Notes › EENG 3421: Probability and Statistics for Engineers Lecture 8
Families of Discrete Random Variables
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Table of Contents
Families of Random Variables
- Certain families of random variables make multiple appearances in experiments
- The PMF has the same mathematical form, but has a different shape or number of parameters
Bernoulli $(p)$ Random Variable
$X$ is a Bernoulli $(p)$ random variable if the PMF of $X$ has the form
$$P_X(x) = \begin{cases} 1-p & x = 0 \\ p & x = 1 \\ 0 & \text{otherwise} \end{cases}$$where $0 < p < 1$
Test one circuit and observe $X$, the number of rejects. Since there are only two outcomes, $X = 1$ with probability $p$ and $X = 0$ with probability $1-p$, so the number of rejects is a Bernoulli $(p)$ random variable
Binomial $(n, p)$ Random Variable
$X$ is a binomial $(n, p)$ random variable if the PMF of $X$ has the form
$$P_X(x) = \binom{n}{x} p^x (1-p)^{n-x}$$where $0 < p < 1$ and $n$ is an integer such that $n \ge 1$
- Counts the number of successes in $n$ independent Bernoulli trials
- Bernoulli $(p)$ is the special case Binomial $(1, p)$
In $n$ independent tests each rejected with probability $p$, let $K$ equal the number of rejects. Then $K$ is Binomial $(n, p)$ with
$$P_K(k) = \binom{n}{k} p^k (1-p)^{n-k}$$A 10-question exam has four choices per question. A student guesses every answer. Let $X$ be the number of correct answers, then $X$ is Binomial with $n = 10$, $p = 0.25$
$$P_X(2) = \binom{10}{2}(0.25)^2(0.75)^8 = 0.281$$Components are defective with probability $0.001$ independently. Let $X$ be the number of defective components in $n$ produced. Then:
$$P[\text{at least one defective}] = 1 - P[X=0] = 1 - (0.999)^n$$20 bulbs each function more than three months with probability $0.2$. Let $X$ be the number that function. Then $X$ is Binomial with $n = 20$, $p = 0.2$
$$P[X \ge 5] = 1 - \sum_{x=0,1,2,3,4} \binom{20}{x} (0.2)^x (0.8)^{20-x} = 1 - (0.0115 + 0.0576 + 0.137 + 0.205 + 0.218) = 0.37$$Geometric $(p)$ Random Variable
$X$ is a geometric $(p)$ random variable if the PMF of $X$ has the form
$$P_X(x) = \begin{cases} p(1-p)^{x-1} & x = 1, 2, \dots \\ 0 & \text{otherwise} \end{cases}$$where $0 < p < 1$
- Counts the number of trials until the first success
- PMF forms a geometric series, hence the name
Proof.
$$P[X > k] = 1 - P[X \le k] = 1 - \sum_{x=1}^{k} p(1-p)^{x-1} = 1 - p \cdot \frac{1-(1-p)^k}{1-(1-p)} = (1-p)^k$$
Five cars ahead each take a space with probability $0.2$. Let $X$ be the number of trials to fill the space, then $X$ is Geometric with $p = 0.2$
$$P_X(5) = (0.8)^4(0.2) = 0.082$$A two-way paging system retransmits until an ACK is received. Each attempt succeeds with probability $p$
(a) $N$ is the number of times the message is sent. Since $N = n$ requires $n-1$ failures then a success, $N$ is Geometric:
$$P_N(n) = \begin{cases}(1-p)^{n-1}p & n = 1, 2, \dots \\ 0 & \text{otherwise}\end{cases}$$(b) The goal $P[N \le 3] \ge 0.95$ requires:
$$P[N \le 3] = 1 - P[N > 3] = 1 - (1-p)^3 \ge 0.95$$$$(1-p)^3 \le 0.05 \implies p \ge 1 - (0.05)^{1/3} \approx 0.6316$$Pascal $(k, p)$ Random Variable
$X$ is a Pascal $(k, p)$ random variable if the PMF of $X$ has the form
$$P_X(x) = \binom{x-1}{k-1} p^k (1-p)^{x-k}$$where $0 < p < 1$ and $k$ is an integer such that $k \ge 1$
- Counts the number of trials until the $k$th success
- Geometric $(p)$ is the special case Pascal $(1, p)$
- Derivation: $X = l$ iff there are $k-1$ successes in the first $l-1$ trials AND a success on trial $l$, and these events are independent, so:
A fair coin is tossed until heads appears four times. Let $X$ be the number of flips, then $X$ is Pascal $(4, 0.5)$
$$P[X = 10] = \binom{9}{3}(0.5)^4(0.5)^6 = 0.082$$Poisson $(\alpha)$ Random Variable
- Models the number of occurrences of an event in a fixed time or space interval
- The parameter $\alpha$ represents the average or expected number of occurrences
- Often written as:
- $\lambda$ is the average rate of occurrences of the event
- $T$ is the length of the time or space interval
$X$ is a Poisson $(\alpha)$ random variable if the PMF of $X$ has the form
$$P_X(x) = \begin{cases} \dfrac{\alpha^x e^{-\alpha}}{x!} & x = 0,1,2,\dots \ 0 & \text{otherwise} \end{cases}$$where the parameter $\alpha$ is such that $\alpha > 0$
The number of hits at a website in a time interval is a Poisson random variable. Suppose a site receives on average $\lambda = 2$ hits per second.
