Cumulative Distribution Func

a. the CDF is defined by F(z) = P(Z =< z) = 1 - P(Z>z)
For z=>1,
P(Z>z) = P(X>zY or Y>zX) = P(X>zY)+P(Y>zX)
The last equality holds because the two events are mutually exclusive.
P(X>zY)=∫(y=0 to 1) P(X>zy) f(y) dy
=∫(y=0 to 1) P(X>zy) dy (since Y is U(0,1), f(y)=1 on [0,1])
= ∫(y=0 to 1/z) 1-zy dy + ∫(y=1/z to 1) 0 dy
=1/z - 1/2z
= 1/2z
By symmetry, P(Y>zX)=1/2z.
Hence P(Z>z) = 1/2z+1/2z = 1/z
Therefore, F(z)=1-1/z for z=>1, and F(z)=0 for z<1.
Using this formular, you can easily sketch the curve of CDF.

b. the PDF is obtained by differentiating the CDF. So
f(z) = 1/z^2 for z=>1 and f(z)=0 for z<1.
Hence you can sketch the PDF according to this formula.

PS. using the notations above, the horizontal axis is z and the vertical axis is F(z) or f(z).

Let X,Y be uniformly distributed U[0,1]
X and Y are independent. Let Z = max(X,Y) / min(X,Y)
Let U=max(X,Y) V=min(X,Y)
Using the formila of Order Statistics, the joint distribution of U, V is
g(u,v)=2 where 0<u<=v<=1
Now let S=U/V and W=V, then U=SW and V=W where 0<sw<w<=1
Consider the Jacobian matrix which is
1/v -u^2/v
0 1
The determinent is 1/v, and so h(s,w)=2v=2w, 0<sw<w<=1
To find out the pdf of S, just integrate with respect to w
∫(w = 0 to 1/s) 2w dw
=1/s^2 where s>1
The CDF is ∫(t = 1 to s) 1/t^2 dt
=-1/t |(1,s)
=1-1/s