Probability question that links between geometric and exponential distribution with limits.Let λ &gt; 0, and for every n ≥ 1, we have X n ∼ Geo ⁡ ( λ n ).We define X ^ n = 1 n X n . Show that for every t ≥ 0: F X ^ n ( t ) → F X ^ ( t ) when n → ∞ .. while X ^ ∼ Exp ⁡ ( λ ).

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Probability question that links between geometric and exponential distribution with limits.Let   λ  &amp;gt;  0, and for every   n  ≥  1, we have       X    n    ∼  Geo  ⁡  (      λ    n    ).We define                     X        ^              n    =      1    n        X    n  . Show that for every   t  ≥  0:      F                                        X            ^                          n              (  t  )  →      F                            X          ^                      (  t  ) when   n  →  ∞  .. while             X      ^        ∼  Exp  ⁡  (  λ  ).

canhaulatlt · Accepted Answer

Step 1Answer:       P    [                    X        n            ^        ≤  t  ]  =      P    [      X    n    ≤  n  ⋅  t  ]  =      P    [      X    n    ≤  ⌊  n  ⋅  t  ⌋  ]  =  1  −  (  1  −      λ    n        )          ⌊      n      ⋅      t      ⌋        →  1  −      e          −      λ      t        =      F                            X          ^                      (  t  )Step 2Explanation:The first equality comes from the definition of                     X        n            ^      . For the second equality we require the floor function, ⌊⋅⌋, this is because, as I mentioned in my comment the Geometric distribution takes only integer values. What is the probability of rolling less than or equal to 4.5 on a dice? The same as rolling less or equal to   ⌊  4.5  ⌋  =  4. This is simply the CDF of a geometric RV with rate       λ    n  . For the limit we use:       lim          n      →      ∞        (  1  +      x    n        )    n    =      e    x  . Recognition of the CDF of Exponential CDF, which I see you have already done.

Chloe Arnold · Accepted Answer

Step 1Let   t  &amp;gt;  0 then,      F                                        X            n                    ^                      (  t  )  =  P  (      X    n        /    n  ≤  t  )  =  P  (      X    n    ≤  t  n  )  =      ∑          k      =      1              ⌊              t        n            ⌋        P  (      X    n    =  k  )  =      p    n        ∑          k      =      1              ⌊              t        n            ⌋            q    n          k      −      1      Where       p    n    =            λ      n       and       q    n    =  1  −      p    n  .Therefore,       F                                        X            n                    ^                      (  t  )  =      p    n                      1        −                  q          n                      ⌊                          t              n                        ⌋                                      1        −                  q          n                      =  1  −      q    n          ⌊              t        n            ⌋      Step 2Now,       q    n          ⌊      t      n      ⌋        =            (      1      −                        λ          n                    )              ⌊      t      n      ⌋        =  exp  ⁡  (  ⌊  t  n  ⌋  ln  ⁡  (  1  −  λ      /    n  )  )But as   t  &amp;gt;  0 we have,  ⌊  t  n  ⌋  ∼  t  nand   ln  ⁡  (  1  −  λ      /    n  )  ∼  −  λ      /    nTherefore,   ⌊  t  n  ⌋  ln  ⁡  (  1  −  λ      /    n  )  ∼  −  λ  tHence,   ⌊  t  n  ⌋  ln  ⁡  (  1  −  λ      /    n  )  →  −  λ  tBy continuity of exp we have,       q    n          ⌊      t      n      ⌋        →            e              −      λ      t      Therefore,       F                                        X            n                    ^                      (  t  )  →  1  −            e              −      λ      t        =      F                            X          ^                      (  t  )This is also true for   t  =  0, indeed       F                                        X            n                    ^                      (  0  )  =  0  =  1  −  exp  ⁡  (  −  λ  0  )  =      F                            X          ^                      (  0  )

Let lambda>0, and for every n ge 1, we have X_n∼Geo(lambda/n). We define hatX_n=1/n X_n. Show that for every t ge 0: F_{hatX_n}(t) rightarrow F_hatX(t) when n rightarrow infty. while hatX∼Exp(lambda).

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