Discrete Random Variables (Further)

This lesson extends your understanding of discrete random variables beyond what is covered in A-Level Mathematics. You will learn to compute expectations of functions of random variables, derive variance using the `E(X²)` method, and work with the algebra of expectations. These skills are essential for the rest of the Further Statistics module and underpin many of the distributions you will encounter.

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Recap: Probability Distributions

A discrete random variable $X$X takes a countable number of values $x_1, x_2, \ldots$x1​,x2​,…, each with an associated probability $P(X = x_i) = p_i$P(X=xi​)=pi​. The probability distribution must satisfy:

Property	Requirement
Non-negativity	$p_i \geq 0$pi​≥0 for all $i$i
Normalisation	$\sum_i p_i = 1$∑i​pi​=1

Example: A random variable $X$X has the following distribution:

$x$x	1	2	3	4
$P(X = x)$P(X=x)	0.1	0.3	0.4	0.2

Check: $0.1 + 0.3 + 0.4 + 0.2 = 1$0.1+0.3+0.4+0.2=1. Valid distribution.

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Expectation $E(X)$E(X)

The expected value (or mean) of a discrete random variable is:

$E(X) = \mu = \sum_x x \cdot P(X = x)$E(X)=μ=∑x​x⋅P(X=x)

This gives the long-run average value of $X$X if the experiment were repeated many times.

Example: Using the distribution above:

$E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7$E(X)=1(0.1)+2(0.3)+3(0.4)+4(0.2)=0.1+0.6+1.2+0.8=2.7

Exam Tip: The expected value does not have to be a value that $X$X can actually take. It is a weighted average, not a mode or a median.

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Expectation of a Function of $X$X: $E(g(X))$E(g(X))

If $g(X)$g(X) is any function of the random variable $X$X, then:

$E(g(X)) = \sum_x g(x) \cdot P(X = x)$E(g(X))=∑x​g(x)⋅P(X=x)

This is one of the most important results in Further Statistics. You do not need to find the distribution of $g(X)$g(X) separately — you apply $g$g to each value and weight by the corresponding probability.

Example: Find $E(X^2)$E(X2) for the distribution above.

$E(X^2) = 1^2(0.1) + 2^2(0.3) + 3^2(0.4) + 4^2(0.2)$E(X2)=12(0.1)+22(0.3)+32(0.4)+42(0.2) $= 0.1 + 1.2 + 3.6 + 3.2 = 8.1$=0.1+1.2+3.6+3.2=8.1

Example: Find $E(3X + 2)$E(3X+2).

$E(3X + 2) = \sum_x (3x + 2) \cdot P(X = x)$E(3X+2)=∑x​(3x+2)⋅P(X=x) $= (3 \cdot 1 + 2)(0.1) + (3 \cdot 2 + 2)(0.3) + (3 \cdot 3 + 2)(0.4) + (3 \cdot 4 + 2)(0.2)$=(3⋅1+2)(0.1)+(3⋅2+2)(0.3)+(3⋅3+2)(0.4)+(3⋅4+2)(0.2) $= 5(0.1) + 8(0.3) + 11(0.4) + 14(0.2) = 0.5 + 2.4 + 4.4 + 2.8 = 10.1$=5(0.1)+8(0.3)+11(0.4)+14(0.2)=0.5+2.4+4.4+2.8=10.1

Note that $3E(X) + 2 = 3(2.7) + 2 = 10.1$3E(X)+2=3(2.7)+2=10.1, confirming the linearity of expectation.

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Properties of Expectation

The expectation operator is linear:

Rule	Formula
Constant multiple	$E(aX) = aE(X)$E(aX)=aE(X)
Addition of a constant	$E(X + b) = E(X) + b$E(X+b)=E(X)+b
Linear combination	$E(aX + b) = aE(X) + b$E(aX+b)=aE(X)+b
Sum of functions	$E(g(X) + h(X)) = E(g(X)) + E(h(X))$E(g(X)+h(X))=E(g(X))+E(h(X))

However, in general:

$E(g(X)) \neq g(E(X))$E(g(X))=g(E(X))

For example, $E(X^2) \neq (E(X))^2$E(X2)=(E(X))2 in general. The difference between these two quantities is fundamental to the definition of variance.

