Random Variables and Probability Distributions

Understanding Random Variables

A Random Variable is a real-valued function defined on the sample space of a random experiment. In simple words, it assigns a numerical value to each outcome of an experiment.

Example: If two coins are tossed, the sample space is $$S = \{HH, HT, TH, TT\}$$ If we define $X$ as the number of heads, then $$X(HH)=2, \quad X(HT)=1, \quad X(TH)=1, \quad X(TT)=0$$ Thus, $X$ is a random variable taking values $$0, 1, 2$$

Types of Random Variables:

• Discrete Random Variable: Takes only a finite or countably infinite number of values.
• Continuous Random Variable: Takes any value in an interval.

At this level, the main focus is on discrete random variables.

Probability Distribution of a Random Variable

The list of all possible values of a discrete random variable together with their corresponding probabilities is called its probability distribution.

If a discrete random variable $X$ takes values $$x_1, x_2, \dots, x_n$$ with probabilities $$p_1, p_2, \dots, p_n$$ then the following conditions must hold:

1. $$p_i \ge 0 \quad \text{for all } i$$
2. $$\sum_{i=1}^{n} p_i = 1$$

Probability Mass Function (PMF): For a discrete random variable, the function $$P(X=x)$$ is called its probability mass function.

Mean (Mathematical Expectation)

The Mean or Expected Value of a random variable $X$, denoted by $E(X)$ or $\mu$, is the probability-weighted average of all possible values of $X$.

It is given by $$E(X)=\mu=\sum_{i=1}^{n} x_i p_i$$

It represents the long-run average value of the random variable.

Variance and Standard Deviation

The Variance measures the spread of the random variable around its mean.

It is given by $$Var(X)=\sigma^2 = E(X^2) - [E(X)]^2$$ where $$E(X^2)=\sum x_i^2 p_i$$

The Standard Deviation is the positive square root of the variance: $$\sigma = \sqrt{Var(X)}$$

Standard deviation is always non-negative and has the same units as the random variable.

Example 1: Probability Distribution of Heads

Two coins are tossed simultaneously. Find the probability distribution of the number of heads.

Solution: Let $X$ denote the number of heads.

The sample space is $$S = \{HH, HT, TH, TT\}$$ So the possible values of $X$ are $$0, 1, 2$$

Now, $$P(X=0)=P(TT)=\frac{1}{4}$$ $$P(X=1)=P(HT,TH)=\frac{2}{4}=\frac{1}{2}$$ $$P(X=2)=P(HH)=\frac{1}{4}$$

Hence the distribution is:

$X$	0	1	2
$P(X)$	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

Check: $$\frac{1}{4}+\frac{1}{2}+\frac{1}{4}=1$$ So this is a valid probability distribution.

Answer: The probability distribution is $$P(X=0)=\frac{1}{4}, \quad P(X=1)=\frac{1}{2}, \quad P(X=2)=\frac{1}{4}$$

Example 2: Finding Constant 'k'

The probability distribution of a random variable $X$ is given by $P(X=x)=kx$ for $x=1,2,3$ and $P(X=x)=0$ otherwise. Find the value of $k$ and $P(X \le 2)$.

Solution: Since the total probability is 1, $$P(X=1)+P(X=2)+P(X=3)=1$$ $$k(1)+k(2)+k(3)=1$$ $$6k=1$$ $$k=\frac{1}{6}$$

Now, $$P(X\le 2)=P(X=1)+P(X=2)$$ $$= k+2k = 3k$$ $$= 3\left(\frac{1}{6}\right)=\frac{1}{2}$$

Answer: $$k=\frac{1}{6}, \qquad P(X\le 2)=\frac{1}{2}$$

Example 3: Mean of a Die Roll

Find the mean of the number of points obtained in a single throw of a fair die.

Solution: Let $X$ be the number obtained on the die. Then $$X \in \{1,2,3,4,5,6\}$$ and for a fair die, $$P(X=x)=\frac{1}{6} \quad \text{for each } x=1,2,3,4,5,6$$

Therefore, $$E(X)=\sum xP(X=x)$$ $$E(X)=1\cdot\frac{1}{6}+2\cdot\frac{1}{6}+3\cdot\frac{1}{6}+4\cdot\frac{1}{6}+5\cdot\frac{1}{6}+6\cdot\frac{1}{6}$$ $$E(X)=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=\frac{7}{2}=3.5$$

Answer: $$E(X)=\frac{7}{2}=3.5$$

Example 4: Variance and Standard Deviation Calculation

Find the variance of the number of heads in two tosses of a coin.

Solution: From the distribution of the number of heads in two tosses, $$P(X=0)=\frac{1}{4}, \quad P(X=1)=\frac{1}{2}, \quad P(X=2)=\frac{1}{4}$$

First, compute the mean: $$E(X)=0\cdot\frac{1}{4}+1\cdot\frac{1}{2}+2\cdot\frac{1}{4}$$ $$E(X)=0+\frac{1}{2}+\frac{1}{2}=1$$

Now, $$E(X^2)=0^2\cdot\frac{1}{4}+1^2\cdot\frac{1}{2}+2^2\cdot\frac{1}{4}$$ $$E(X^2)=0+\frac{1}{2}+1=\frac{3}{2}$$

Hence, $$Var(X)=E(X^2)-[E(X)]^2$$ $$Var(X)=\frac{3}{2}-1^2=\frac{3}{2}-1=\frac{1}{2}$$

Answer: $$Var(X)=\frac{1}{2}$$

Example 5: Defective Items Distribution

From a lot of 10 items containing 3 defectives, a sample of 2 items is drawn at random without replacement. Find the probability distribution of the number of defective items.

