4-2. Probability Distributions for Discrete Random Variables

离散随机变量的概率分布

1. Probability Distributions

The probability distribution of a discrete random variable X is a list of each possible value of X together with the probability that X takes that value in one trial of the experiment.

The probabilities (Probability Mass Function) in the probability distribution of a random variable $X$ must satisfy the following two conditions:

1. Each probability $P(x)$ must be between 0 and 1: $0≤P(x)≤1$ .

2. The sum of all the probabilities is 1 : $ΣP(x)=1.$

EXAMPLE 1. A fair coin is tossed twice. Let $X$ be the number of heads that are observed.

1. Construct the probability distribution of $X$ .

2. Find the probability that at least one head is observed. $P(X \ge 1)$

[ Solution ]

Probability Distribution for Tossing a Fair Coin Twice

R Source

library(Rstat)

# 0. Sample Space
S <- tosscoin2(2); S 

# 1. Define X
# 1-1. define function to count the number of Heads
countH <- function(x) sum(x=="H")

# 1-2. Define X
X <- apply(S, 1, countH)

# 2. Prob. Distribution Table
y <- table(X) / nrow(S)

# 3. plot table
barplot(y)

Result

> # 0. Sample Space
> S <- tosscoin2(2); S 
##   X1 X2
## 1  H  H
## 2  T  H
## 3  H  T
## 4  T  T
> 
> # 1. Define X
> # 1-1. define function to count the number of Heads
> countH <- function(x) sum(x=="H")
> 
> # 1-2. Define X
> X <- apply(S, 1, countH); X
## [1] 2 1 1 0
> 
> # 2. Prob. Distribution Table
> y <- table(X) / nrow(S); y
## X
##    0    1    2 
## 0.25 0.50 0.25 
> 
> # 3. plot table
> barplot(y)

• $P(X\ge 1) = 0.5 + 0.25 = 0.75$

Plot

EXAMPLE 2. A pair of fair dice is rolled. Let $X$ denote the sum of the number of dots on the top faces.

1. Construct the probability distribution of $X$ .

2. Find $P(X ≥ 9)$ .

3. Find the probability that $X$ takes an even value.

[ Solution ]

Probability Distribution for Tossing Two Fair Dice

R Source

library(Rstat)

# probability distribution of X
rolldie.sum(2)

Probability Distribution : Table

> rolldie.sum(2)
## X
##   2   3   4   5   6   7   8   9  10  11  12 Sum 
##   1   2   3   4   5   6   5   4   3   2   1  36 
## X
##      2      3      4      5      6      7      8      9     10     11     12 
## 0.0278 0.0556 0.0833 0.1111 0.1389 0.1667 0.1389 0.1111 0.0833 0.0556 0.0278 
##    Sum 
## 1.0000 

• $P(X ≥ 9) = 0.1111 + 0.0833 + 0.0556 + 0.0278 = 0.2778$

• $P(X=2) + P(X=4) + P(X=6)+P(X=8)+P(X=10)+P(X=12)$ $= 0.0278 + 0.0833 + 0.1389 + 0.1389 + 0.0833 + 0.0278 = 0.5$

Probability Distribution : Plot

The Mean and Standard Deviation of a Discrete Random Variable

The mean (also called the expected value) of a discrete random variable X is the number

$μ=E(X)=Σx P(x)$

The mean of a random variable may be interpreted as the average of the values assumed by the random variable in repeated trials of the experiment.

EXAMPLE 3. Find the mean of the discrete random variable $X$ whose probability distribution is

x       -2    1    2  3.5
P(x)  0.21 0.34 0.24 0.21

[ Solution ]

R Source

library(Rstat)

x <- c(-2, 1, 2, 3.5)
px <- c(0.21, 0.34, 0.24, 0.21)

mean <- sum(x * px) ; mean

# disc.exp(x, p) : E(X), V(X), Std Dev(X)
disc.exp(x, px)

# Draw the graph
disc.exp(x, px, prt=FALSE, plot=TRUE)

mean

> mean <- sum(x * px) ; mean
[1] 1.135

disc.exp()

> disc.exp(x, px)
## E(X) = 1.135 
## V(X) = 4.712 - 1.135² = 3.424 
## D(X) = √(3.424) = 1.85

Plot

EXAMPLE 4. A service organization in a large town organizes a raffle each month. One thousand raffle tickets are sold for $1 each. Each has an equal chance of winning. First prize is $300, second prize is $200, and third prize is $100. Let $X$ denote the net gain from the purchase of one ticket.

