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Assume that $N_k, k = 1,2,\ldots$ are independent random variables distributed as $\mathrm{Bin}(k,p)$, respectively. Let $X_1,X_2,\ldots$ be independent $\mathrm{Exp}(a)$-distributed random variables of $N_k, k = 1,2,\ldots$

Find the limit in probability of $Y_k = \dfrac{ \sum_{i=1}^k (X_i-a)}{N_k}$ as $k \rightarrow \infty $

All the examples I've found on this kind of questions have been pretty simple. But this one I can't figure out how to deal with.

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  • $\begingroup$ Just commenting to say Trout Mask Replica is an awesome album $\endgroup$ Mar 3 '16 at 19:42
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Hint :

$Y_k = \dfrac{ \sum_{i=1}^k (X_i-a)}{N_k}=\dfrac{ \sum_{i=1}^k (X_i-a)} k \dfrac k {N_k}$

And $N_k=\sum_{i=1}^k Z_i$ with $Z_i \sim B(1,p)$ (A Bernouilli variable).

Then use the weak version of a famous theorem in probability...

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  • $\begingroup$ So if I've got this right: $\frac{ \sum_{i=1}^k (X_i - a )}{k} \frac{k}{N_k}$ = $\frac{k(\frac{1}{a} -a)}{pk} $ = $\frac{\frac{1}{a} -a}{p} $ as $k \rightarrow \infty $ $\endgroup$
    – K B
    Mar 4 '16 at 10:21
  • $\begingroup$ In probability yes. It would be $\frac{ \sum_{i=1}^k (X_i - a )}{k} \frac{k}{N_k} \underset{\mathbb{P}}\to \frac{\frac{1}{a} -a}{p}$ $\endgroup$
    – nicomezi
    Mar 4 '16 at 15:15

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