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

Mbalisikerc 2022-07-17 Answered
Let X n be a geometric random variable with parameter p = λ / n. Compute P ( X n / n > x )
x > 0 and show that as n approaches infinity this probability converges to P ( Y > x ), where Y is an exponential random variable with parameter λ. This shows that X n / n is approximately an exponential random variable.
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Answers (1)

Monastro3n
Answered 2022-07-18 Author has 15 answers
Step 1
For a geometric random variable X, we have that P { X > k } = q k assuming that k is an integer. This can not be used directly for P { X n > 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 > n x } = { X n > i } where i is a unique integer such that i n x < i + 1, then P { X n > n x } = q i . For P { X n / n > x } we have:
P { X n / n > x } = P { X n > n x } ( 1 λ n ) n x = ( 1 λ x n x ) n x ( 1 λ x i ) i
Step 2
Since ( 1 λ x i ) i = e λ x as i and the probability of the exponential random variable, Y, is P { Y > x } = 1 F ( x ) = e λ x , this shows that P { X n / n > x } converges to P { Y > x } as i . This reveals that X n / n is approximately an exponential random variable.
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