# Discrete Distributions¶

A discrete distribution is a distribution of data in statistics that has discrete values. Discrete values are countable, finite, non-negative integers, such as 1, 10, 15, etc.

## Bernoulli Distribution¶

The Bernoulli distribution is a univariate discrete distribution used to model random experiments that have binary outcomes.

### Bernoulli Random Variable¶

A Bernoulli RV $$X \sim Bern(p)$$ is a random variable that is either 0 or 1 with probability $$p$$ or $$1-p$$ respectively. Suppose that you perform an experiment with two possible outcomes: either success or failure.

Let X be a discrete random variable. $$x \in {0,1}$$

$\begin{split} f_x(x)=P(X=x)=\begin{cases} 1-p, & \text{ if x = 0 } \\ p, & \text{if x = 1 } \\ 0, & \text{otherwise} \end{cases} \end{split}$

#### P.M.F¶

\begin{split} \begin{align} \large P(X=1) &=p \\ \large P(X=0)&=1-p \end{align} \end{split}

Using the indicator function notation

$$I_{A}(x)= \begin{cases}1, & \text { if } x \in A \\ 0, & \text { if } x \notin A\end{cases}$$

$P(X=x)=p^{x}(1-p)^{1-x} \cdot I_{\{0,1\}}(x)$

### Mean (Expected Value)¶

The expected value of a Bernoulli random variable X is

$\large E[X]=p$

Proof

\begin{split} \begin{align} \large \\ E[X] &=\sum_{x} k P(X=x) \\ &= 0 * P(x=0) + 1 * P(x=1) \\ &= 0 * (1-p) + 1 * (p) \\ &= p \end{align} \end{split}

### Variance¶

The variance of a Bernoulli random variable X is

$Var[X] = p(1-p)$

#### Proof¶

$$E(X^2) =\sum_{k} k^2 P(X=k) = 1^2 * p = p$$

$\begin{split} V(X) &= {E}[X^2] - {E}[X]^2 \\ &= p - p^2 \\ &= p(1-p) \end{split}$

## Geometric Distribution¶

The geometric distribution is a discrete probability distribution that calculates the probability of the first success occurring during a specific trial.

The geometric distribution is the probability distribution of the number of failures we get by repeating a Bernoulli experiment until we obtain the first success.

### Geometric Random Variable¶

Definition 1

A geometric rv $$X \sim Geom(p)$$ consists of

• independent Bernoulli trials,

• each with the same probability of success p or Failure (1-p),

• repeated until the first success is obtained.

Definition 2

The geometric rv is the distribution of the number of trials needed to get the first success in repeated independent Bernoulli trials.

Example If we toss a coin until we obtain head, the number of tails before the first head has a geometric distribution.

Attention

It can also define as number of failures before first success.

#### Parameter¶

The geometric distribution has one parameter, p = the probability of success for each trial. You denote the distribution as G(p), which indicates a geometric distribution with a success probability of p.

### Uses¶

• Six in a series of die rolls?

• Person to support a law during a repeated sampling for an interview?

• Product to have a defect in a random sample from an assembly line?

• Successful attempt for a project or task?

### Properties¶

1. Each trial is identical, and can result in a success or failure.

2. The probability of success, p, is constant from one trial to the next.

3. The trials are independent, so the outcome on any particular trial does not influence the outcome of any other trial.

4. Trials are repeated until the first success.

### P.M.F¶

The sample space of geometric random variable is

$S = \{1,01,001,0001,00001,000001,\dots\}$

Bernoulli trail success = 1 = P
Bernoulli trail failure = 0 = 1 - P

$$P(X=1)=p$$
$$P(X=2)=(1-p)p$$
$$P(X=3)=(1-p)(1-p)p$$
$$P(X=4)=(1-p)(1-p)(1-p)p$$ = failure, failure, failure, Success
$$P(X=5)=(1-p)^{4}p$$
$$P(X=x)=(1-p)^{x-1}p$$

$\begin{split} f(x) = P(X=x)&=(1-p)^{x-1}p \quad for \enspace x = {1,2,3,4,5,\dots} \\ f(x) =P(X=x)&=(1-p)^{x-1} \cdot p \cdot I_{\{1,2,3, \ldots\}}(x) \end{split}$

### Mean (Expected Value)¶

The expected value of a geometric random variable X is

$\begin{split} E[X] &= \sum_{k=1}^{\infty} k P(X=k) \\ &= \sum_{k=1}^{\infty} k (1-p)^{k-1}p \\ &= \frac{1} p \end{split}$

### Variance¶

The expected value of a geometric random variable X is

$\begin{split} V(X) &= E[X^2] - E[X]^2 \\ &= \frac{1-p}{p^{2}} \end{split}$

Interview Question

Q: On each day we play a lottery in which the probability of winning is $$1%$$.
What is the expected value of the number of days that will elapse before we win for the first time?

