# Joint cumulative distribution

• Jun 7th 2008, 05:17 AM
peterpan
Joint cumulative distribution
How does one write the joint cumulative distribution

$\displaystyle F(X_1 \leq x_1, X_2 \leq x_2)$

as a combination of marginal distributions?
• Jun 7th 2008, 05:24 AM
Moo
Hello,

Quote:

Originally Posted by peterpan
How does one write the joint cumulative distribution

$\displaystyle F(X_1 \leq x_1, X_2 \leq x_2)$

as a combination of marginal distributions?

Have a look here : Marginal distribution - Wikipedia, the free encyclopedia ?

I think this would lead to :

$\displaystyle P(X_1 \le x_1, X_2 \le x_2)=\sum_{k \le x_2} P(X_1 \le x_1, X_2=k)=\sum_{k \le x_2} \ \sum_{l \le x_1} P(X_1=l, X_2=k)$

(assuming that $\displaystyle X_1$ and $\displaystyle X_2$ are discrete random variables ; for continuous ones, see the link, the part with the integrals)
• Jun 7th 2008, 05:35 AM
peterpan
Thanks Moo.

In this case I am actually considering continuous random variables which are independent, identically distributed.
• Jun 7th 2008, 05:39 AM
Moo
Quote:

Originally Posted by peterpan
Thanks Moo.

In this case I am actually considering continuous random variables which are independent, identically distributed.

If they are independent, $\displaystyle P(X_1 \le x_1, X_2 \le x_2)=P(X_1 \le x_1) \cdot P(X_2 \le x_2)$

Isn't it ?
• Jun 7th 2008, 06:06 AM
peterpan
Thats what I initially thought too, but wasnt sure. Thanks for the clarification!
• Jun 7th 2008, 07:47 AM
peterpan
Is it also fair to say that

$\displaystyle 1 - P(X \leq x) = P(X > x)$

this would then mean that

$\displaystyle 1 - P(X_1 > x_1, X_2 > x_2) = P(X_1 \leq x_1, X_2 \leq x_2)$
• Jun 7th 2008, 08:03 AM
Moo
Hi again :)

Quote:

Originally Posted by peterpan
Is it also fair to say that

$\displaystyle 1 - P(X \leq x) = P(X > x)$

Yes

Quote:

this would then mean that

$\displaystyle 1 - P(X_1 > x_1, X_2 > x_2) = P(X_1 \leq x_1, X_2 \leq x_2)$
No :)

Because $\displaystyle P(X_1 > x_1, X_2 > x_2)$ is the simultaneous probability that $\displaystyle X_1 > x_1$ and $\displaystyle X_2 > x_2$.

So its contrary is that $\displaystyle X_1 \le x_1$ or $\displaystyle X_2 \le x_2$.
If you have done a bit of logic, you can understand...

$\displaystyle 1 - P(X_1 > x_1, X_2 > x_2) =P(X_1 \le x_1, X_2 > x_2)$$\displaystyle +P(X_1 \le x_1, X_2 \le x_2)+P(X_1 > x_1, X_2 \le x_2)$

Is it clear ? (Wink)
• Jun 7th 2008, 09:00 AM
peterpan
Hey Moo (Hi)

Im starting to get it...

Really what im trying to do here is define a joint distribution $\displaystyle F(x,y)$.

I am trying to do this using the result:

$\displaystyle F(x, y) = 1 - P(X > x) - P(Y>y) + P(X > x, Y > y)$

and the fact that I know the marginal distributions:

$\displaystyle P(X \leq x), P(Y \leq y)$

Still a little unclear on how this works.

Thanks

Peter (Cool)