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).

Danika Mckay 2022-10-25 Answered
Probability question that links between geometric and exponential distribution with limits.
Let λ > 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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Answers (2)

canhaulatlt
Answered 2022-10-26 Author has 17 answers
Step 1
Answer: 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 2
Explanation:
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.
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Chloe Arnold
Answered 2022-10-27 Author has 6 answers
Step 1
Let t > 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 2
Now, q n t n = ( 1 λ n ) t n = exp ( t n ln ( 1 λ / n ) )
But as t > 0 we have,
t n t n
and ln ( 1 λ / n ) λ / n
Therefore, t n ln ( 1 λ / n ) λ t
Hence, t n ln ( 1 λ / n ) λ t
By 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 )
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