Answer: Each time we play the lottery, the outcome is a Bernoulli random variable (equal to 1 if we win), with parameter $$p=0.01$$. Therefore, the number of days before winning is a geometric random variable with parameter $$p=0.01$$. Its expected value is

$E[X] = \frac{1} p = \frac{1} {0.01} = 100$

## Binomial Distribution¶

The binomial distribution is a discrete probability distribution that calculates the probability an event will occur a specific number of times in a set number of opportunities.

### Binomial Random Variable¶

Definition 1

A binomial rv $$X \sim Bin(n,p)$$ is a random variable that is the number of successes in n independent Bernoulli trials, each with probability p. The probability of success is p. The probability of failure is 1-p. The number of trials is n.

Definition 2

The binomial distribution is the distribution of the number of successes = X in a fixed number = n of independent Bernoulli trials.

#### Parameters¶

The binomial distribution has two parameters, n and p.

• n: the number of trials.

• p: the event or success probability.

### Uses¶

Use the binomial distribution when your outcome is binary. Binary outcomes have only two possible values that are mutually exclusive.

• Six heads when you toss the coin ten times?

• 12 women in a sample size of 20?

• Three defective items in a batch of 100?

• Two flu infections over 20 years?

### Properties¶

1. Experiment is n trials (n is fixed in advance)

2. Trials are identical and result in a success or a failure (i.e. Bernoulli trials) with P(success) = p and P(failure) = 1 - p.

3. Trials are independent (outcome of one trial does not influence any other)

### P.M.F¶

The sample space of binomial random variable is

$\begin{split} S = \left\{\left(x_{1}, x_{2}, \ldots, x_{n}\right) \mid x_{i}\right. =\left\{\begin{array}{l} 1 \text { if } \text { success } \\ 0 \text { if failure }\end{array}\right. \end{split}$

$$f(x)=P(X=0)=P(\{00 \cdots 0\})=(1-p)^{n}$$
$$f(x)=P(X=1)=P(\{10 \cdots 0,0100 \ldots,0 \cdots 01\}) = n*p*(1-p)^{n-1}$$
$$f(x)=P(X=2)=P(\{11 \cdots 0,0110 \ldots,00 \cdots 11\}) = \binom{n}{2}p^2(1-p)^{n-2}$$

Explanation P(X=2): Among n number of fixed trials, we have 2 bernoulli trials successes with probability P and rest are failures bernoulli trails with probability (1-p). So, we need to choose 2 from n to get the exact probability of success.

$f(x)=P(X=x)= \binom{n}{x}p^x(1-p)^{n-x} \cdot I_{\{1,2,3, \ldots\}}(x)$

Where k = 1 (success) and n-k = 0 (failure).

Suppose n = 4

$$\mathrm{P}(X=3)=\mathrm{P}(\mathrm{SSSF} \text { or } \mathrm{SSFS} \text { or SFSS or FSSS })$$

### Binomial Theorem¶

$$\sum_{k=0}^n {n \choose k}p^{k}(1-p)^{n-k} = 1$$

### Mean (Expected Value)¶

The expected value of a binomial random variable X is

$\begin{split} E[X] &=\sum_{k} k P(X=k) \\ &=\sum_{k=0}^n k {n \choose k}p^{k}(1-p)^{n-k} \\ &= n * p \end{split}$

Proof

$\begin{split} \begin{array}{ll} =\mathrm{E}\left[\sum_{i=1}^{n} Y_{i}\right] \quad \text { (representation as a sum of } n \text { independent Bernoulli r.v.) } \\ =\sum_{i=1}^{n} \mathrm{E}\left[Y_{i}\right] \quad \text { (linearity of the expected value) } \\ =\sum_{i=1}^{n} p \quad \quad \text { (expected value of a Bernoulli r.v.) } \\ =n p \end{array} \end{split}$

RECALL: Bern(p) has expected value p. x1, x2 … xn are independent bern p. so $$sum_{k=1}^n X_n = sum_{k=1}^n E[X_n] = n * p$$

### Variance¶

The variance of a binomial random variable X is

$$V(X)= E(X^2) - E(X)^2 = n * p * (1-p)$$

Recall: Bern(p) has variance p * (1-p).

## Negative Binomial Distribution¶

The negative binomial distribution is almost the same as a binomial distribution with one difference

• Binomial distribution has a fixed number of trials.

Repeat independent Bernoulli trials until a total of r successes is obtained. The negative binomial random variable X counts the number of failures before the rth success.

### Negative Binomial Random Variable¶

The negative binomial rv $$X \sim NB(r,p)$$ is the distribution of the number of trials = X needed to get a fixed number of successes = r.

