# Probability of $P(X_1<X_2|X_1<2X_2)$

Find the $P(X_1<X_2|X_1<2X_2)$ Given: $$f(x) = e^{-x}, \qquad 0<x<\infty$$ zero elsewhere. The rvs have same pdf and they are independent variables. Here is my attempt: \begin{align*}P(X_1<X_2|X_1<2X_2)&= \frac{\int_0^\infty\int_{x_1}^{\infty} e^{-x_1-x_2} dx_2\ dx_1} {\int_0^\infty\int_{\frac{x_1}{2}}^{\infty} e^{-x_1-x_2} dx_2\ dx_1} \end{align*} However for these two probabilities, I can not insert the solution for having the conditions. Please help .Or this one can be solved by this $\frac {P(X_1<X_2) \cap P(X_1<2X_2)}{P(X_1<2X_2)}$

Indeed, as you suggest: $P(X_1<X_2 \cap X_1 < 2X_2)=P(X_1<X_2)$, so that $$P(X_1<X_2 \cap X_1 < 2X_2)=\frac{P(X_1<X_2)}{P(X_1<2X_2)}$$ The probabilities on the right side can be calculated as follows \begin{align*} P(X_1<X_2)&=\int_{X_2}P(X_1<x)f_{X_2}(x)dx=\int_{X_2}F_{X_1}(x)f_{X_2}(x)dx=\\&=\int_{0}^{+\infty}(1-e^{-x})e^{-x}dx \end{align*} and similarly \begin{align*} P(X_1<2X_2)&=\int_{X_2}P(X_1<2x)f_{X_2}(x)dx=\int_{X_2}F_{X_1}(2x)f_{X_2}(x)dx=\\&=\int_{0}^{+\infty}(1-e^{-2x})e^{-x}dx \end{align*}

• Am I corret if I arrived at$\frac34$ for the first probability? – Jonarie Ramos Vergara Oct 14 '14 at 13:43
• I did not do it, so I do not know, but it seems correct – Jimmy R. Oct 14 '14 at 13:45