On the mixing advantage Kais Hamza Monash University ANZAPW, July - PowerPoint PPT Presentation

On the mixing advantage Kais Hamza Monash University ANZAPW, July 2013, University of Queensland Joint work with Aidan Sudbury, Peter Jagers & Daniel Tokarev Kais Hamza On the mixing advantage

Introduction Kais Hamza On the mixing advantage

Introduction ◮ X j i , X i are the lifetime of an individual/unit and, max( X 1 i , . . . , X n i ) and max( X 1 , X 2 , . . . , X n ) represent the lifetime of population/system. All random variables are assumed to be non-negative. X i , X 1 i , . . . , X n i are identically distributed. Kais Hamza On the mixing advantage

Introduction ◮ X j i , X i are the lifetime of an individual/unit and, max( X 1 i , . . . , X n i ) and max( X 1 , X 2 , . . . , X n ) represent the lifetime of population/system. All random variables are assumed to be non-negative. X i , X 1 i , . . . , X n i are identically distributed. ◮ X 1 X 1 X 2 X n M 1 = E [max( X 1 1 , . . . , X n . . . → 1 )] 1 1 1 Kais Hamza On the mixing advantage

Introduction ◮ X j i , X i are the lifetime of an individual/unit and, max( X 1 i , . . . , X n i ) and max( X 1 , X 2 , . . . , X n ) represent the lifetime of population/system. All random variables are assumed to be non-negative. X i , X 1 i , . . . , X n i are identically distributed. ◮ X 1 X 1 X 2 X n M 1 = E [max( X 1 1 , . . . , X n . . . → 1 )] 1 1 1 X 1 X 2 M 2 = E [max( X 1 X n 2 , . . . , X n X 2 . . . → 2 )] 2 2 2 Kais Hamza On the mixing advantage

Introduction ◮ X j i , X i are the lifetime of an individual/unit and, max( X 1 i , . . . , X n i ) and max( X 1 , X 2 , . . . , X n ) represent the lifetime of population/system. All random variables are assumed to be non-negative. X i , X 1 i , . . . , X n i are identically distributed. ◮ X 1 X 1 X 2 X n M 1 = E [max( X 1 1 , . . . , X n . . . → 1 )] 1 1 1 X 1 X 2 M 2 = E [max( X 1 X n 2 , . . . , X n X 2 . . . → 2 )] 2 2 2 . . . Kais Hamza On the mixing advantage

Introduction ◮ X j i , X i are the lifetime of an individual/unit and, max( X 1 i , . . . , X n i ) and max( X 1 , X 2 , . . . , X n ) represent the lifetime of population/system. All random variables are assumed to be non-negative. X i , X 1 i , . . . , X n i are identically distributed. ◮ X 1 X 1 X 2 X n M 1 = E [max( X 1 1 , . . . , X n . . . → 1 )] 1 1 1 X 1 X 2 M 2 = E [max( X 1 X n 2 , . . . , X n X 2 . . . → 2 )] 2 2 2 . . . X 1 X 2 X n M n = E [max( X 1 n , . . . , X n . . . → n )] X n n n n Kais Hamza On the mixing advantage

Introduction ◮ X j i , X i are the lifetime of an individual/unit and, max( X 1 i , . . . , X n i ) and max( X 1 , X 2 , . . . , X n ) represent the lifetime of population/system. All random variables are assumed to be non-negative. X i , X 1 i , . . . , X n i are identically distributed. ◮ X 1 X 1 X 2 X n M 1 = E [max( X 1 1 , . . . , X n . . . → 1 )] 1 1 1 X 1 X 2 M 2 = E [max( X 1 X n 2 , . . . , X n X 2 . . . → 2 )] 2 2 2 . . . X 1 X 2 X n M n = E [max( X 1 n , . . . , X n . . . → n )] X n n n n ◮ We wish to compare E [max( X 1 , X 2 , . . . , X n )] to M i = E [max( X 1 i , . . . , X n i )], i = 1 , . . . , n . Kais Hamza On the mixing advantage

Introduction A B Reliability – Warm Duplication Method Kais Hamza On the mixing advantage

Introduction A B A A Reliability – Warm Duplication Method Kais Hamza On the mixing advantage

Introduction A B A B A B Reliability – Warm Duplication Method Kais Hamza On the mixing advantage

Introduction A B A A B A B B Reliability – Warm Duplication Method Kais Hamza On the mixing advantage

Introduction Kais Hamza On the mixing advantage

Introduction ◮ Question: Is it better to mix or go with a single type? Kais Hamza On the mixing advantage

Introduction ◮ Question: Is it better to mix or go with a single type? ◮ Obviously, if one type dominates all others, then choosing that type only is optimum. Kais Hamza On the mixing advantage

Introduction ◮ Question: Is it better to mix or go with a single type? ◮ Obviously, if one type dominates all others, then choosing that type only is optimum. ◮ Question: What if all types are similar (no dominant type); i.e. E [max( X 1 1 , . . . , X n 1 )] = . . . = E [max( X 1 n , . . . , X n n )]? Kais Hamza On the mixing advantage

Introduction Kais Hamza On the mixing advantage

Introduction Assume all random variables are independent. Kais Hamza On the mixing advantage