### Properties¶

1. The number of successes r is fixed in advance.

2. Trials are identical and result in a success or a failure (Bernoulli trials with P(success) = p and P(failure) = 1-p.

3. Trials are independent (outcome of one trial does not influence any other)

### PMF¶

$$S = \left\{\left(x_{1}, x_{2}, \ldots, x_{n}\right) \mid x_{i}\right. =\left\{\begin{array}{l} 1 \text { if } \text { success on ith trail } \\ 0 \text { if failure ith trail }\end{array}\right. and \sum_{i=1} = r$$

$$P(y=0)=P(\{11111\})=(p)^{5}$$

$$P(Y=1)=P(\{011111,101111,110111,111011,111101\}) = \binom{5}{4}p^5(1-p)^{5-4}$$

$$P(Y=2) = \binom{6}{4}p^5(1-p)^{5-4}$$

$$P(X = k) = \binom{k+r-1}{r-1} (1-p)^kp^r$$

### Mean (Expected Value)¶

$$E(X)=\sum_{k} k P(X=k)$$

$$E(X)= \frac{r(1-p)}{p}$$

### Variance¶

$$V(X)= \frac{r(1-p)}{p^2}$$

### Relationship between Geometric and Negative Binomial rv¶

$$X \sim Geom(p)$$

= Repeated, independent, identical, Bernoulli trails util first successes.

$$Y \sim NB(1,p)$$

= Count the number of failure until first success util first successes. =

$$\underbrace{}_{Failure} \underbrace{}_{Failure} success$$

Note: Y = X - 1. then E(Y) = E(X) - 1 = 1/p - 1 = $$\frac{1-p}{p}$$

$$NB(r,p)$$ = $$\underbrace{}_{Failure} \underbrace{}_{Failure} success \underbrace{}_{Failure} \underbrace{}_{Failure} success \underbrace{}_{Failure} \underbrace{}_{Failure} rth success$$

means we have stack geometric rv in a row rth time. that’s why we multiply by r in expected value and variance in NB rv.

## Poisson Distribution¶

The Poisson distribution is a discrete probability distribution that describes probabilities for counts of events that occur in a specified observation space. It is named after Siméon Denis Poisson.

Suppose that an event can occur several times within a given unit of time. When the total number of occurrences of the event is unknown, we can think of it as a random variable.

### Poisson Random Variable¶

A Poisson rv $$X \sim Poisson(\lambda)$$ is a discrete rv that describes the total number of events that happen in a certain time period.

#### Parameter¶

The Poisson distribution is defined by a single parameter, lambda (λ),

which is the mean number of occurrences during an observation unit. A rate of occurrence is simply the mean count per standard observation period. For example, a call center might receive an average of 32 calls per hour.

### Uses¶

1. # of vehicles crossing a bridge in one day

2. # of gamma rays hitting a satellite per hour

3. # of cookies sold at a bake sale in one hour

4. # of customers arriving at a bank in a week

### PMF¶

A discrete random variable X has Poisson distribution with parameter ($$\lambda$$ > 0) if the probability mass function of X is

$\begin{split} f(x)=P(X=x)= \begin{cases}\frac{e^{-\lambda} \lambda^{x}}{x !} & , x=0,1,2, \ldots \\ 0 & , \text { otherwise }\end{cases} \end{split}$

which may also be written as

$f(x)=\frac{e^{-\lambda} \lambda^{x}}{x !} I_{\{0,1,2, \ldots\}}(x)$

where

• k is the number of occurrences ($$k = 0,1,2\dots$$) It could be zero because nothing happened in that time period.

• e} is (e = 2.71828..)

While this pmf might appear to be highly structured, it really is the epitome of randomness. Imagine taking a 20 acre plot of land and dividing it into 1 square foot sections. (There are 871,200 sections!) Suppose you were able to scatter 5 trillion grass seeds on this land in a completely random way that does not favor one section over another. One can show that the number of seeds that fall into any one section follows a Poisson distribution with some parameter λ. More specifically, one can show that the Poisson distribution is a limiting case of the binomial distribution when n gets really large and p get really small. “Success” here is the event that any given seed falls into one particular section. We then want to count the number of successes in 5 trillion trials.

In general, the Poisson distribution is often used to describe the distribution of rare events in a large population.

All probabilities sum to 1

$$\sum_{k=0}^{\infty} P(X=k)=\sum_{k=0}^{\infty} \frac{\lambda^{k}}{k !} e^{-\lambda}=e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!} = e^{-\lambda} * e^{\lambda} = 1$$

### Mean (Expected Value)¶

$$E(X)=\sum_{k=0}^{\infty} k P(X=k)=\sum_{k=0}^{\infty} k \frac{\lambda^{k}}{k !} e^{-\lambda}=\lambda \sum_{k=1}^{\infty} \frac{\lambda^{k-1}}{(k-1) !} e^{-\lambda} = \lambda$$

$$E\left(X^{2}\right)=\sum_{k=0}^{\infty} k^{2} P(X=k)=\sum_{k=0}^{\infty} k^{2} \frac{\lambda^{k}}{k !} e^{-\lambda}=\lambda(\lambda+1)^{e}$$

### Variance¶

$$V(X)=E\left(X^{2}\right)-(E(X))^{2}=\lambda(\lambda+1)-\lambda^{2}=\lambda$$