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Let $X_n$ be a geometric random variable with parameter $p=\lambda/n$. 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 $\lambda$. This shows that $X_n/n$ is approximately an exponential random variable.

So I know for a geometric r.v. the $P(X>k)=q^k$ and $q=1-p$. So I think that this means for $X_n$, $q=1-\lambda/n$. What is throwing me off is the $X_n/n$ part. I am not sure how this will affect the probability calculation. For $P(Y>x)$ with parameter $\lambda$ I know this equals $1-F(x)$ where $F(x)=1-e^{-\lambda x}$. Now how will this help me be able to show that $X_n$ converges to $P(Y>x)$?

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    Hint: $P\{X_n/n > x\} = P\{X_n > nx\} \approx \left(1-\frac{\lambda}{n}\right)^{nx} = \left(1-\frac{\lambda x}{nx}\right)^{nx} \approx \left(1-\frac{\lambda x}{m}\right)^{m}$. What can you say about the rightmost expression as $m \to \infty$ as it must do when $n \to \infty$?2012-09-26
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    @Dilip Sarwate- The $\frac{\lambda x}{m}$ would be 0 as $m \to \infty$? And does that leave you with $1^m$ as $m \to \infty$? Which would be 1?2012-09-26
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    Don't forget that $(1-\lambda x/m)$ is also being raised to the $m$-th power and _that_ $m$ in the exponent is also going to $\infty$. You can't let one of the $m$'s in $(1-\lambda x/m)^m$ go to $\infty$ and not the other. So try again.2012-09-26
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    $e^{-\lambda x}$.2012-09-26
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    Good! Now compare your answer to $P\{Y > x\} = 1 - F(x)$ where $F(x)$ is known to you. Then clean up your calculations a bit to take into account the fact that $nx$ is not necessarily an integer, and so you cannot write $P\{X_n > nx\} = q^{nx}$, you need to use something like $\lfloor nx\rfloor$ etc., and **post your own nicely-written solution as an answer to this question.** Such self-answers are absolutely encouraged on this site.2012-09-26
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    @Dilip Sarwate-So compare the fact that the limit of $P\{X_n/n > x\}$ equals $P(Y>x)$ and given the fact that the latter is an exponential r.v. this means that the former is approximately an exponential r.v.? I'm not sure what you mean by using $\lfloor nx\rfloor$ though. I havent seen that notation around variables before. Are those brackets?2012-09-27
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    Suppose $x$ is an irrational number. Then, $nx$ is not an integer for any choice of $n$. Now, the formula $P\{X>k\}=q^k$ for a geometric random variable assumes $k$ is an integer and cannot be used directly for $P\{X_n>nx\}$. But since $X_n$ is a geometric random variable and since the event $\{X_n>nx\}$ is the same as the event $\{X_n>i\}$ _where $i$ is the integer part of $nx$, that is, $i$ is the unique integer such that $i\leq nx,_ we can write that $P\{X_n>nx\}=q^i$, but not $P\{X_n>nx\}=q^{nx}$. It is conventional to use $\lfloor nx\rfloor$ to mean the integer part of $nx$.2012-09-27

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So here is what I think the answer is with help from Dilip Sarwate:

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 $P\{X_n > nx\}$ 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 $\{X_n>nx\}=\{X_n>i\}$ where $i$ is a unique integer such that $i\leq nxnx\}=q^i$. For $P\{X_n/n > x\}$ we have:

$$P\{X_n/n > x\} = P\{X_n > nx\} \approx \left(1-\frac{\lambda}{n}\right)^{nx} = \left(1-\frac{\lambda x}{nx}\right)^{nx} \approx \left(1-\frac{\lambda x}{i}\right)^{i}$$

Since $\left(1-\frac{\lambda x}{i}\right)^{i}=e^{-\lambda x}$ as $i \to \infty$ and the probability of the exponential random variable, $Y$, is $P\{Y > x\} = 1 - F(x)=e^{-\lambda x}$, this shows that $P\{X_n/n > x\}$ converges to $P\{Y > x\}$ as $i \to \infty$. This reveals that $X_n/n$ is approximately an exponential random variable.