In an interval of $T = 0.25$ seconds,
$$\alpha = \lambda T = (2)(0.25) = 0.5$$Let $H$ be the number of hits in $0.25$ seconds. Then
$$P_H(h) = \frac{(0.5)^h e^{-0.5}}{h!}, \quad h = 0,1,2,\dots$$The probability of no hits is
$$P[H=0] = \frac{(0.5)^0 e^{-0.5}}{0!} = e^{-0.5} = 0.607$$If the same website receives $\lambda = 2$ hits per second, let $J$ be the number of hits in one second. Then
$$\alpha = \lambda T = (2)(1) = 2$$$$P_J(j) = \frac{2^j e^{-2}}{j!}, \quad j = 0,1,2,\dots$$The probability of no more than two hits is
$$P[J \le 2] = P[J=0] + P[J=1] + P[J=2]$$$$= e^{-2} + 2e^{-2} + \frac{2^2}{2!}e^{-2} = 0.677$$The number of database queries processed in a 10-second interval is Poisson with $\alpha = 5$. Let $K$ be the number of queries.
$$P_K(k) = \frac{5^k e^{-5}}{k!}, \quad k = 0,1,2,\dots$$The probability of no queries in 10 seconds is
$$P[K=0] = e^{-5} = 0.0067$$The average rate is
$$\lambda = \frac{\alpha}{T} = \frac{5}{10} = 0.5 \text{ queries/sec}$$In a 2-second interval
$$\alpha = \lambda T = (0.5)(2) = 1$$Let $N$ be the number of queries in 2 seconds:
$$P_N(n) = \frac{1^n e^{-1}}{n!}, \quad n = 0,1,2,\dots$$The probability of at least two queries is
$$P[N \ge 2] = 1 - P[N=0] - P[N=1]$$$$= 1 - e^{-1} - e^{-1} = 0.264$$Discrete Uniform $(k,l)$ Random Variable
- All outcomes in the range ${k,k+1,\dots,l}$ are equally likely
$X$ is a discrete uniform $(k,l)$ random variable if the PMF of $X$ has the form
$$P_X(x) = \begin{cases} \dfrac{1}{l-k+1} & x = k, k+1, k+2, \dots, l \\ 0 & \text{otherwise} \end{cases}$$where the parameters $k$ and $l$ are integers such that $k < l$
Roll a fair die. Let $N$ be the number on the face that appears. Then $N$ is a discrete uniform $(1,6)$ random variable.
$$P_N(n) = \begin{cases} \frac{1}{6} & n = 1,2,3,4,5,6 \\ 0 & \text{otherwise} \end{cases}$$In a package of M&Ms, let $Y$ be the number of yellow M&Ms, uniformly distributed between 5 and 15.
The PMF is
$$P_Y(y) = \begin{cases} \frac{1}{11} & y = 5,6,7,\dots,15 \\ 0 & \text{otherwise} \end{cases}$$(a) Probability that $Y < 10$:
$$P[Y < 10] = P_Y(5)+P_Y(6)+P_Y(7)+P_Y(8)+P_Y(9) = \frac{5}{11}$$(b) Probability that $Y > 12$:
$$P[Y > 12] = P_Y(13)+P_Y(14)+P_Y(15) = \frac{3}{11}$$(c) Probability that $8 \le Y \le 12$:
$$P[8 \le Y \le 12] = P_Y(8)+P_Y(9)+P_Y(10)+P_Y(11)+P_Y(12) = \frac{5}{11}$$References
- Course lecture slides 11–12
- Course lecture slides 13–14
Sources
- Course lecture slides 11–12
- Course lecture slides 13–14