Find $P (\frac{1}{6}<X_{1}<X_{2})$?

Let $X_1$ and $X_2$ have jointly density function $f$ given by

$f(x_1,x_2)=\left\{\begin{array}{cc}2;&0<x_1<1&0<x_2<1&0<x_1+x_2<1\\0;&\text{otherwise}\end{array}\right.$

Find $P (\frac{1}{6}<X_{1}<X_{2})$.

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Corrected some notations confusing random variables and arguments of their PDF. –  Did Dec 16 '12 at 9:21
Any luck with the answers? –  Did Dec 19 '12 at 16:24

The probability you're looking for is given by $$P(1/6<X_1<X_2) = \int_{\mathbb{R}^2}{dx\,dy\, I_{\{1/6<x<y\}}f(x,y)}.$$ Your definition of the joint density function $f$ imposes the constraints $0<x<1$, $0<y<1$ and $y<1-x$; the indicator function of the event you're interested in imposes the constraints $x>1/6$ and $y>x$. These can only be satisfied simultaneously if $x<1/2$. One thus gets $$P(1/6<X_1<X_2) = 2\int_{1/6}^{1/2}dx\int_x^{1-x}dy\\ = 2\int_{1/6}^{1/2}dx(1-2x)\\ =2\left\{\left[1/2-1/6\right]-2(1/2)\left[(1/2)^2-(1/6)^2\right]\right\}\\ =2/9.$$

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Since $(X_1,X_2)$ is uniformly distributed on the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$, the answer is direct through a drawing of the triangle $T$ and of the region $R\subset T$ defined as $R=\{(x_1,x_2)\mid\frac16\leqslant x_1\leqslant x_2\leqslant1\}$. Then the answer is the ratio of the area of $R$ by the area of $T$, that is, twice the area of $R$.

Since $R$ is the triangle with vertices $(\frac16,\frac16)$, $(\frac12,\frac12)$ and $(\frac16,\frac56)$, by symmetry, twice $R$ is the triangle with vertices $(\frac16,\frac16)$, $(\frac56,\frac16)$ and $(\frac16,\frac56)$. Again by symmetry, four times $R$ is the square $S$ with vertices $(\frac16,\frac16)$, $(\frac56,\frac16)$, $(\frac56,\frac56)$ and $(\frac16,\frac56)$.

The area of $S$ is $(\frac56-\frac16)^2=\frac49$ hence the probability one is looking for is $\frac29$.

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$$\Pr[\frac{1}{6} < X_1 < X_2] = \int\limits_{\frac{1}{6}}^{\infty} dX_1 \int\limits_{X_1}^{\infty} dX_2 f(X_1, X_2)$$ $$= \int\limits_{\frac{1}{6}}^{\frac{1}{2}} dX_1 \int\limits_{X_1}^{1-X_1} dX_2 2$$

Why the upper-limit on X_1 integration is $\frac{1}{2}$: The conditions that $X_1 + X_2 < 1$ and $X_1<X_2$ are enforced by the limits on $X_2$ only if $X_1<1-X_1$ or $X_1<\frac{1}{2}$.

Evaluating this gives $\frac{2}{9}$.

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Wrong: $X_2$ cannot go over $[X_1,1]$ because of the constraint that $X_1+X_2\leqslant1$. –  Did Dec 15 '12 at 14:59
Sorry about that. Edited. –  S Prasanth Dec 15 '12 at 15:04
Isn't this now an exact duplicate (with some steps omitted) of Eckhard's answer? –  Did Dec 15 '12 at 15:06