3 Random variable: Discrete
In many situations, it is desirable to assign a numerical value to each outcome of a random experiment. Such an assignment is called a random variable.
Mathematically, a random variable is a real-valued function of the experimental outcome.
Definition: A random variable is a function that associates a real number with each element in the sample space.
We shall use a capital letter, say \(X\), to denote a random variable and its corresponding small letter, \(x\) in this case, for one of its values.
Illustration: Two balls are drawn in succession without replacement from an urn containing 4 red balls and 3 black balls.
Let, \(Y\) is the number of red balls.
The small letter \(y\) is the numerical value for each possible outcomes.
The possible outcomes and the values of \(y\) of the random variable \(Y\) are:
| Sample space | \(y\) |
|---|---|
| \(RR\) | 2 |
| \(RB\) | 1 |
| \(BR\) | 1 |
| \(BB\) | 0 |
Since the values of \(Y\) is determined by a random experiment so, \(Y\) is a random variable (discrete).
3.1 Types of random variable
There are two important types of random variables, discrete and continuous.
A discrete random variable is one whose possible values form a discrete set; in other words, the values can be ordered, and there are gaps between adjacent values. The random variable Y, just described, is discrete.
In contrast, the possible values of a continuous random variable always contain an interval, that is, all the points between some two numbers.
3.2 Discrete random variable and Probability mass function
Suppose \(X\) is a discrete random variable. The probability mass function (PMF) of \(X\) can be denoted as \(f(x)\) where
\[ f(x)=P(X=x) \]For each possible outcome \(x\) ; \(f(x)\) must satisfies:
\[f(x) \ge 0\]
\[\sum _x f(x)=1\]
3.2.1 The Cumulative Distribution Function of a Discrete Random Variable
A function called the cumulative distribution function (CDF) specifies the probability that a random variable is less than or equal to a given value. The cumulative distribution function of the random variable \(X\) is the function
- \(F(x) = P(X \le x)\)
Example 3.1: Calculating probabilities A certain industrial process is brought down for recalibration whenever the quality of the items produced falls below specifications. Let \(X\) represent the number of times the process is recalibrated during a week, and assume that \(X\) has the following probability mass function.
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(f(x)\) | 0.35 | 0.25 | 0.20 | 0.15 | 0.05 |
Compute the following :
i) \(P(X=2)\);
ii) \(P(X<3)\) and \(P(X>2)\);
iii) \(F(2)\)
Example (Walpole et al. 2017, Example 3.8)A shipment of 20 similar laptop computers to a retail outlet contains 3 that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the number of defectives.
Solution: Let \(X\) be a random variable whose values x are the possible numbers of defective computers purchased by the school. Then \(x\) can only take the numbers 0, 1, and 2. Now ,
\(P(X=0)=f(0)=P(N,N)=\frac{\binom{17}{2}\binom{3}{0}}{\binom{20}{2}}=\frac{136}{190}\)
\(P(X=1)=f(1)=P(D,N)=\frac{\binom{3}{1}\binom{17}{1}}{\binom{20}{2}}=\frac{51}{190}\)
\(P(X=2)=f(2)=P(D,D)=\frac{\binom{3}{2}\binom{17}{0}}{\binom{20}{2}}=\frac{3}{190}\)
Thus, the probability distribution of X is
| \(x\) | 0 | 1 | 2 |
|---|---|---|---|
| \(f(x)\) | \(\frac{136}{190}\) | \(\frac{51}{190}\) | \(\frac{3}{190}\) |
Example (Baron 2019, Exercise 3.1 (a)) A computer virus is trying to corrupt two files. The first file will be corrupted with probability 0.4. Independently of it, the second file will be corrupted with probability 0.3.
Compute the probability mass function (PMF) of \(X\), the number of corrupted files.
Solution: (Will be discussed in class)
3.2.2 Expectation (Population Mean) of discrete random variable
Let \(X\) be a discrete random variable with probability mass function \(f(x) = P(X = x)\).
The mean of \(X\) is given by
\[\mu=\sum_x x.f(x)\]
The mean of \(X\) is sometimes called the expectation, or expected value, of X and may also be denoted by \(E(X)\) or by \(\mu\).
Example 3.2 : Let \(X\) represent the number of times the process is recalibrated during a week, and assume that \(X\) has the following probability mass function.
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(f(x)\) | 0.35 | 0.25 | 0.20 | 0.15 | 0.05 |
Compute the mean or expected value of \(X\).
Solution:
The expected value of \(X\) is:
\[ \mu =E[X]=\sum_{x=0}^4 x.f(x) \]
\[ =0(0.35)+1(0.25)+2(0.20)+3(0.15)+4(0.05)=1.30 \]
3.2.3 Variance (population) of discrete random variable
Let \(X\) be a discrete random variable with probability distribution \(f(x)\) and mean \(\mu\). The variance of \(X\) is
\[ var(X)=\sigma^2 =E[(X-\mu)^2]=E(X^2)-\mu^2 \] Where,
\[E(X^2)=\sum_{x} x^2.f(x)\]
Example 3.2 : Let \(X\) represent the number of times the process is recalibrated during a week, and assume that \(X\) has the following probability mass function.
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(f(x)\) | 0.35 | 0.25 | 0.20 | 0.15 | 0.05 |
Solution: From Example 3.1, we have
Expected value of \(X\) is \(\mu =E(X)=1.30\)
Now, \[E(X^2)=\sum_{x=0}^4 x^2.f(x)\]
\(=0^2(0.35)+1^2 (0.25)+2^2 (0.20)+3^2 (0.15)+4^2 (0.05)\)
\(=3.20\)
Hence, \(var(X)=\sigma^2 =E(X^2)-\mu^2=3.20-(1.30)^2=1.51\)
The standard deviation is the square root of the variance:
\(\sigma=\sqrt {var(X)}\)
If \(a\) and \(b\) are constants, then
a) \(E(b)=b\)
b) \(E(aX+b)=aE(X)+b\)
c) \(var(b)=0\)
d) \(var(aX+b)=a^2 \ \ var (X)\)
Example Computer chips often contain surface imperfections. For a certain type of computer chip, the probability mass function of the number of defects X is presented in the following table.
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(f(x)\) | 0.4 | 0.3 | 0.15 | 0.1 | 0.05 |
a. Find \(P(X \le 2)\).
b. Find \(P(X > 1)\).
c. Find expected value/average of \(X\).
d. Find variance and standard deviation of \(X\).
e. Also find \(E(2X+5)\) and \(var(2X+5)\).