Basic probability distribution basic Concept of probability distributions 3:geometric distribution

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PMF

Suppose that independent trials, each having a probability $p $, $ < P < 1$, of being a success, is performed until A success occurs. If we let $X $ equal the number of failures required and then the geometric distribution mass function is $ $f (X; p) =\pr (X=X) = (1-p) ^{x}p$$ for $x =0, 1, 2, \cdots$.

Proof:

$$ \begin{align*} \sum_{x=0}^{\infty}f (x; p) &= \sum_{x=0}^{\infty} (1-p) ^{x}p\\ &= p\sum_{x=0}^{\infty} (1-p) ^ {X}\\ & = P\cdot {1\over (1-p)}\\ & = 1 \end{align*} $$

Mean

The expected value is $$\mu = e[x] = {1-p\over p}$$

Proof:

Firstly, we know that $$\sum_{x=0}^{\infty}p^x = {1\over 1-p}$$ where < P < 1$. Thus $$ \begin{align*} {d\over dp}\sum_{x=0}^{\infty}p^x &= \sum_{x=1}^{\infty}xp^{x-1}\\ &= {1\over (1-p) ^2} \ end{align*} $$ The expected value is $$ \begin{align*} e[x] &= \sum_{x=0}^{\infty}x (1-p) ^{x}p\\ &=p (1-p) \sum_{x=1 }^{\infty}x (1-p) ^{x-1}\\ &= p (1-p) {1\over (n (1-p)) ^2}\\ &= {1-p\over p} \end{align*} $$

Variance

The variance is $$\sigma^2 = \mbox{var} (X) = {1-p\over p^2}$$

Proof:

$$ \begin{align*} E\left[x^2\right] &=\sum_{x=0}^{\infty}x^2 (1-p) ^{x}p\\ &= (1-p) \sum_{x=1}^{\infty}x^2 (1-p ) ^{x-1}p \end{align*} $$ Rewrite the right hand summation as $$ \begin{align*} \sum_{x=1}^{\infty} x^2 (1-p) ^{x-1}p&= \ Sum_{x=1}^{\infty} (x-1+1) ^2 (1-p) ^{x-1}p\\ &= \sum_{x=1}^{\infty} (x-1) ^2 (1-p) ^{x-1}p + \sum_{x=1}^{\infty} 2 (x-1 ) (1-p) ^{x-1}p + \sum_{x=1}^{\infty} (1-p) ^{x-1}p\\ &= E\left[x^2\right] + 2e[x] + 1\\ &= E\left[X^2\right] + {2-p\ Over P} \end{align*} $$ Thus $ $E \left[x^2\right] = (1-p) E\left[x^2\right] + {(1-p) (2-p) \over p}$$ This is $ $E \left[x^2\ri ght]= {(1-p) (2-p) \over p^2}$$ so the variance is $$ \begin{align*} \mbox{var} (X) &= E\left[x^2\right]-e[x]^2\\ & = {(1-p) (2-p) \over p^2}-{(1-p) ^2\over p^2}\\ &= {1-p\over p^2} \end{align*} $$

Examples

1. Let $X $ is geometrically distributed with probability parameter $p ={1\over2}$. Determine the expected value $\mu$, the standard deviation $\sigma$, and the probability $P \left (| x-\mu| \geq 2\sigma\right) $. Compare with Chebyshev ' s inequality.

Solution:

The geometric distribution mass function is $ $f (x; p) = (1-p) ^{x}p,\ x=0, 1, 2, \cdots$$ the expected value is $$\mu = {1- P\over P} = 1$$ the standard deviation are $$\sigma = \sqrt{1-p\over p^2} = 1.414214$$ the probability that $X $ takes a Val UE more than-deviations from $\mu$ is $ $P \left (| x-1| \geq 2.828428\right) = P (x\geq 4) = 0.0625$$ R code:

Chebyshev ' s inequality gives the weaker estimation $ $P \left (| X-\mu| \geq 2\sigma\right) \leq {1\over4} = 0.25$$

2. A die is thrown until one gets a 6. Let $V $ is the number of throws used. What is the expected value of $V $? What is the variance of $V $?

Solution:

The PMF of geometric distribution is $ $f (x; p) = (1-p) ^xp,\ = 0, 1, 2, \cdots$$ where $p = {1\over 6}$. Let $X = v-1$, so the expected value of $V $ is $$ \begin{align*} e[v] &= e[x+1]\\ &= e[x] + 1\\ &= {1-p\over p } + 1\\ &= {1-{1\over6} \over {1\OVER6}} + 1\\ &= 6 \end{align*} $$ the variance of $V $ is $$ \begin{align*} \mbox {Var} (V) &= \mbox{var} (x+1) \ &= \mbox{var} (X) \ &= {1-p\over p^2}\\ &= {1-{1\over 6} \over \left ({1\over6}\rig HT) ^2}\\ &= \end{align*} $$ Note that this is another form of the geometric distribution which is so-called the Shi fted geometric distribution (i.e. $X $ equals to the number of trials required). By the above process we can see the expected value of the shifted geometric distribution are $$\mu = {1\over p}$$ and The variance of the shifted geometric distribution is $$\sigma^2 = {1-p\over p^2}$$

3. Assume $W $ is geometrically distributed with probability parameter $p $. What is $P (W < n) $?

Solution:

$$ \begin{align*} P (W < N) &= 1-p (w \geq n) \ &= (1-p) ^n \end{align*} $$

4. In order to test whether a given die is fair, it's thrown until a 6 appears, and the number $n $ of throws is counted. How great should $n $ being before we can reject the null hypothesis $ $H _0: \mbox{the die is fair}$$ against the alternative H Ypothesis $ $H _1: \mbox{the probability of having a 6 are less than 1/6}$$ at significance level $5\%$?  

Solution:

The probability of have to use at least $n $ throws given $H _0$ (i.e. the significance probability) is $ $P = \left (1-{1 \over 6}\right) ^n$$ We'll reject $H _0$ if $P < 0.05$. R Code:

n = 1while (n > 0) {+   p = (5/6) ^ n+   if (P < 0.05) break+   

That's, we have the to reject $H _0$ if $n $ are at least 17.

Reference

1. Ross, S. (2010). A first Course in probability (8th Edition). Chapter 4. Pearson. Isbn:978-0-13-603313-4.
2. Brink, D. (2010). Essentials of Statistics:exercises. Chapter 5 & 10. isbn:978-87-7681-409-0.

Basic probability distribution basic Concept of probability distributions 3:geometric distribution