Solution: Let $X$ be the number of defective items in the sample of 2. Then possible values are $$X=0,1,2$$

Total number of ways of choosing 2 items from 10 is $$\binom{10}{2}=45$$

For $X=0$: Both selected items are non-defective. There are 7 non-defective items, so $$P(X=0)=\frac{\binom{7}{2}}{\binom{10}{2}}=\frac{21}{45}=\frac{7}{15}$$

For $X=1$: Choose 1 defective from 3 and 1 non-defective from 7: $$P(X=1)=\frac{\binom{3}{1}\binom{7}{1}}{\binom{10}{2}}=\frac{3\cdot 7}{45}=\frac{21}{45}=\frac{7}{15}$$

For $X=2$: Choose 2 defectives from 3: $$P(X=2)=\frac{\binom{3}{2}}{\binom{10}{2}}=\frac{3}{45}=\frac{1}{15}$$

Thus the distribution is:

$X$	0	1	2
$P(X)$	$\frac{7}{15}$	$\frac{7}{15}$	$\frac{1}{15}$

Check: $$\frac{7}{15}+\frac{7}{15}+\frac{1}{15}=1$$

Answer: $$P(X=0)=\frac{7}{15}, \quad P(X=1)=\frac{7}{15}, \quad P(X=2)=\frac{1}{15}$$

Example 6: Expected Gain in a Game

A player tosses two coins. He wins Rs 5 if 2 heads appear, Rs 2 if 1 head appears and Rs 1 if no head appears. Find his expected gain.

Solution: Let $X$ denote the gain of the player. Then $$X=5 \text{ if 2 heads appear}$$ $$X=2 \text{ if 1 head appears}$$ $$X=1 \text{ if no head appears}$$

From the distribution of heads in two tosses, $$P(2 \text{ heads})=\frac{1}{4}, \qquad P(1 \text{ head})=\frac{1}{2}, \qquad P(0 \text{ heads})=\frac{1}{4}$$

Hence, $$E(X)=5\cdot\frac{1}{4}+2\cdot\frac{1}{2}+1\cdot\frac{1}{4}$$ $$E(X)=\frac{5}{4}+1+\frac{1}{4}=\frac{5}{4}+\frac{4}{4}+\frac{1}{4}=\frac{10}{4}=\frac{5}{2}$$

Answer: $$\text{Expected gain}=\frac{5}{2}=\text{Rs }2.50$$

Example 7: Using E(X) to find Variance

A random variable $X$ has mean 3 and variance 2. Find $E(X^2)$.

Solution: We know that $$Var(X)=E(X^2)-[E(X)]^2$$ Given, $$E(X)=3, \qquad Var(X)=2$$ So, $$2=E(X^2)-3^2$$ $$2=E(X^2)-9$$ $$E(X^2)=11$$

Answer: $$E(X^2)=11$$

Example 8: Dice Difference Distribution

Two dice are thrown. Let $X$ be the absolute difference of the numbers appearing. Find the probability distribution of $X$.

Solution: When two dice are thrown, total outcomes are $$36$$ The possible values of the absolute difference are $$0,1,2,3,4,5$$

Now count the number of outcomes for each value:

For $X=0$: $$(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$$ So frequency is $6$, hence $$P(X=0)=\frac{6}{36}=\frac{1}{6}$$

For $X=1$: There are 10 outcomes, so $$P(X=1)=\frac{10}{36}=\frac{5}{18}$$

For $X=2$: There are 8 outcomes, so $$P(X=2)=\frac{8}{36}=\frac{2}{9}$$

For $X=3$: There are 6 outcomes, so $$P(X=3)=\frac{6}{36}=\frac{1}{6}$$

For $X=4$: There are 4 outcomes, so $$P(X=4)=\frac{4}{36}=\frac{1}{9}$$

For $X=5$: There are 2 outcomes, so $$P(X=5)=\frac{2}{36}=\frac{1}{18}$$

Hence the probability distribution is:

$X$	0	1	2	3	4	5
$P(X)$	$\frac{1}{6}$	$\frac{5}{18}$	$\frac{2}{9}$	$\frac{1}{6}$	$\frac{1}{9}$	$\frac{1}{18}$

Check: $$\frac{1}{6}+\frac{5}{18}+\frac{2}{9}+\frac{1}{6}+\frac{1}{9}+\frac{1}{18}=1$$

Answer: The probability distribution is as shown above.

Example 9: Expectation of a Constant and Linear Property

Given $E(X)=5$, find $E(3X+4)$.

Solution: Using linearity of expectation, $$E(aX+b)=aE(X)+b$$ So, $$E(3X+4)=3E(X)+4$$ $$=3(5)+4=15+4=19$$

Answer: $$19$$

Example 10: Step-by-Step Variance of a Distribution

A random variable $X$ has the following distribution. Find its variance.

$X$	1	2	3
$P(X)$	0.2	0.5	0.3

Solution: First, compute the mean: $$E(X)=1(0.2)+2(0.5)+3(0.3)$$ $$E(X)=0.2+1.0+0.9=2.1$$

Now compute $$E(X^2)=1^2(0.2)+2^2(0.5)+3^2(0.3)$$ $$E(X^2)=0.2+4(0.5)+9(0.3)$$ $$E(X^2)=0.2+2.0+2.7=4.9$$

Hence, $$Var(X)=E(X^2)-[E(X)]^2$$ $$Var(X)=4.9-(2.1)^2$$ $$Var(X)=4.9-4.41=0.49$$

Answer: $$Var(X)=0.49$$