Mean in terms of reliability function

[claudiocamera]

It is possible to express the mean E{x) of a Random Variable in terms of the Reliability Function. In equations:

E{X} = ∫ R(x) dx - ∫ F(x) dx where the limits of the first integral are from 0 to +∞ ; the limits of the second integral are from -∞ to 0; F(x) is the cumulative distribution function of X and R(x) is the reliability function which equals 1 - F(x).

Well, if R(x) = 1- F(x) we can write the first integral with limits 0 to +∞ as :

∫ [1 -F(x)] dx = ∫ dx - ∫ F(x) dx. If it is correct, since we have +∞ as a limit, the first integral on the left side diverges !!! So, what's wrong with my analysis

 

[LouisSheffield]

It is true that the integral on the left diverges - so does (equivalently) the first to the right of the "=".

 

[claudiocamera]

It is true that the integral on the left diverges - so does (equivalently) the first to the right of the "=".

Dear Louis,

Since we are dealing with an equality, we should expect that if one side diverges, the other also diverges. Unfortunately, it is not the answer to the question raised:

E(x) = ∫ R(x) dx with the limits 0 to +≈ . So, ∫ R(x) dx can not diverge! As a conclusion, there is something either wrong or missing in the analysis that I presented . I'm trying to find out what it is. Can you help me ?