Given X i i.i.d random variables, | X i | ≤ l for some l &gt; 0, p k = P ( X i = k ) &gt; 0 ∀ k.Let N n ( k ) = ∑ i = 1 n 1 X i = k and let ∏ n = ∏ k = 1 l ( p k ) ( N n ( k ) ) .Prove the sequence 1 n log ⁡ ∏ n converges almost surely and compute the limit.I try this using the 3 Series theorem but I am unsure how to define the sequence to prove a.s. convergence using the 3 series theorem. Breaking down the sequence I arrive at the following: A n = 1 n ∑ k = 1 l ∑ i = 1 n 1 X i = k log ⁡ ( p ( X 1 = k ) )I'm not sure how to create the sequence though to prove the 3 criteria of the 3 series theorem if that is what is required for the proof.

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Given       X    i   i.i.d random variables,       |        X    i        |    ≤  l for some   l  &amp;gt;  0,       p    k    =  P  (      X    i    =  k  )  &amp;gt;  0     ∀  k.Let       N    n    (  k  )  =      ∑          i      =      1              n            1                  X        i            =      k       and let       ∏    n    =      ∏          k      =      1        l    (      p    k        )          (              N        n            (      k      )      )      .Prove the sequence       1    n    log  ⁡      ∏    n   converges almost surely and compute the limit.I try this using the 3 Series theorem but I am unsure how to define the sequence to prove a.s. convergence using the 3 series theorem. Breaking down the sequence I arrive at the following:      A    n    =      1    n        ∑          k      =      1              l            ∑          i      =      1        n        1                  X        i            =      k        log  ⁡  (  p  (      X    1    =  k  )  )I&#039;m not sure how to create the sequence though to prove the 3 criteria of the 3 series theorem if that is what is required for the proof.

Arcatuert3u · Accepted Answer

In the proof I assumed that       X    i    &amp;gt;  0.Firstly, by boundness of       X    i   there exists   m  ≤  l such that  P  (      X    i    =  m  )  &amp;gt;  0  ,        P  (      X    i    =  m  +  1  )  =  0.Let&#039;s denote      ∏    n    =      p    1                  N        n            (      1      )        …      p    m                  N        n            (      m      )        ,now by property of log                              1          n                log        ⁡        (                  ∏          n                )                            =        log        ⁡        (                  p          1                                    N              n                        (            1            )                          /                        n                          …                  p          m                                    N              n                        (            m            )                          /                        n                          )                                          =                  ∑                      i            =            1                    m                log        ⁡        (                  p          i                                    N              n                        (            i            )                          /                        n                          )        .            For each   i  ∈  {  1  ,  2  ,  …  ,  m  }                                            lim                      n            →            ∞                          log        ⁡        (                  p          i                                    N              n                        (            i            )                          /                        n                          )                                          =        log        ⁡        (                  p          i                                    lim                              n                →                ∞                                                    N              n                        (            i            )                          /                        n                          )        .            By strong law of large numbers                                            lim                      n            →            ∞                                    N          n                (        i        )                  /                n                                          =                  lim                      n            →            ∞                                                              ∑                              k                =                1                            n                                      1                              {                                  X                  k                                =                                  p                  i                                }                                              n                =                  p          i                ,                                            a.s        .            Hence      lim          n      →      ∞            1    n    log  ⁡  (      ∏    n    )  =  log  ⁡  (      p    1                  p        1              …      p    m                  p        m              )  .

Ayanna Trujillo · Accepted Answer

I agree we need SLLN argument here for the       N    n    (  k  ) but can&#039;t we simply set each       X    i    =      1                  X        i            =      k       and thus   E      |        X    i        |    ≤  inf and where each       X    i   is i.i.d as stated and so by SLLN       N    n    (  k  )      /    n =       S    n        /    n =   E  (      S    n    )      /    n = n      p    k  /n =       p    k   and then we&#039;re done since the log sum is a finite sum of bounded terms?

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