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Let $X$ and $Y$ be independent random variables, each uniformly distributed on the interval $[0,2]$. I am trying to find ${\bf P}(XY\geq 1)$.

$${\bf P}(XY\geq 1) = \int_{x}f_X(x)P(Y\geq \frac{1}{X}|X=x)dx=\int_{x}\frac{1}{2}(1-F_Y(\frac{1}{x}))dx=\int_{x}\frac{1}{2}(1-\frac{1}{2x})dx$$

I am stuck here, how to define the scope of $x$ for this integration? It didn't work if I just set the range $[0,2]$ for $x$.Any help will be appreciated, thanks a lot.

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Draw the coordinate axes as usual, and the $2\times2$ square on which the joint density function "lives." Draw the first quadrant half of the hyperbola $xy=1$. We want the probability of landing in the part $K$ of the square which is "above" the hyperbola. Our probability is the double integral over $K$ of $\frac{1}{4}$ over $K$. More informally, it is one-quarter of the area of $K$.

Note that $xy=1$ meets the top boundary of the square (the line $y=2$) at $x=1/2$. It follows that our probability is $$\int_{x=1/2}^2 \left(\int_{y=1/x}^2 \frac{dy}{4}\right) \,dx.$$

We used the language of several variable calculus. But the area of $K$ can be computed using one-variable techniques. It is $$\int_{x=1/2}^2 \left(2-\frac{1}{x}\right)\,dx.$$

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  • $\begingroup$ Thanks a lot. Drawing the coordinate graph is really helpful. And the solution using one variable techniques is quite illuminating. $\endgroup$ – Q Yang May 21 '15 at 11:55
  • $\begingroup$ You are welcome. As you can imagine, I have solved this kind of problem many times. However, I still always make a sketch. Without a sketch, I might write down the correct integral. With a sketch, I am sure that I have written down the right thing. $\endgroup$ – André Nicolas May 21 '15 at 12:03

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