Binomial Random Variable

Examples:

  • Suppose you toss a fair coin 12 times. What is the probability that you’ll get 5 heads?

  • Suppose you pick a random sample of 25 circuit boards used in the manufacture of a particular cell phone. You know that the long run percentage of defective boards is 5%. What is the probability that 3 or more boards are defective?

  • Suppose 40% of online purchasers of a particular book would like a new copy and 60% want a used copy. What is the probability that amongst 100 random purchasers, 50 or more used books are sold?

These three situations, and many more, can be modeled by a binomial random variable.

Key elements of these examples above:

  • $n$ Bernoulli trials.
  • Same probability of success on each trial $(p=0.5, p=0.05, p=0.6)$.
  • These are independent Bernoulli trials (one trial does not affect the outcome of another).

Properties of a binomial random variable:

  • Experiment is $n$ trials ($n$ is fixed in advance).
  • Trials are identical and resultin a success or a failure (i.e. Bernoulli trials) with $P(\text{success})=p$ and $P(\text{failure})=1-p$. *Trials are independent (outcome of one trial does not influence any other).

If $X$ is the number of success in the $n$ independent and identical trials, $X$ is a binomial random variable.

Notation: $\color{red}{X\sim\text{Bin}(n,p)}$

Find the pmf, expectation, and variance for a binomial random variable, $X\sim\text{Bin}(n,p)$

What is the sample space for a binomial experiment?

\[S=\{(x_1,x_2,\ldots,x_n\}\\ x_i=\begin{cases} \begin{align} &1&&\text{if success on i^{th} trial}\\ &0&&\text{if failure}\\ \end{align} \end{cases}\\ \vert S\vert=2^n\] PMF: \[\begin{align} P(X=0)&=P(\{0...\})=(1-p)^n\\ P(X=1)&=P(\{10..0,010..0,...,00..01\})=np(1-p)^{n-1}\\ P(X=2)&=P(\{\underbrace{110..0}_{\text{2 1's + (n-2) 0's}},...\})=\binom{n}{2}p^2(1-p)^{n-2}\\ P(X=k)&=P({\underbrace{...}_{\text{k 1's}\\\text{n-k 0's}},...})=\color{red}{\binom{n}{k}p^k(1-p)^{n-k}},\quad k=0,1,...\\ \end{align}\] Expected value:

Definition: The expected value of a discrete random variable, $E(X)$, is given by

\[E(X)=\sum_{k}kP(X=k)\]

$X\sim\text{Bin}(n,p)$

\[E(X)=\sum_{k=0}^{n}k\binom{n}{k}p^k(1-p)^{n-k}=\color{red}{np}\]

Recall:

$\text{Bern}(p)$ has expected value $p$, $X_1,X_2,\ldots X_n$ as independent$\text{Bern}(p)$

\[X = \sum_{k=1}^{n}X_n\\ E(X) = \sum_{k=1}^{n}E(X_n) = np\] Variance:

Definition: The variance of a random variable is given by

$\sigma_X^2=V(X)=E[(X-E(X))^2]$.

Computational formula: $V(X)=E(X^2)-(E(X))^2$.

$X\sim\text{Bin}(n,p)$

\[\begin{align} V(X)&=\sum_{k=0}^{n}\bigg(k-E(X)\bigg)^2\binom{n}{k}p^k(1-p)^{n-k}\\ &=\color{red}{n\times\underbrace{p(1-p)}_{\text{variance of Bern($p$)}}} \end{align}\]

Negative Binomial Random Variable

Examples:

  • Suppose you toss a fair coin until you obtain 5 heads. How many tails before the fifth head?

  • Suppose you randomly choose circuit boards until you find 3 defectives. You know that the long run percentage of defective boards is 5%. How many must you examine?

  • Suppose 40% of online purchasers of a particular book would like a new copy and 60% want a used copy. How many new books are sold before the fiftieth used book?

These three situations can be modeled by a negative binomial random variable.

Key elements for these examples above:

  • Independent Bernoulli trials until $r$ successes.
  • Count the number of failure until $r^{th}$ success.

Definition: Repeate independent Bernoulli trials until a total of $r$ successes is obtained. The negative binomial random variable $Y$ counts the number of failures before the $r^{th}$ success. Notation: $Y\sim\text{NB}(r,p)$.

  • The number of successes $r$ is fixed in advance.
  • Trials are identical and result in a success or a failure (i.e. Bernoulli trials) with $P(\text{success})=p$ and $P(\text{failure})=1-p$.
  • Trials are independent (outcome of one trial does not influence any other).

Compare to $X\sim\text{Bin}(n,p)$: $X$ is the number of successes in the $n$ independent and identical trials and $n$ is fixed in advance.

Example: A physician wishes to recruit 5 people to participate in a medical study. Let $p=0.2$ be the probability that a randomly selected person agrees to participate. What is the probability that 15 people must asked before 5 are found who agree to participate.

$Y$ is the number of failures before the 5 people are found.

\[S=\{(x_1,x_2,\ldots)\}\\ \begin{cases} \begin{align} &1&&\text{if $S$ on the $i^{th}$ trial}\\ &0&&\text{if failure}\\ \end{align} \end{cases}\\ \text{and }\sum x_i=5\]

We have

\[\begin{align} P(Y=0)&=P(\{\})=(0.2)^5\\ P(Y=1)&=P(\{011111,101111,...,111101\})=\binom{5}{4}(0.2)^5(0.8)^1\\ P(Y=2)&=\binom{6}{4}(0.2)^5(0.8)^2 \end{align}\]

In general

\[P(Y=K)=\binom{k+5-1}{4}(0.2)^5(0.8)^k\]

Summary

If $Y\sim\text{NP}(r,p)$, we have

\[\color{red}{P(Y=k)=\binom{k+r-1}{r-1}p^r(1-p)^k\quad\text{for }k=0,1,2...}\] \[\begin{align} E(Y)&=\frac{r(1-p)}{p}\\ V(Y)&=\frac{r(1-p)}{p^2} \end{align}\]

Relationship between geometric and negative binomial random variables?

$X\sim\text{Geom}(p)\leftarrow$ repeat independent identical Bernoulli trials until $\color{red}{1^{st}}$ success.

$Y\sim\text{NP}(1,p)\leftarrow$ count the number of failures until $\color{red}{1^{st}}$ success.

Because $Y$ counts the number of failures until the first success, so $Y$ is the same as $X - 1$.

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

For the $\text{NB}(r,p)$, we have $r$ $\text{Geom}(p)$ stack one to another.

\[\text{NB}(r,p)=\underbrace{....}_{F}1\underbrace{.....}_{F}1\underbrace{.....}_{F}\overbrace{\color{red}{1}}^{r^{th}\text{ success}}\]

Then we have

\[\begin{align} E(Y)&=\frac{r(1-p)}{p}\\ V(Y)&=\frac{r(1-p)}{p^2} \end{align}\]