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! >