Find the distribution of $X=\frac{\max\{U,V\}}{\min\{U,V\}}$

Question

Suppose that $U,V$ are two independent uniform $(0,1)$ random variables. Find the distribution of $X=\dfrac{\max\{U,V\}}{\min\{U,V\}}$.

Ok to find $\max(U,V)$ we have $P(\max(U,V)\leq r)=P(U\leq r)P(V \leq r)= r^2$ so $P(\max(U,V)=r)=2r$. I can do the same trick with $\min(u,v)$ but then I don't know how to deal with the fraction of the two?

Does this answer your question? $X$ and $Y$ are uniformly ditributed on $(0,1)$. distribution of $\max(X,Y)/\min(X,Y)$ — StubbornAtom, Apr 22 '20 at 08:16

score 1 · Accepted Answer · answered Oct 15 '19 at 15:07

1

Hint:

If you consider $(U,V)$ as a uniform distribution on a unit square,

then $\mathbb P\left(\dfrac{\max\{U,V\}}{\min\{U,V\}} \le x\right)$ involves find the area of the square less two congruent right-angled triangles

answered Oct 15 '19 at 15:07

Henry

157,058

Find the distribution of $X=\frac{\max\{U,V\}}{\min\{U,V\}}$

1 Answers1