If $X_i$, $i=1,2,3$ are independent exponential random variable with rates $\lambda_i$, find $$E[\max(X_i) \mid X_1<X_2<X_3]$$
I really did not understand this exercise, because if $$X_1<X_2<X_3\Rightarrow X_3=\max(X_i)$$ right?
So it should not simply be $$E[\max(X_i) \mid X_1<X_2<X_3]=E[X_3]$$ because the memorylessness?
EDIT: Let $X_{(n)}$ the maximum $$F_{X_{(n)}}=F_{X_1}(x)F_{X_2}(x)F_{X_3}(x)=(1-e^{-\lambda_1x})(1-e^{-\lambda_2x})(1-e^{-\lambda_3x})$$ $$=1-e^{-\lambda_2x}-e^{-\lambda_1x}+e^{-(\lambda_1+\lambda_2)x}-e^{-\lambda_3x}+e^{-(\lambda_2+\lambda_3)x}+e^{-(\lambda_1+\lambda_3)x}-e^{-(\lambda_1+\lambda_2+\lambda_3)x}$$ $$E[X_{(n)}]=\int_0^\infty 1-F_{X_{(n)}}dx$$ $$=\int_0^\infty e^{-\lambda_2x}+e^{-\lambda_1x}-e^{-(\lambda_1+\lambda_2)x}+e^{-\lambda_3x}-e^{-(\lambda_2+\lambda_3)x}-e^{-(\lambda_1+\lambda_3)x}+e^{-(\lambda_1+\lambda_2+\lambda_3)x}$$ $$=\frac{1}{\lambda_2}+\frac{1}{\lambda_1}-\frac{1}{\lambda_1+\lambda_2}+\frac{1}{\lambda_3}-\frac{1}{\lambda_2+\lambda_3}-\frac{1}{\lambda_1+\lambda_3}+\frac{1}{\lambda_1+\lambda_2+\lambda_3}$$
I tried some simplifications, but could not get the answer
EDIT: The book answer is $$E[\max(X_i) \mid X_1<X_2<X_3]=\frac{1}{\lambda_1+\lambda_2+\lambda_3}+\frac{1}{\lambda_2+\lambda_3}+\frac{1}{\lambda_3}$$
Hint: $\mathsf E(\max_i X_i \mid X_1<X_2<X_3) = \mathsf E(X_3 \;\mathbf 1_{X_3>X_2>X_1})\big/ \mathsf P(X_3>X_2>X_1) \\ = \dfrac{\int_0^\infty \int_0^{x_3}\int_0^{x_2} x_3f_{X_3}(x_3)f_{X_2}(x_2)f_{X_1}(x_1)\operatorname d x_1 \operatorname d x_2\operatorname d x_3}{\int_0^\infty \int_0^{x_3}\int_0^{x_2} f_{X_3}(x_3)f_{X_2}(x_2)f_{X_1}(x_1)\operatorname d x_1 \operatorname d x_2\operatorname d x_3}$