Compute $P(X_n/n>x)$ for $x>0$ where $X_n$ is a geometric random variable with parameter $p=\lambda/n$ 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 $P(Y>x)$?  
 A: 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 $i\leq nx<i+1$, then $P\{X_n>nx\}=q^i$. 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.
