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 $$P\{X_1<X_3<X_2<\alpha-X_3\}.$$

Is it $$ P\{X_1<X_3<X_2<\alpha-X_3\} \\ =\int_{x_3=0}^{\infty} \int_{x_2=0}^{\alpha-x_3} \int_{x_3=0}^{x_2} \int_{x_1=0}^{x_3} f_{X_1}(x_1)\, \mathrm{d}x_1 f_{X_3}(x_3)\, \mathrm{d}x_3 f_{X_2}(x_2)\, \mathrm{d}x_2 f_{X_3}(x_3)\, \mathrm{d}x_3~, $$ or $$ P\{X_1<X_3<X_2<\alpha-X_3\}= \int_{x_3=0}^{\infty} \int_{x_2=0}^{\alpha-x_3} \int_{x_1=0}^{x_3} f_{X_1}(x_1) \,\mathrm{d} x_1 f_{X_2}(x_2) \,\mathrm{d} x_2 f_{X_3}(x_3) \,\mathrm{d} x_3 ~? $$ Thanks.

  • $\begingroup$ 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. $\endgroup$ – herb steinberg Mar 17 at 21:32
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    $\begingroup$ The inequalities imply $2 X_3 < \alpha$, therefore $$\operatorname{P}(X_1 < X_3 < X_2 < \alpha - X_3) = \\ \int_0^{\alpha/2} \int_{x_3}^{\alpha - x_3} \int_0^{x_3} f_{X_1}(x_1) f_{X_2}(x_2) f_{X_3}(x_3) \,dx_1 dx_2 dx_3 \,[\alpha > 0].$$ $\endgroup$ – Maxim Mar 19 at 0:17
  • $\begingroup$ Why, how do you did simplify? $\endgroup$ – Monir Mar 19 at 0:33
  • $\begingroup$ Ignore $x_1$ for now and draw the region $x_3 < x_2 < \alpha - x_3$ on the plane. $\endgroup$ – Maxim Mar 19 at 3:42

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