Let X n be a geometric random variable with parameter p = λ / n. Compute P ( X n / n &gt; x ) x &gt; 0 and show that as n approaches infinity this probability converges to P ( Y &gt; x ), where Y is an exponential random variable with parameter λ. This shows that X n / n is approximately an exponential random variable.

Question

Let       X    n   be a geometric random variable with parameter   p  =  λ      /    n. Compute   P  (      X    n        /    n  &amp;gt;  x  )  x  &amp;gt;  0 and show that as n approaches infinity this probability converges to   P  (  Y  &amp;gt;  x  ), where Y is an exponential random variable with parameter   λ. This shows that       X    n        /    n is approximately an exponential random variable.

Monastro3n · Accepted Answer

Step 1For a geometric random variable X, we have that   P  {  X  &amp;gt;  k  }  =      q    k   assuming that k is an integer. This can not be used directly for   P  {      X    n    &amp;gt;  n  x  } because x can be an irrational number. If this is the case, then nx is not an integer for any choice of n. However, since       X    n   is a geometric random variable and   {      X    n    &amp;gt;  n  x  }  =  {      X    n    &amp;gt;  i  } where i is a unique integer such that   i  ≤  n  x  &amp;lt;  i  +  1, then   P  {      X    n    &amp;gt;  n  x  }  =      q    i  . For   P  {      X    n        /    n  &amp;gt;  x  } we have:  P  {      X    n        /    n  &amp;gt;  x  }  =  P  {      X    n    &amp;gt;  n  x  }  ≈            (      1      −              λ        n            )              n      x        =            (      1      −                        λ          x                          n          x                    )              n      x        ≈            (      1      −                        λ          x                i            )              i      Step 2Since             (      1      −                        λ          x                i            )              i        =      e          −      λ      x       as   i  →  ∞ and the probability of the exponential random variable, Y, is   P  {  Y  &amp;gt;  x  }  =  1  −  F  (  x  )  =      e          −      λ      x      , this shows that   P  {      X    n        /    n  &amp;gt;  x  } converges to   P  {  Y  &amp;gt;  x  } as   i  →  ∞. This reveals that       X    n        /    n is approximately an exponential random variable.

Compute P(X_n/n>x) for x>0 where X_n is a geometric random variable with parameter p=lambda/n

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