# Definitions

Let X and Y be random variables defined on the same random experiment. We define:

### Joint cdf of X and Y

 F(s,t) = P([X l.e. s] and [Y l.e. t]) It follows from this definition that, F(infinity,t) = P(Y l.e. t) = Fy(t) F(s,infinity) = P(X l.e. s) = Fx(t)

### Joint pdf of X and Y

 f(s,t) ds dt = P(s

> 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```

## 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)).

#### 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:

#### 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.

#### 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!

Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>