Independent solution

How to solve this Order Statistics question

Setup

Setup

Let M be the maximum of three independent losses with cumulative distribution 1 minus the inverse cube of x above 1.

F(x)=1−x−3,x>1F(x)=1-x^{-3},\qquad x>1
M=max⁡(X1,X2,X3)M=\max(X_1,X_2,X_3)

Model

Model

Independence makes the maximum's cumulative distribution the cube of the single-loss cumulative distribution. Differentiate it to obtain the maximum density.

FM(m)=F(m)3=(1−m−3)3F_M(m)=F(m)^3=(1-m^{-3})^3
fM(m)=9m−4(1−m−3)2f_M(m)=9m^{-4}(1-m^{-3})^2

Compute

Compute

Expanding the density inside the first-moment integral produces three elementary powers. Their weighted integral is 2.025 thousand.

E[M]=9∫1∞(m−3−2m−6+m−9) dmE[M]=9\int_1^{\infty}(m^{-3}-2m^{-6}+m^{-9})\,dm
E[M]=9(12−25+18)=2.025E[M]=9\left(\frac12-\frac25+\frac18\right)=2.025

Answer

Answer

Thus the expected maximum loss is 2025 in the stated units, corresponding to choice A.

2025(A)\boxed{2025\quad\text{(A)}}