TITLE: Joint cdf, pdf, etc..
# !b1 Definitions
Let X and Y be random variables defined on the
same random experiment. We define:
!b3 Joint cdf of X and Y
!eq F(s,t) = P([X l.e. s] and [Y l.e. t])
It follows from this definition that,
!eq F(infinity,t) = P(Y l.e. t) = Fy(t)
!eq F(s,infinity) = P(X l.e. s) = Fx(t)
!b3 Joint pdf of X and Y
!eq f(s,t) ds dt = P(s<X<s+ds, t<Y<t+dt)
It follows from this definition that,
> f is obtainable from F by taking derivatives
2
d
f(s, t) = ----- F(s, t)
dt ds
> F is obtainable from f by integration
t s
/ /
| |
F(s, t) = | | f(x, y) dx dy
| |
/ /
-infinity -infinity
# !b2 Problem
Let (X,Y) be uniformly distributed over the unit circle
centered at the origin. Compute the joint pdf, joint cdf,
marginal density of X (i.e. fx(s)) and the marginal cdf of Y
(i.e. Fy(t)).
!b4 Solution:
The joint pdf is constant over the circle and since the
total area of the circle is Pi we get that for (x,y) inside
the circle,
> f is
1
f := (x, y) -> ----
Pi
# The joint cdf is computed from f by integration. When
(s,t) is a point inside the circle the answer will depend
on which cuadrant (s,t) is.
> when (s,t) is in the III or IV cuadrant we have,
s t
/ /
| | 1
F(s, t) = | | ---- dy dx
| | Pi
/ /
-1 2 1/2
-(1 - x )
> integrating over y,
s
/ 2 1/2
| t + (1 - x )
F(s, t) = | --------------- dx
| Pi
/
-1
# and computing this integral (e.g. with MAPLE)
> F(s,t) = int(int(f(x,y),y=-sqrt(1-x^2)..t),x=-1..s);
2 1/2
4 t s + 2 s (1 - s ) + 2 arcsin(s) + 4 t + Pi
F(s, t) = 1/4 ------------------------------------------------
Pi
# The other two simmilar cases: i.e. when (s,t) is in the II or in the I
quadrant, are left as exercises. HINT: draw the pictures of the
regions.
Now, is this all?

no-no. There is still one more case! To find it try to compute
F(0.9,0.8) or F(1/2,1). Actually by thinking about this case
you would find the answer to the second question:
!b4 The marginal cdf of Y:
This is given by FY(t) where,
> FY(t) = 2*int(int(1/Pi,x=0..sqrt(1-y^2)),y=-1..t);
2 1/2
2 t (1 - t ) + 2 arcsin(t) + Pi
FY(t) = 1/2 ----------------------------------
Pi
# where t varies from -1 to 1.
!b4 The marginal pdf of X:
By symmetry FY = FX. Thus, the marginal pdf of X is just the
derivative w.r.t. t of FY(t) above. This is
> given by
2
2 1/2 t 2
2 (1 - t ) - 2 ----------- + -----------
2 1/2 2 1/2
(1 - t ) (1 - t )
1/2 -------------------------------------------
Pi
# Done!
>