Deriving Joint Distribution Function from Joint Density (check my working pls)

[Legendre]

Homework Statement

Given:

f(x,y) = x + y, for 0<x<1 and 0<y<1
f(x,y) = 0, otherwise

Derive the joint distribution function of X and Y.

Homework Equations

N.A.

The Attempt at a Solution

Using the definition, I obtained part of the joint distribution F(x,y) = (1/2)(xy)(x+y) for 0<x<1, 0<y<1. Leaving out the working for this as it is pretty standard. I am not sure if my next step is correct and so is hoping you guys can help check my logic...

1) F(x,y) has to be defined for all (x,y) in R^2. So I have to consider all possible points. Correct?

2) When 0<x<1 and y > or = 1, F(x,y) = left limit of F(x,y) as y tend to 1 = F(x,1) = (1/2)(x)(x+1)? I am guessing that values of y above 1 do not affect the joint distribution F(x,y) so it takes the same value as its left limit at the "boundary" of possible y values?

3) By symmetry, when x > or = 1 and 0<y<1, F(x,y) = (1/2)(y)(y+1).

4) Finally, for x > or = 1 and y > or = 1, F(x,y) = 1. And F(x,y) = 0 for all other values.

Phew. I am most concern about step 2). The rest were included for completeness. Thanks!

 

[Legendre]

Making it shorter because I think the length is the reason why there are no replies...

f(x,y) = x + y, for 0<x<1 and 0<y<1
f(x,y) = 0, otherwise

f = joint density. we want the joint distribution function, F.

I only require asistance in verifying my logic for the expression for F over { 0<x<1 and y > or = 1} = B. I think that over region B, F(x,y) = left limit of F(x,y) as y tend to 1. Is this correct?

Thanks!

 

[lanedance]

i think its because the question isn't very clear - I'm not exactly sure what you mean by joint distribution function?

the way I read your description, you have 2 dependent random vairbales X & Y, whose joint distribution function is:
$f_{X,Y}(x,y) = x + y$ , for 0<x<1 and 0<y<1
$f_{X,Y}(x,y) = 0$ , otherwise

Do you mean joint cumulative distribution function? defined by:
$F_{X,Y}(x,y) = P(X\leq x, Y\leq y)$

I haven't looked at those much, but if so I would start by settting up an integral... start by looking at the area defined by your constraints...

 

[lanedance]

if this is the case, the bounding bahviour of F as x (y) respectively go to 1 will be the marginal cumulative distributions of Y (X)

 

[Legendre]

i think its because the question isn't very clear - I'm not exactly sure what you mean by joint distribution function?

the way I read your description, you have 2 dependent random vairbales X & Y, whose joint distribution function is:
$f_{X,Y}(x,y) = x + y$ , for 0<x<1 and 0<y<1
$f_{X,Y}(x,y) = 0$ , otherwise

Do you mean joint cumulative distribution function? defined by:
$F_{X,Y}(x,y) = P(X\leq x, Y\leq y)$

I haven't looked at those much, but if so I would start by settting up an integral... start by looking at the area defined by your constraints...

Oops, sorry! I forgot to adjust for notation differences. In my course, they call the joint cumulative distribution function the distribution function and etc. But YES, what you laid out was precisely the question I require assistance on.

I think I am fine with 90% of the working required for this question. Just a small part that I am not sure about:

What is F(x,y) over the area { 0 < x < 1 and y > or = 1 } ?

The double integral of f(x,y) that gives F(x,y) over 0 < x < 1 and 0 < y < 1 is (1/2)(xy)(x+y). The book I am using says for the area I am asking about, all we do is to find F(x,1). So it is saying that (1/2)(x)(x+1) is the answer to my question in bold.

But it doesn't explain why. I am guessing we are taking the limit of F(x,y) as y tend to 1 from the left? (i.e. " left limit of F(x,y) as y tend to 1")

 

[lanedance]

remember F(x,y) = P(X<=x, Y<=y)

so for y>=1 F(x, y) will just reduce to the marginal cumulative distribution for x as P(Y<=1) = 1

if you set up the integral it should be more clear

 

[Legendre]

remember F(x,y) = P(X<=x, Y<=y)

so for y>=1 F(x, y) will just reduce to the marginal cumulative distribution for x as P(Y<=1) = 1

if you set up the integral it should be more clear

Oh...I see. Thanks a lot man, you saved my *** several times on questions related to probability! Setting up the integral...?

f(x,y) is defined for {0<x<1,0<y<1}

by definition,$$F(x,y) = \int_{-\infty}^y \int_{-\infty}^x \! f(u,v) \, du dv$$

$$F(x,y) = \int_{0}^y \int_{0}^x \! f(x,y) \, dx dy$$Right? But the result only works for 0<x<1 and 0<y<1? OR have I specified the limits wrongly?

But...I go the answer I was looking for : for the region {0<x<1,y>or=1}, F(x,y) = cumulative distribution of x. Thanks!EDIT: Corrected terrible mistake with the double integral.

 

[lanedance]

Oh...I see. Thanks a lot man, you saved my *** several times on questions related to probability! Setting up the integral...?

f(x,y) is defined for {0<x<1,0<y<1}

by definition,


$$F(x,y) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \! f(x,y) \, dx dy$$

$$F(x,y) = \int_0^1 \int_0^1 \! f(x,y) \, dx dy$$


Right? But the result only works for 0<x<1 and 0<y<1? OR have I specified the limits wrongly?

But...I go the answer I was looking for : for the region {0<x<1,y>or=1}, F(x,y) = cumulative distribution of x. Thanks!

no worries - but be careful, x and y have a depndency, so its actually the marginal distribution of x see http://en.wikipedia.org/wiki/Marginal_distribution

your integral is not quiet right either, note
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \! f(x,y) dx dy = \int_{0}^{1} \int_{0}^{1} f(x,y) dx dy = P(X\leq 1, Y \leq 1) = 1$$

so have another think about the reigion of itegration to evaluate $F_{X,Y}(x,y)= P(X\leq x, Y \leq x)$

 

[Legendre]

OMG terrible terrible mistake there. Not sure what I was thinking at the time of posting!

Suppose to be:$$F(x,y) = \int_{-\infty}^y \int_{-\infty}^x \! f(u,v) \, du dv$$

$$F(x,y) = \int_{0}^y \int_{0}^x \! f(u,v) \, du dv$$But the 2nd expression for F(x,y) only holds for the region {0<x<1,0<y<1} right? Beyond that, its either 0 or the respective marginal distribution?