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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 $P(X^2+Y^2 \leq 1)$?


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


\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. –  PEV Nov 1 '10 at 14:40
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 \>. –  Arturo Magidin Nov 1 '10 at 14:41
@GeorgeS Please avoid minor edits on questions more than 4 years old (especially to edit incorrectly). –  Did 2 days ago
Okay sure. Will do –  George S 2 days ago

2 Answers 2

up vote 2 down vote accepted

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, $P(X^2+Y^2 \leq 1) = \int_{0}^{0.5} \int_{y}^{\sqrt{1-y^2}} 2 \ dx \ dy$? –  PEV Nov 1 '10 at 15:42
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.) –  Mike Spivey Nov 1 '10 at 15:44
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 $x \leq 1-y$? –  PEV Nov 1 '10 at 17:46
@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. –  Mike Spivey Nov 1 '10 at 19:59
@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). –  Mike Spivey Nov 1 '10 at 19:59

No, $P(X+Y \leq 1 ) = \int_0^1 \int_0^{\min (x,1-x)} 2 dy dx = \int_0^{1/2} \int_0^x 2 dy dx + \int_{1/2}^1 \int_0^{1-x} 2 dy dx = 1/4+1/4=1/2$

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