What are the Properties of Joint Density Functions?

[Dassinia]

Hello,

Homework Statement

The joint probability density function of X and Y is given by
f(x,y)=c*1(|y|<x<1) 1 is the indicator function

-Find c
-Find the marginal densities of X and Y
-Find the means, variances and the covariance
-Find conditional densities, means and variances of X given Y and of Y given X

Homework Equations

3. The Attempt at a Solution [/B]
The thing is I don't know how to deal with the |y| to find the limits of integration
For example to find c :
1=c∫∫f(x,y) dx dy
y goes from -∞ to 1
But for x ?

Thanks

 

[LCKurtz]

Hello,

Homework Statement

The joint probability density function of X and Y is given by
f(x,y)=c*1(|y|<x<1) 1 is the indicator function

-Find c
-Find the marginal densities of X and Y
-Find the means, variances and the covariance
-Find conditional densities, means and variances of X given Y and of Y given X

Homework Equations

3. The Attempt at a Solution [/B]
The thing is I don't know how to deal with the |y| to find the limits of integration
For example to find c :
1=c∫∫f(x,y) dx dy
y goes from -∞ to 1
But for x ?

Thanks

Have you drawn a picture in the $xy$ plane of the region where $|y|<x<1$?

 

[Dassinia]

Have you drawn a picture in the $xy$ plane of the region where $|y|<x<1$?

No,
It gives a triangle with x from 0 to 1 and y from -1 to 1, so these would be the limits right ?
Thanks

 

[LCKurtz]

No,
It gives a triangle with x from 0 to 1 and y from -1 to 1, so these would be the limits right ?
Thanks

It's a triangle alright. But constant limits like that would describe a rectangle.

 

[Dassinia]

Of course
x from 0 to 1 and y from -x to x

 

[LCKurtz]

Of course
x from 0 to 1 and y from -x to x

Yes, assuming integration order $dydx$, that would cover the whole triangle. Of course, you will need to take a bit more care with the limits when you calculate the marginal densities.

 

[Dassinia]

Yes, assuming integration order $dydx$, that would cover the whole triangle. Of course, you will need to take a bit more care with the limits when you calculate the marginal densities.

For the marginal densities:
f_X(x)=∫ f(x,y) dy from -infinity to x
f_X(x)=∫ c dy from -x to x
and
f_Y(y)=∫ f(x,y) dx from -infinity to y
f_Y(y)=∫ c dx from 0 to 1 it'd give 1 ?

 

[LCKurtz]

For the marginal densities:
f_X(x)=∫ f(x,y) dy from -infinity to x
f_X(x)=∫ c dy from -x to x
and
f_Y(y)=∫ f(x,y) dx from -infinity to y
f_Y(y)=∫ c dx from 0 to 1 it'd give 1 ?

Before you start the marginal densities you should figure out the value of $c$ by setting $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)~dydx = 1$. Of course, you don't need infinite limits for this particular $f$, as you have noted above.

Then, when you do the marginal densities you need actual formulas for $f_X(x)$ and $f_Y(y)$. You need to work out the integrals and give formulas in $x$ or $y$ and indicating where they are zero. Show your work for them.

 

[Dassinia]

Before you start the marginal densities you should figure out the value of $c$ by setting $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)~dydx = 1$. Of course, you don't need infinite limits for this particular $f$, as you have noted above.

Then, when you do the marginal densities you need actual formulas for $f_X(x)$ and $f_Y(y)$. You need to work out the integrals and give formulas in $x$ or $y$ and indicating where they are zero. Show your work for them.

1=∫∫ c dy dx with y from -x to x and x from 0 to 1
1=∫ 2x dx from 0 to 1
1=c
Marginal densities
fX(x)=∫ f(x,y) dy
fX(x)=∫ 1 dy from -x to x
fX(x)=2x
and
fY(y)=∫ f(x,y) dx
fY(y)=∫ 1 dx from 0 to 1
fy(y)=1 ? Or my limits of integration are false ?
Thanks

 

[LCKurtz]

1=∫∫ c dy dx with y from -x to x and x from 0 to 1
1=∫ 2x dx from 0 to 1
1=c

OK.

Marginal densities
fX(x)=∫ f(x,y) dy
fX(x)=∫ 1 dy from -x to x
fX(x)=2x

You have to define $f_X(x)$ for all $x$. It is only $2x$ for certain values of $x$. $f(x,y)$ isn't $1$ for all $x,y$.

and
fY(y)=∫ f(x,y) dx
fY(y)=∫ 1 dx from 0 to 1
fy(y)=1 ? Or my limits of integration are false ?
Thanks

Again, you have to define it for all $y$. And the integrand isn't $1$ for all $y$, and the limits depend on the value of $y$.

 

[Dassinia]

OK.
You have to define $f_X(x)$ for all $x$. It is only $2x$ for certain values of $x$. $f(x,y)$ isn't $1$ for all $x,y$.
Again, you have to define it for all $y$. And the integrand isn't $1$ for all $y$, and the limits depend on the value of $y$.

fX(x)=2x when x ∈ [0,1]
and fX(x)=0 elsewhere

For the limits of the marginal density of Y, is it
fY(y)=∫ 1 dx from |y| to 1
fy(y)=1-|y| ? when y ∈ [-1;1]
fY(y)=0 elsewhere

 

[LCKurtz]

fX(x)=2x when x ∈ [0,1]
and fX(x)=0 elsewhere

For the limits of the marginal density of Y, is it
fY(y)=∫ 1 dx from |y| to 1
fy(y)=1-|y| ? when y ∈ [-1;1]
fY(y)=0 elsewhere

Yes. Those are correct.

 

[Dassinia]

Yes. Those are correct.

I just have one last question
For the calculation of the conditional mean
E[X|Y]=∫ x * f_X|Y(x|y) dx
=∫ x * f(x,y)/f_Y(y) dx
=∫ x * 1/(1-|y|) dx
Here for the limits of integration I took |y| to 1, is it correct ?
Thanks

 

[LCKurtz]

I just have one last question
For the calculation of the conditional mean
E[X|Y]=∫ x * f_X|Y(x|y) dx
=∫ x * f(x,y)/f_Y(y) dx
=∫ x * 1/(1-|y|) dx
Here for the limits of integration I took |y| to 1, is it correct ?
Thanks

That looks like a correct setup. What did you get for your final answer? By the way, it is easy to write integrals in Tex and you should learn to do it. Just right click on the expression below to see how it is done$$
E[y|x] = \int_{|y|}^{1} \frac{x}{1-|y|}dx$$ Just surround it with a pair of ## or $$ depending on whether you want it on the same line or a separate line.