Mbalisikerc

2022-07-17

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.

$x>0$ 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.

Monastro3n

Beginner2022-07-18Added 15 answers

Step 1

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 $i\le nx<i+1$, then $P\{{X}_{n}>nx\}={q}^{i}$. For $P\{{X}_{n}/n>x\}$ we have:

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

Step 2

Since ${(1-\frac{\lambda x}{i})}^{i}={e}^{-\lambda x}$ as $i\to \mathrm{\infty}$ 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 $i\to \mathrm{\infty}$. This reveals that ${X}_{n}/n$ is approximately an exponential random variable.

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 $i\le nx<i+1$, then $P\{{X}_{n}>nx\}={q}^{i}$. For $P\{{X}_{n}/n>x\}$ we have:

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

Step 2

Since ${(1-\frac{\lambda x}{i})}^{i}={e}^{-\lambda x}$ as $i\to \mathrm{\infty}$ 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 $i\to \mathrm{\infty}$. This reveals that ${X}_{n}/n$ is approximately an exponential random variable.

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