1. Construct the probability distribution of $X$ .

2. Find the probability of winning any money in the purchase of one ticket.

3. Find the expected value of $X$ , and interpret its meaning.

[ Solution ]

R Source

library(Rstat)

x <- c(299, 199, 99, -1)
px <- c(0.001, 0.001, 0.001, 0.997)

# 2.
sum(px[1:3])

# 3.
sum(x * px)

# E(X), V(X), Std Dev(X)
disc.exp(x, px)

# Draw the graph
disc.exp(x, px, prt=FALSE, plot=TRUE)

2. Winning Prob.

> # 2.
> sum(px[1:3])
[1] 0.003

3. Expected Value

> # 3.
> sum(x * px)
[1] -0.4

disc.exp()

> disc.exp(x, px)
## E(X) = -0.4 
## V(X) = 139.8 - 0.4² = 139.64 
## D(X) = √(139.64) = 11.817 

Plot

EXAMPLE 5. A life insurance company will sell a $200,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $195. Find the expected value to the company of a single policy if a person in this risk group has a 99.97% chance of surviving one year.

[ Solution ]

R Source

library(Rstat)

x <- c(195, -199805)
px <- c(0.9997, 0.0003)

# 2.
sum(x * px)

# E(X), V(X), Std Dev(X)
disc.exp(x, px)

# Draw the graph
disc.exp(x, px, prt=FALSE, plot=TRUE)

Expected Value

> # 2.
> sum(x * px)
[1] 135

disc.exp()

> disc.exp(x, px)
## E(X) = 135 
## V(X) = 12014625 - 135² = 11996400 
## D(X) = √(11996400) = 3463.582 

Plot

The variance, $σ^2$ , of a discrete random variable X is the number

$σ^2=Σ(x−μ)^2 P(x)$

which by algebra is equivalent to the formula

$σ^2= Σ(x^2* P(x))−μ^2$ .

The standard deviation, $σ$ , of a discrete random variable X is the square root of its variance, hence is given by the formulas

$σ=\sqrt{Σ(x−μ)^2 P(x)} = \sqrt{Σx^2 P(x) − μ^2}$

EXAMPLE 6. A discrete random variable X has the following probability distribution:

x      -1    0  1    4
P(x)  0.2  0.5  a  0.1

Compute each of the following quantities.

1. a.

2. $P(0)$

3. $P(X > 0)$ .

4. $P(X ≥ 0)$ .

5. $P(X≤−2)$ .

6. The mean $μ$ of $X$ .

7. The variance $σ^2$ of $X$ .

8. The standard deviation $σ$ of $X$ .

[ Solution ] which()

R Source

library(Rstat)

# 1. a
a <- 1 - sum(0.2, 0.5, 0.1); a
x <- c(-1, 0, 1, 4)
px <- c(0.2, 0.5, a, 0.1)


# 2. P(0)
px[which(x==0)]

# 3. P(X>0)
sum(px[which(x>0)])

# 4. P(X>=0)
sum(px[which(x>=0)])

# 5. P(X<=-2)
sum(px[which(x<=-2)])

# 6.
mean <- sum(x*px) ; mean

# 7.
variance <- sum(x^2*px) - mean^2 ; variance

# 8.
std <- srqt(variance) ; std

# E(X), V(X), Std Dev(X)
disc.exp(x, px)

# Draw the graph
disc.exp(x, px, prt=FALSE, plot=TRUE)

1-5.

> # 1. a
> a <- 1 - sum(0.2, 0.5, 0.1); a
## [1] 0.2
> x <- c(-1, 0, 1, 4)
> px <- c(0.2, 0.5, a, 0.1)
> 
> 
> # 2. P(0)
> px[which(x==0)]
## [1] 0.5
> 
> # 3. P(X>0)
> sum(px[which(x>0)])
## [1] 0.3
> 
> # 4. P(X>=0)
> sum(px[which(x>=0)])
## [1] 0.8
> 
> # 5. P(X<=-2)
> sum(px[which(x<=-2)])
## [1] 0

6. mean

> # 6.
> mean <- sum(x*px) ; mean
[1] 0.4

7. variance

> # 7.
> variance <- sum(x^2*px) - mean^2 ; variance
[1] 1.84

8. standard deviation

> stdev <- sqrt(variance) ; stdev
[1] 1.356466

disc,exp()

> disc.exp(x, px)
## E(X) = 0.4 
## V(X) = 2 - 0.4² = 1.84 
## D(X) = √(1.84) = 1.356

Plot

概率质量函数 Probability Mass Function 期望值 Expected Value