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edited Apr 23, 2017 at 3:48

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Assuming that $K$ takes on values $1,2,3,\ldots$ with $P\{X=k\} = (1-\beta)^{k-1}\beta$, $k > 0$, and not $(1-\beta)^{k}\beta$ as the problem states, then $P\{K > k\} = (1-\beta)^{k}$, either by

recognizing that $K$ is the number of repeated independent trials to have an event of probability $\beta$ occur for the first time, and so, $K > k$ if and only if the event did not occur on the first $k$ trials

or by

brute-force adding up \begin{align}P\{X \leq k\} &= \sum_{i=1}^k P\{K = i\}\\ &= \sum_{i=1}^k (1-\beta)^{i-1}\beta\\ &= \beta\big(1 + (1-\beta) + (1-\beta)^2 + \cdots + (1-\beta)^{k-1}\big)\\ &= \beta\cdot \frac{1-(1-\beta)^k}{1 -(1-\beta)}\\ &= 1-(1-\beta)^k \end{align} and so, $P\{K > k\} = 1 - P\{X \leq k\} = (1-\beta)^k$, as before.

$K$ is called a geometric random variable with parameter $\beta$.

For $n \geq 2$, let $K_n$ denote a geometric random variable with parameter $\frac 1n$ and define $X_n = \frac 1n K_n$. Note that we can think of $X_n$ as $\beta K$ where $K$ is a geometric random variable with parameter $\beta = \frac 1n$. We have that for any fixed positive real number $x$ $$P\{X_n > x\} = P\left\{\frac 1n K_n > x\right\} = P\{K_n > nx\}\approx \left(1 - \frac 1n\right)^{nx},$$ that is, $$\lim_{n\to\infty} P\{X_n > x\} = \lim_{n\to\infty}F_{X_n}(x)= e^{-x}.$$$$\lim_{n\to\infty} P\{X_n > x\} = \lim_{n\to\infty}1 - F_{X_n}(x)= e^{-x}.$$ The sequence of random variables $X_n$ is thus converging in distribution to an exponential random variable with parameter $1$.

Assuming that $K$ takes on values $1,2,3,\ldots$ with $P\{X=k\} = (1-\beta)^{k-1}\beta$, $k > 0$, and not $(1-\beta)^{k}\beta$ as the problem states, then $P\{K > k\} = (1-\beta)^{k}$, either by

recognizing that $K$ is the number of repeated independent trials to have an event of probability $\beta$ occur for the first time, and so, $K > k$ if and only if the event did not occur on the first $k$ trials

or by

brute-force adding up \begin{align}P\{X \leq k\} &= \sum_{i=1}^k P\{K = i\}\\ &= \sum_{i=1}^k (1-\beta)^{i-1}\beta\\ &= \beta\big(1 + (1-\beta) + (1-\beta)^2 + \cdots + (1-\beta)^{k-1}\big)\\ &= \beta\cdot \frac{1-(1-\beta)^k}{1 -(1-\beta)}\\ &= 1-(1-\beta)^k \end{align} and so, $P\{K > k\} = 1 - P\{X \leq k\} = (1-\beta)^k$, as before.

$K$ is called a geometric random variable with parameter $\beta$.

For $n \geq 2$, let $K_n$ denote a geometric random variable with parameter $\frac 1n$ and define $X_n = \frac 1n K_n$. Note that we can think of $X_n$ as $\beta K$ where $K$ is a geometric random variable with parameter $\beta = \frac 1n$. We have that for any fixed positive real number $x$ $$P\{X_n > x\} = P\left\{\frac 1n K_n > x\right\} = P\{K_n > nx\}\approx \left(1 - \frac 1n\right)^{nx},$$ that is, $$\lim_{n\to\infty} P\{X_n > x\} = \lim_{n\to\infty}F_{X_n}(x)= e^{-x}.$$ The sequence of random variables $X_n$ is thus converging in distribution to an exponential random variable with parameter $1$.

