\dfrac{1}{2}x^2 - \dfrac{1}{2}x)|^1_0 \\
    &= \dfrac{1}{2} - \dfrac{1}{2} \\
    &= 0 \end{align*}


    What did I mess up now?
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  9. #9
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    Re: Multivariate Probability Distribution

    Quote Originally Posted by Greenmango View Post
    \begin{align*}P(X-Y>0.5) &= \int_{-\infty}^\infty \int_{-\infty}^{x-0.5}1dydx \\
    &= \int_{0}^1 \int_{0}^{x-0.5}1dydx \\
    &= \int_{0}^1 (x - \frac{1}{2}-0)dx \\
    &= (\dfrac{1}{2}x^2 - \dfrac{1}{2}x)|^1_0 \\
    &= \dfrac{1}{2} - \dfrac{1}{2} \\
    &= 0 \end{align*}


    What did I mess up now?
    The lower bound for x needs to change. If $y<x-0.5$, then $0.5+y<x$, which means $0.5<x<1$.
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    Re: Multivariate Probability Distribution

    Ah okay gotcha, thanks so much!
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