Let m,n be natural numbers and let α i j be reals. For each i = 1 , 2 , . . . , m; j = 1 , 2 , . . . , n. Show that: ∏ j = 1 n ( 1 − ∏ i = 1 m cos 2 ⁡ α i j ) + ∏ i = 1 m ( 1 − ∏ j = 1 n sin 2 ⁡ α i j ) ≥ 1

Question

Let m,n be natural numbers and let       α          i      j       be reals. For each   i  =  1  ,  2  ,  .  .  .  ,  m;   j  =  1  ,  2  ,  .  .  .  ,  n. Show that:            ∏              j        =        1                    n                    (      1      −              ∏                  i          =          1                          m                            cos        2            ⁡              α                  i          j                    )        +          ∏              i        =        1                    m                    (      1      −              ∏                  j          =          1                          n                            sin        2            ⁡              α                  i          j                    )        ≥    1

Asa Austin · Accepted Answer

Let       X          i      j       be independent random variables with  P  (      X          i      j        =  0  )  =      cos    2    ⁡      α          i      j        ,  P  (      X          i      j        =  1  )  =      sin    2    ⁡      α          i      j        .Let A be the event   (  ∃  i  )  (  ∀  j  )  (      X          i      j        =  1  ) and B the event   (  ∃  j  )  (  ∀  i  )  (      X          i      j        =  0  ). These are mutually exclusive, so  P  (  A  ∩  B  )  =  0.We have                    P        (        A        )                            =                    1        −        P        (        ¬        A        )                                          =                    1        −        P                  (          (          ∀          i          )          (          ∃          j          )          (                      X                          i              j                                =          0          )          )                                                  =                    1        −                  ∏          i                P                  (          (          ∃          j          )          (                      X                          i              j                                =          0          )          )                                                  =                    1        −                  ∏          i                          (          1          −          P          (          (          ∀          j          )          (                      X                          i              j                                =          1          )          )          )                                                  =                    1        −                  ∏          i                          (          1          −                      ∏            j                    P          (                      X                          i              j                                =          1          )          )                                                  =                    1        −                  ∏          i                          (          1          −                      ∏            j                                sin            2                    ⁡                      α                          i              j                                )                .            Similarly  P  (  B  )  =  1  −      ∏    j        (    1    −          ∏      i              cos      2        ⁡          α              i        j              )    .Hence      ∏    j        (    1    −          ∏      i              cos      2        ⁡          α              i        j              )    +      ∏    i        (    1    −          ∏      j              sin      2        ⁡          α              i        j              )    =  2  −  P  (  A  )  −  P  (  B  )  .But   1  ≥  P  (  A  ∪  B  )  =  P  (  A  )  +  P  (  B  )  −  P  (  A  ∩  B  )  =  P  (  A  )  +  P  (  B  ), so the RHS is   ≥  1

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