Introduction Assume all random variables are independent. ◮ It is easy to show (direct consequence of the arithmetic-geometric mean inequality) that E [max( X 1 , . . . , X n )] ≥ E [max( X 1 i , . . . , X n i )] . In fact, the same arithmetic-geometric mean inequality shows that n E [max( X 1 , . . . , X n )] ≥ 1 � E [max( X 1 i , . . . , X n i )] . n i =1 In other words, mixing is advantageous. Kais Hamza On the mixing advantage

Introduction Assume all random variables are independent. ◮ It is easy to show (direct consequence of the arithmetic-geometric mean inequality) that E [max( X 1 , . . . , X n )] ≥ E [max( X 1 i , . . . , X n i )] . In fact, the same arithmetic-geometric mean inequality shows that n E [max( X 1 , . . . , X n )] ≥ 1 � E [max( X 1 i , . . . , X n i )] . n i =1 In other words, mixing is advantageous. ◮ If M i = E [max( X 1 i , . . . , X n i )], i = 1 , . . . , n , we call mixing factor θ = E [max( X 1 , . . . , X n )] max( M 1 , . . . , M n ) . We show that when M i = M , θ ≤ 2 − 1 / n < 2. Kais Hamza On the mixing advantage

Existing literature Kais Hamza On the mixing advantage

Existing literature ◮ An extensive literature exists on E [max( X 1 , . . . , X n )] in the iid case – see David and Nagaraja (2003). However, very little work exists for the non-identically distributed case. Kais Hamza On the mixing advantage

Existing literature ◮ An extensive literature exists on E [max( X 1 , . . . , X n )] in the iid case – see David and Nagaraja (2003). However, very little work exists for the non-identically distributed case. ◮ Arnold and Groeneveld (1979) obtain upper and lower bounds on E [max( X 1 , . . . , X n )] even when X 1 , . . . , X n are not independent and not identically distributed, but in terms of E [ X 1 ] and var( X i ), not M 1 , . . . , M n . This generalises Hartley and David (1954) and Gumbel (1954) who deal with the iid case. Kais Hamza On the mixing advantage

Existing literature Kais Hamza On the mixing advantage

Existing literature ◮ Sen (1970) shows that max( X 1 , . . . , X n ) stochastically dominates max( Y 1 , . . . , Y n ), where Y 1 , . . . , Y n are iid equally-weighted probability mixtures of X 1 , . . . , X n : P (max( X 1 , . . . , X n ) ≤ z ) ≤ P (max( Y 1 , . . . , Y n ) ≤ z ) . In particular n 1 � E [max( X 1 i , . . . , X n i )] n i =1 E [max( Y 1 , . . . , Y n )] ≤ E [max( X 1 , . . . , X n )] . ≤ However, E [max( Y 1 , . . . , Y n )] cannot be expressed in terms of M 1 , . . . , M n . Kais Hamza On the mixing advantage

Unbounded independent case Kais Hamza On the mixing advantage

Unbounded independent case Theorem (H., Jagers, Sudbury & Tokarev, 2009) If X 1 , . . . , X n are independent random variables with the property that E [max( X 1 i , . . . , X n i )] = M i , i = 1 , 2 , ..., n , then n 1 � ≤ E [max( X 1 , . . . , X n )] M i n i =1 n 1 M i + n − 1 � ≤ max( M 1 , . . . , M n ) . n n i =1 In particular, if M i = M , i = 1 , ..., n , M ≤ E [max( X 1 , . . . , X n )] ≤ (2 − 1 / n ) M . Kais Hamza On the mixing advantage

Unbounded independent case Theorem (H., Jagers, Sudbury & Tokarev, 2009) If X 1 , . . . , X n are independent random variables with the property that E [max( X 1 i , . . . , X n i )] = M i , i = 1 , 2 , ..., n , then n 1 � ≤ E [max( X 1 , . . . , X n )] M i n i =1 n 1 M i + n − 1 � ≤ max( M 1 , . . . , M n ) . n n i =1 In particular, if M i = M , i = 1 , ..., n , M ≤ E [max( X 1 , . . . , X n )] ≤ (2 − 1 / n ) M . The upper bound is obtained by letting some of the random variables be concentrated on 0 and x and letting x → ∞ . Kais Hamza On the mixing advantage

Bounded independent case Kais Hamza On the mixing advantage

Bounded independent case Theorem (H. & Sudbury, 2011) If a set of random variables X 1 , . . . , X n are independent, concen- trated on [0 , b ] and s.t. E [max( X 1 i , . . . , X n i )] = M i , i = 1 , . . . , n , then, putting M n = max( M 1 , . . . , M n ), n � ( b − M i ) 1 / n b − ≤ E [max( X 1 , . . . , X n )] i =1 n − 1 � (1 − M i / b ) 1 / n . ≤ b − ( b − M n ) i =1 Kais Hamza On the mixing advantage

Bounded independent case Kais Hamza On the mixing advantage

Bounded independent case Corollary In the case M i = M , i = 1 , . . . , n we have M ≤ E [max( X 1 , . . . , X n )] ≤ b − b (1 − M / b ) 2 − 1 / n where the latter expression approaches (2 − 1 / n ) M as b → + ∞ and M (2 − M / b ) as n → + ∞ . Kais Hamza On the mixing advantage

Bounded independent case Kais Hamza On the mixing advantage

Bounded independent case Changing X into b − X transforms maxima into minima immediately yielding the following result. Kais Hamza On the mixing advantage