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I have a joint distribution of 2 random variables $X$ and $Y$: $$ f(x, y) = \frac{3}{2}x + \frac{y}{2} $$ if $x,y \in [0;1]$ and $0$ otherwise. What's the following probability:$$ P\left(XY < \dfrac{1}{2}\right) $$ I thought about sum of probabilities: $P(X < \frac{1}{2Y}) + P(Y < \frac{1}{2X})$, but appropriate integrals don't converge: $$ \int^{1}_{0}\int^{\frac{1}{2y}}_{0}\left(\frac{3}{2}x + \frac{y}{2}\right) dxdy $$ $$ \int^{1}_{0}\int^{\frac{1}{2x}}_{0}\left(\frac{3}{2}x + \frac{y}{2}\right) dydx $$

Then I imagined space of $X, Y$ as a square on a plane with center in $(0.5, 0.5)$. Then the area when $X > 0.5$ and $Y > 0.5$ violate $ XY < \frac{1}{2}$, therefore I calculated $$ 1 - P(0.5 < x < 1, 0.5 < y < 1)$$ as $$ 1 - \int_{\frac{1}{2}}^{1}\int_{\frac{1}{2}}^{1}\left(\frac{3}{2}x + \frac{y}{2}\right) dxdy = \frac{5}{8}$$

But I'm not sure that it's right and therefore I'm here. I will be grateful for any help you can provide.

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Actually, if $X>\frac{1}{2}$ and $Y>\frac{1}{2}$, you may still have $XY<\frac{1}{2}$. For instance, if $X=Y=\frac{5}{8}$, $XY=\frac{25}{64}<\frac{1}{2}$.

What you really need is $Y<\frac{1}{2X}$. So you can integrate the joint density of $X$ and $Y$ over the region $R$ where $0 \leq x \leq 1$, $0 \leq y \leq 1$, and $y < \frac{1}{2x}$. The reason your integrals did not converge is because you did not account for the fact that $X$ and $Y$ are always $ \leq 1$. You will have to break $R$ up into two regions to write down the integral, for instance $$P\left(Y< \frac{1}{2X}\right)=\int_R f(x,y) \ dy \ dx = \int_0^\frac{1}{2}\int_0^1 f(x,y) \ dy \ dx + \int_\frac{1}{2}^1 \int_0^\frac{1}{2x} f(x,y) \ dy \ dx.$$

Finally, you do NOT want $P\left(Y<\frac{1}{2X}\right)+P\left(X< \frac{1}{2Y}\right)$ because these are the same event; if $Y<\frac{1}{2X}$ then $X<\frac{1}{2Y}$ and vice versa. So just calculate one of them.

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To be found is:$$\Pr\left(XY<\frac12\right)=\int^1_0\int^1_0f(x,y)g(x,y)dxdy$$ where $g$ denotes the indicator function of set $\{\langle x,y\rangle\in[0,1]^2\mid xy<\frac12\}$.

Observe that:

$$\int^1_0\int^1_0f(x,y)g(x,y)dydx=\int^{\frac12}_0\int^1_0f(x,y)dydx+\int^1_{\frac12}\int^{\frac1{2x}}_0f(x,y)dydx$$

Try it out.

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