# Math Help - Density Function

1. ## Density Function

Consider the following joint probability density function of the random variable X and Y:

f(x,y) = (3x - y) / 9 for 1 < x < 3 and 1 < y < 2

a) Find the marginal distributions of X and Y.
b) Are X and Y independent?
C) Find P(X > 2)

Not sure how to do this. I missed the class that went over it. Can someone show me how to do this? Thanks

2. Originally Posted by kenshinofkin
Consider the following joint probability density function of the random variable X and Y:

f(x,y) = (3x - y) / 9 for 1 < x < 3 and 1 < y < 2

a) Find the marginal distributions of X and Y.
b) Are X and Y independent?
C) Find P(X > 2)

Not sure how to do this. I missed the class that went over it. Can someone show me how to do this? Thanks
For a joint distribution of $X$ and $Y$ with density $f(x,y)$ , the marginal distributions are:
$

f_x(u)=\int_{sup(y)} f(u,y) dy
$

$f_y(u)=\int_{sup(x)} f(x,u) dx$

where $sup(y)$ is the set on which $f(u,y)$ is defined (may be dependent on $u$) and similarly for $sup(x)$

For $X$ and $Y$ to be independent we require that:

$f(u,v)=f_x(u)f_y(v)$

or slightly more confusingly:

$f(x,y)=f_x(x)f_y(y)$

RonL

3. Well I was wondering if someone could show me how to do it. I have alot of them i can do in my book just wanted to see one done. I found what you typed in my book but am having a problem doing it.

4. Originally Posted by kenshinofkin
Consider the following joint probability density function of the random variable X and Y:

f(x,y) = (3x - y) / 9 for 1 < x < 3 and 1 < y < 2

a) Find the marginal distributions of X and Y.
b) Are X and Y independent?
C) Find P(X > 2)

Not sure how to do this. I missed the class that went over it. Can someone show me how to do this? Thanks
Originally Posted by CaptainBlack
For a joint distribution of $X$ and $Y$ with density $f(x,y)$ , the marginal distributions are:
$

f_x(u)=\int_{sup(y)} f(u,y) dy
$
$f_x(u)=\int_{sup(y)} f(u,y) dy =\int_{y=1}^2 (3u - y) / 9 dy=\left([3uy-y^2/2]_1^2\right)/9=\frac{u}{3}-\frac{1}{6}$

$f_y(u)=\int_{sup(x)} f(x,u) dx$
$f_y(u)=\int_{sup(x)} f(x,u) dx=\int_{x=1}^3 (3x - u) / 9 dx=\frac{4}{3}-\frac{2u}{9}$

RonL

5. So for finding P(X > 2) what do I do. Plug 2 into something?

6. Originally Posted by kenshinofkin
So for finding P(X > 2) what do I do. Plug 2 into something?
Since:

$
P(X>2)=\int_{v=1}^2\int_{u=1}^2 f(u,v) du dv = \int_{u=1}^2 f_x(u) du
$

It is the integral of the marginal distrtribution of $x$ from $1$ to $2$.

$
P(X>2)=\int_{u=1}^2 f_x(u) du = \int_{u=1}^2 \left(\frac{u}{3}-\frac{1}{6}\right) du
$

RonL