Given ( X i ) i = 1 ∞ a series of independent random variables, X i ∼ G ( 1 2 ) (Geometric distribution) for all i ≥ 1.For n ≥ 2, we mark the following: X ¯ n = 1 n ∑ i = 1 n X i Y n = ln ⁡ ( 1 + ( X ¯ n ) 2 ) T n = ( Y n − ln ⁡ 5 ) 2 And we need to find the value C n and the distribution of T, which satisfies: 1 C n ∑ n = 1 12 T n → d TWell, i'll save you the trouble. The final and correct answer is C n = 32 25 n , T n ∼ χ ( 12 ) 2 (chi distribution with 12 degrees of freedom)But i can't understand why. This is all I know so far:Let ( Z i ) i = 1 n be a series of independent random variable with the same distribution Z i ∼ N ( μ , σ 2 ), then we know that Z i − μ σ ∼ N ( 0 , 1 ) for all i, and we know that the distribution of the following sum is: ∑ i = 1 n ( Z i − μ σ ) 2 ∼ χ ( n ) 2 My guess is that in my question: Z i := Y n = ln ⁡ ( 1 + ( X ¯ n ) 2 ) μ := E ( Y n ) = ln ⁡ 5 σ 2 := V ( Z i ) = C n and thus: ∑ i = 1 n ( Z i − μ σ ) 2 = ∑ i = 1 n ( Z i − μ ) 2 σ 2 := ∑ n = 1 12 T n C n → d TBut i can't understand how is that E ( Y n ) = ln ⁡ 5, or even what is the distribution of Y n ?Am I on the right track? What are your thoughts about this question? How would you solve it?

Question

Given             (            X    i                      )                    i      =      1              ∞       a series of independent random variables,       X    i    ∼  G  (            1      2        ) (Geometric distribution) for all   i  ≥  1.For   n  ≥  2, we mark the following:                    X        ¯              n    =      1    n        ∑          i      =      1              n            X    i              Y    n    =  ln  ⁡            (        1  +  (                    X        ¯              n        )    2              )                  T    n    =  (      Y    n    −  ln  ⁡  5      )    2  And we need to find the value       C    n   and the distribution of T, which satisfies:      1          C      n            ∑          n      =      1              12            T    n        →    d    TWell, i&#039;ll save you the trouble. The final and correct answer is      C    n    =      32          25      n        ,        T    n    ∼      χ          (      12      )        2     (chi distribution with 12 degrees of freedom)But i can&#039;t understand why. This is all I know so far:Let             (            Z    i                      )                    i      =      1              n       be a series of independent random variable with the same distribution       Z    i    ∼  N  (  μ  ,      σ    2    ), then we know that                     Z        i            −      μ        σ    ∼  N  (  0  ,  1  ) for all i, and we know that the distribution of the following sum is:       ∑          i      =      1        n              (                          Z        i            −      μ        σ                      )              2    ∼      χ          (      n      )        2  My guess is that in my question:      Z    i    :=      Y    n    =  ln  ⁡            (        1  +  (                    X        ¯              n        )    2              )        μ  :=  E  (      Y    n    )  =  ln  ⁡  5      σ    2    :=  V  (      Z    i    )  =      C    n  and thus:      ∑          i      =      1        n              (                          Z        i            −      μ        σ                      )              2    =                    ∑                  i          =          1                n            (              Z        i            −      μ              )        2                            σ            2        :=                    ∑                  n          =          1                          12                            T        n                    C      n            →    d    TBut i can&#039;t understand how is that   E  (      Y    n    )  =  ln  ⁡  5, or even what is the distribution of       Y    n  ?Am I on the right track? What are your thoughts about this question? How would you solve it?

Stella Andrade · Accepted Answer

Step 1There is the so called delta-method. In this case,   f  (  x  )  =  log  ⁡  (  1  +      x    2    ) and since we have       n    ∗  (                    X        n            ¯        −  2  ) converges to N(0,2),Step 2      n    ∗  (  f  (                    X        n            ¯        )  −  f  (  2  )  ) converges to   N  (  0  ,  2  ∗  (      f    ′    (  2  )      )    2    )  =  N  (  0  ,  32      /    25  )From this,             T              n                    C              n             converges to a chi-squared distribution.

Given (X_i)_{i=1}^{infty} a series of independent random variables, X_i ~ G(1/2) (Geometric distribution) for all i >= 1.

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