Results 1 to 7 of 7

Thread: help for continous joint distribution

  1. #1
    Newbie pupupanda's Avatar
    Joined
    Mar 2017
    From
    new zealand
    Posts
    19

    help for continous joint distribution

    Thank you so much for helping me out with this problem
    if I have a joint distribution
    fXY (x,y) =24(xy) x > 0; y > 0; x + y < 1
    0 otherwise
    how can I calculate the FX,Y (x, y)?
    ∫∫24xydydx
    I can't figure out the limit for this calculation.
    and how can I plot this in R
    Thank you
    Last edited by pupupanda; Mar 24th 2017 at 06:44 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,366
    Thanks
    2684
    Awards
    1

    Re: help for continous joint distribution

    Quote Originally Posted by pupupanda View Post
    Thank you so much for helping me out with this problem
    if I have a joint distribution
    fXY (x,y) =24(xy) x > 0; y > 0; x + y < 1
    0 otherwise
    how can I calculate the FX,Y (x, y)?
    ∫∫24xydydx I can't figure out the limit for this calculation.
    and how can I plot this in R
    This very problem is completely worked out on pp202-3 of Probability & Statistics by Jay L Devore, 4TH Ed.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie pupupanda's Avatar
    Joined
    Mar 2017
    From
    new zealand
    Posts
    19

    Re: help for continous joint distribution

    Hello, do you have the link to this? Thanks a lot
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    6,577
    Thanks
    1711

    Re: help for continous joint distribution

    Hey pupupanda.

    Can you show us your attempt at finding the integral and keeping the x and y variables as variables?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    5,774
    Thanks
    2417

    Re: help for continous joint distribution

    One thing to notice about this problem before getting into it.

    It is entirely symmetric in $x$ and $y$

    Thus $F_{XY}(x,y) = F_{XY}(y,x)$

    and we only have to consider cases where $x \geq y$

    there are 5 cases

    a) $x< 0 \vee y < 0$

    b) $0<x<1 \wedge 0<y<1 \wedge x+y \leq 1$

    c) $0<x<1 \wedge 0<y<1 \wedge x+y > 1$

    d) $1<x \wedge 0 < y < 1$

    e) $1 < x \wedge 1 < y$

    (a) and (e) are trivial, (a) has probability 0, and (e) has probability 1

    (b) is also simple $F_{XY}(x,y) = \displaystyle \int_0^x \int_0^y ~f(u,v)~dv~du$

    (c) is a bit trickier but one can see that

    $F_{XY}(x,y) = \displaystyle \int _0^{1-y}\int _0^yf(u,v)~dv~du + \int _{1-y}^x\int _0^{1-u}f(u,v)~dv~du$

    and finally for (d) the integral in (c) has to be slightly modified to


    $F_{XY}(x,y) = \displaystyle \int _0^{1-y}\int _0^yf(u,v)~dv~du + \int _{1-y}^1\int _0^{1-u}f(u,v)~dv~du$

    so gathering all this together we get

    $F_{XY}(x,y) = \begin{cases}

    0 &x< 0 \vee y < 0 \\

    \displaystyle \int_0^x \int_0^y ~f(u,v)~dv~du &0<x<1 \wedge 0<y<1 \wedge x \geq y \wedge x+y \leq 1 \\

    \displaystyle \int _0^{1-y}\int _0^yf(u,v)~dv~du + \int _{1-y}^x\int _0^{1-u}f(u,v)~dv~du &0<x<1 \wedge 0<y<1 \wedge x\geq y\wedge x+y > 1 \\

    \displaystyle \int _0^{1-y}\int _0^yf(u,v)~dv~du + \int _{1-y}^1\int _0^{1-u}f(u,v)~dv~du &1<x \wedge 0 < y < 1 \\

    1 &x>1 \wedge y > 1 \\

    F_{XY}(y,x) &y > x

    \end{cases}$

    I leave all the integrations to you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie pupupanda's Avatar
    Joined
    Mar 2017
    From
    new zealand
    Posts
    19

    Re: help for continous joint distribution

    Thank you very much,
    but I can only understand (b), and I am lost
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,366
    Thanks
    2684
    Awards
    1

    Re: help for continous joint distribution

    $ \begin{align*}\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {f(x,y)dydx} } &= \int_0^1 {\left\{ {\int_0^{1 - x} {24xydy} } \right\}}dx\\ &= \int_0^1 {12x\left\{ {{y^2}} \right\}dx} \\&= \int_0^1 {12x{{\left( {1 - x} \right)}^2}dx = 1}\end{align*}$

    Let $A=\{(x,y) : 0\le x\le 1,~0\le y\le 1~ x+y<0.5\}.
    $ \begin{align*}P(X,Y) \in A &= \iint\limits_A {f(x,y)} \\&=\int_0^{0.5} {\int_0^{0.5 - x} {(24)~xy} } dydx\\&=0.0625 \end{align*}$

    Marginal pfd
    ${f_X}(x) = \int_{ - \infty }^\infty {f\left( {x,y} \right)} dy =\begin{cases}\int_0^{1 - x} {((24)~xy} dy = 12x{(1 - x)^2} &: 0\le x\le 0 \\0 & \text{elsewise}\end{cases}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Mar 23rd 2013, 04:35 PM
  2. Replies: 1
    Last Post: Mar 21st 2013, 02:59 AM
  3. Joint distribution, conditional distribution
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Dec 3rd 2011, 08:44 AM
  4. Percentiles in a Continous Random Variable distribution
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: Aug 5th 2009, 10:23 PM
  5. Mean and Variance of Continous distribution
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: May 10th 2008, 02:57 PM

Search Tags


/mathhelpforum @mathhelpforum