Assuming that $K$ takes on values $1,2,3,\ldots$ with $P\{X=k\} = (1-\beta)^{k-1}\beta$, $k > 0$, and not $(1-\beta)^{k}\beta$ as the problem states, then $P\{K > k\} = (1-\beta)^{k}$, either by

recognizing that $K$ is the number of repeated independent trials to have an event of probability $\beta$ occur for the first time, and so, $K > k$ if and only if the event did not occur on the first $k$ trials

or by

brute-force adding up \begin{align}P\{X \leq k\} &= \sum_{i=1}^k P\{K = i\}\\ &= \sum_{i=1}^k (1-\beta)^{i-1}\beta\\ &= \beta\big(1 + (1-\beta) + (1-\beta)^2 + \cdots + (1-\beta)^{k-1}\big)\\ &= \beta\cdot \frac{1-(1-\beta)^k}{1 -(1-\beta)}\\ &= 1-(1-\beta)^k \end{align} and so, $P\{K > k\} = 1 - P\{X \leq k\} = (1-\beta)^k$, as before.

$K$ is called a geometric random variable with parameter $\beta$.

For $n \geq 2$, let $K_n$ denote a geometric random variable with parameter $\frac 1n$ and define $X_n = \frac 1n K_n$. Note that we can think of $X_n$ as $\beta K$ where $K$ is a geometric random variable with parameter $\beta = \frac 1n$. We have that for any fixed positive real number $x$ $$P\{X_n > x\} = P\left\{\frac 1n K_n > x\right\} = P\{K_n > nx\}\approx \left(1 - \frac 1n\right)^{nx},$$ that is, $$\lim_{n\to\infty} P\{X_n > x\} = \lim_{n\to\infty}1 - F_{X_n}(x)= e^{-x}.$$ The sequence of random variables $X_n$ is thus converging in distribution to an exponential random variable with parameter $1$.

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created Apr 22, 2017 at 22:49

Dilip Sarwate

47.8k
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235

Assuming that $K$ takes on values $1,2,3,\ldots$ with $P\{X=k\} = (1-\beta)^{k-1}\beta$, $k > 0$, and not $(1-\beta)^{k}\beta$ as the problem states, then $P\{K > k\} = (1-\beta)^{k}$, either by

recognizing that $K$ is the number of repeated independent trials to have an event of probability $\beta$ occur for the first time, and so, $K > k$ if and only if the event did not occur on the first $k$ trials

or by

brute-force adding up \begin{align}P\{X \leq k\} &= \sum_{i=1}^k P\{K = i\}\\ &= \sum_{i=1}^k (1-\beta)^{i-1}\beta\\ &= \beta\big(1 + (1-\beta) + (1-\beta)^2 + \cdots + (1-\beta)^{k-1}\big)\\ &= \beta\cdot \frac{1-(1-\beta)^k}{1 -(1-\beta)}\\ &= 1-(1-\beta)^k \end{align} and so, $P\{K > k\} = 1 - P\{X \leq k\} = (1-\beta)^k$, as before.

$K$ is called a geometric random variable with parameter $\beta$.

For $n \geq 2$, let $K_n$ denote a geometric random variable with parameter $\frac 1n$ and define $X_n = \frac 1n K_n$. Note that we can think of $X_n$ as $\beta K$ where $K$ is a geometric random variable with parameter $\beta = \frac 1n$. We have that for any fixed positive real number $x$ $$P\{X_n > x\} = P\left\{\frac 1n K_n > x\right\} = P\{K_n > nx\}\approx \left(1 - \frac 1n\right)^{nx},$$ that is, $$\lim_{n\to\infty} P\{X_n > x\} = \lim_{n\to\infty}F_{X_n}(x)= e^{-x}.$$ The sequence of random variables $X_n$ is thus converging in distribution to an exponential random variable with parameter $1$.