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What is the probability of $P(X_1 X_2\le 2)$. Both variables are independent and each has the probability density function $f(x)=1, 1<x<2$, zero elsewhere. First I would like to assume that the pdf of $f(x_1,x_2) = 1$ by virtue of independence. I am now stuck with this one.

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The basic idea is to integrate like so:

$$\int_{\Bbb R}\int_{x_2\le 2x_1^{-1}}f(x_1,x_2)\,dx_2\,dx_1$$ This gives

$$\int_{1}^2\int_1^{2x_1^{-1}}1\,dx_2\,dx_1$$ $$=\int_1^2 2x_1^{-1}-1\,dx_1$$ $$=\bigg(2\log(x_1)-x_1\bigg|_1^2$$ $$=2\log 2-2-(2\log 1-1)$$

$$=2\log 2-1= \log\left({4\over e}\right)$$


Note

If you draw the picture, you can see that this is the right idea, because the region where the density function is positive and where $X_2\le 2X_1^{-1}$ is exactly the area under the hyperbola $X_1X_2=2$ and above $X_2=1$ and to the right of $X_1=1$, and since the probability you are in a set, $A$ is just

$$\int_A f(x_1,x_2)$$

it's easy to see our computation is the right one.

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