# Thread: help for continous joint distribution

1. ## 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

2. ## Re: help for continous joint distribution

Originally Posted by pupupanda
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.

3. ## Re: help for continous joint distribution

Hello, do you have the link to this? Thanks a lot

4. ## 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?

5. ## Re: help for continous joint distribution

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.

6. ## Re: help for continous joint distribution

Thank you very much,
but I can only understand (b), and I am lost

7. ## 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}\$