# Find the probability $P\{X_1<X_3<X_2<\alpha-X_3\}$

Let $$X_1, X_2$$ and $$X_3$$ three independent random variables with pdf and cdf $$f_{X_i}(x_i)$$ and $$F_{X_i}(x_i)$$ for $$x_i\geq 0$$, receptively. Using CDFs and PDFs of $$X_i$$, find the probability $$P\{X_1

Is it $$P\{X_1 or $$P\{X_1 Thanks.

• You seem to assume all $X_i\ge 0$, $x_2$ is missing a lower limit on both expressions. The second expression looks right. The first is weird. – herb steinberg Mar 17 at 21:32
• The inequalities imply $2 X_3 < \alpha$, therefore $$\operatorname{P}(X_1 < X_3 < X_2 < \alpha - X_3) = \\ \int_0^{\alpha/2} \int_{x_3}^{\alpha - x_3} \int_0^{x_3} f_{X_1}(x_1) f_{X_2}(x_2) f_{X_3}(x_3) \,dx_1 dx_2 dx_3 \,[\alpha > 0].$$ – Maxim Mar 19 at 0:17
• Why, how do you did simplify? – Monir Mar 19 at 0:33
• Ignore $x_1$ for now and draw the region $x_3 < x_2 < \alpha - x_3$ on the plane. – Maxim Mar 19 at 3:42