# 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 $P\left({X}_{n}/n>x\right)$
$x>0$ and show that as n approaches infinity this probability converges to $P\left(Y>x\right)$, where Y is an exponential random variable with parameter $\lambda$. This shows that ${X}_{n}/n$ is approximately an exponential random variable.
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Monastro3n
Step 1
For a geometric random variable X, we have that $P\left\{X>k\right\}={q}^{k}$ assuming that k is an integer. This can not be used directly for $P\left\{{X}_{n}>nx\right\}$ 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 $\left\{{X}_{n}>nx\right\}=\left\{{X}_{n}>i\right\}$ where i is a unique integer such that $i\le nx, then $P\left\{{X}_{n}>nx\right\}={q}^{i}$. For $P\left\{{X}_{n}/n>x\right\}$ we have:
$P\left\{{X}_{n}/n>x\right\}=P\left\{{X}_{n}>nx\right\}\approx {\left(1-\frac{\lambda }{n}\right)}^{nx}={\left(1-\frac{\lambda x}{nx}\right)}^{nx}\approx {\left(1-\frac{\lambda x}{i}\right)}^{i}$
Step 2
Since ${\left(1-\frac{\lambda x}{i}\right)}^{i}={e}^{-\lambda x}$ as $i\to \mathrm{\infty }$ and the probability of the exponential random variable, Y, is $P\left\{Y>x\right\}=1-F\left(x\right)={e}^{-\lambda x}$, this shows that $P\left\{{X}_{n}/n>x\right\}$ converges to $P\left\{Y>x\right\}$ as $i\to \mathrm{\infty }$. This reveals that ${X}_{n}/n$ is approximately an exponential random variable.