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Suppose $f(x,y) = c$ for $0\lt y\lt x\lt 1$ and $0$ outside. What is $P(X+Y \leq 1)$? What is $P(X^2+Y^2 \leq 1)$?

So

\begin{equation*} P(X+Y \leq 1) = \int_{0}^{1} \int_{0}^{1-x} 2 \ dy \ dx? \end{equation*}

Likewise,

\begin{equation*} $P(X^2+Y^2 \leq 1) = \int_{0}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \ dy \ dx$? \end{equation*}

This is assuming that $c=2$.

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    For some reason the LaTeX is not parsing.2010-11-01
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    Fixed. Both `<` and `>` give problems, because they are initially interpreted as HTML. Try either `\lt` and `\gt`, or use the backslash as an escape character, `\<` and `\>`.2010-11-01
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    @GeorgeS Please avoid minor edits on questions more than 4 years old (especially to edit incorrectly).2015-05-21
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    Okay sure. Will do2015-05-21

2 Answers 2

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Since the joint density is constant on the support region, you can get your answers by considering areas. The answer in each case is the area that corresponds to the event in question divided by the total area of the support region.

Thus, to calculate $P(X + Y \leq 1)$, you want the area of the lower triangular region below (i.e., the region for which $x + y \leq 1$), divided by the total area of the triangle.

alt text

This is $$\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$$

Similarly, to calculate $P(X^2 + Y^2 \leq 1)$, you want the area of the circle sector divided by the total area of the triangle.

alt text

This is $$\frac{\frac{\pi}{8}}{\frac{1}{2}} = \frac{\pi}{4}.$$

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    So $P(X+Y \leq 1)$ also equals $\int_{0}^{1/2} \int_{y}^{1-y} 2 \ dx \ dy$? Likewise, $P(X^2+Y^2 \leq 1) = \int_{0}^{0.5} \int_{y}^{\sqrt{1-y^2}} 2 \ dx \ dy$?2010-11-01
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    Right on the first one. For the second one, though, the $y$-coordinate of the intersection is $\frac{\sqrt{2}}{2}$. (Do the algebra to see why.)2010-11-01
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    Are there any ways of getting the bounds without a picture? How did you go about graphing $x+y \leq 1$? First graph $y \leq 1-x$ and $x \leq 1-y$?2010-11-01
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    @Trevor: Plop's answer shows one way to do the first without a picture. However, I recommend doing the picture. When I'm evaluating a double integral I nearly always find it helpful to draw a picture of the region of integration.2010-11-01
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    @Trevor: As far as graphing $x+y \leq 1$, I graphed $y \leq 1 - x$. You don't need the other, too (as it gives the same graph).2010-11-01
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No, $P(X+Y \leq 1 ) = \int_0^1 \int_0^{\min (x,1-x)} 2 dy dx = \int_0^{1/2} \int_0^x 2 dy dx + \int_{1/2}^1 \int_0^{1-x} 2 dy dx = 1/4+1/4=1/2$