Exam Tip: A very common error is to assume $E(X^2) = (E(X))^2$E(X2)=(E(X))2. This is almost never true. The correct relationship is $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2, which shows that $E(X^2) \geq (E(X))^2$E(X2)≥(E(X))2, with equality only when $X$X is constant.

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Variance $\text{Var}(X)$Var(X)

The variance measures the spread of a distribution:

$\text{Var}(X) = E((X - \mu)^2) = \sum_x (x - \mu)^2 P(X = x)$Var(X)=E((X−μ)2)=∑x​(x−μ)2P(X=x)

The computationally easier formula is:

$\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2

Example (continued): From above, $E(X) = 2.7$E(X)=2.7 and $E(X^2) = 8.1$E(X2)=8.1.

$\text{Var}(X) = 8.1 - 2.7^2 = 8.1 - 7.29 = 0.81$Var(X)=8.1−2.72=8.1−7.29=0.81

The standard deviation is $\sigma = \sqrt{\text{Var}(X)} = \sqrt{0.81} = 0.9$σ=Var(X)​=0.81​=0.9.

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Properties of Variance

Rule	Formula
Constant multiple	$\text{Var}(aX) = a^2 \text{Var}(X)$Var(aX)=a2Var(X)
Addition of a constant	$\text{Var}(X + b) = \text{Var}(X)$Var(X+b)=Var(X)
Linear transformation	$\text{Var}(aX + b) = a^2 \text{Var}(X)$Var(aX+b)=a2Var(X)

Note that adding a constant shifts the distribution but does not change its spread, so the variance is unchanged.

Example: If $\text{Var}(X) = 0.81$Var(X)=0.81, find $\text{Var}(3X + 2)$Var(3X+2).

$\text{Var}(3X + 2) = 3^2 \times 0.81 = 9 \times 0.81 = 7.29$Var(3X+2)=32×0.81=9×0.81=7.29

Exam Tip: Remember that variance scales by the square of the constant, while expectation scales linearly. This is the most common source of errors in Further Statistics calculations.

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Worked Example: Finding an Unknown Parameter

A random variable $X$X has the following distribution, where $k$k is a constant:

$x$x	0	1	2	3
$P(X = x)$P(X=x)	$k$k	$2k$2k	$3k$3k	$4k$4k

Step 1: Find $k$k.

$k + 2k + 3k + 4k = 1 \implies 10k = 1 \implies k = 0.1$k+2k+3k+4k=1⟹10k=1⟹k=0.1

Step 2: Find $E(X)$E(X).

$E(X) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.4) = 0 + 0.2 + 0.6 + 1.2 = 2.0$E(X)=0(0.1)+1(0.2)+2(0.3)+3(0.4)=0+0.2+0.6+1.2=2.0

Step 3: Find $E(X^2)$E(X2).

$E(X^2) = 0(0.1) + 1(0.2) + 4(0.3) + 9(0.4) = 0 + 0.2 + 1.2 + 3.6 = 5.0$E(X2)=0(0.1)+1(0.2)+4(0.3)+9(0.4)=0+0.2+1.2+3.6=5.0

Step 4: Find $\text{Var}(X)$Var(X).

$\text{Var}(X) = 5.0 - 2.0^2 = 5.0 - 4.0 = 1.0$Var(X)=5.0−2.02=5.0−4.0=1.0

Step 5: Find $E(2X^2 - 3X + 1)$E(2X2−3X+1).

$E(2X^2 - 3X + 1) = 2E(X^2) - 3E(X) + 1 = 2(5.0) - 3(2.0) + 1 = 10 - 6 + 1 = 5$E(2X2−3X+1)=2E(X2)−3E(X)+1=2(5.0)−3(2.0)+1=10−6+1=5

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Mode and Median of a Discrete Distribution

The mode is the value of $x$x with the highest probability. A distribution can be multimodal if two or more values share the maximum probability.

The median is the value $m$m such that $P(X \leq m) \geq 0.5$P(X≤m)≥0.5 and $P(X \geq m) \geq 0.5$P(X≥m)≥0.5.

Example: For the distribution with $k = 0.1$k=0.1 above, the mode is $x = 3$x=3 (probability 0.4). The cumulative probabilities are: $P(X \leq 0) = 0.1$P(X≤0)=0.1, $P(X \leq 1) = 0.3$P(X≤1)=0.3, $P(X \leq 2) = 0.6$P(X≤2)=0.6, $P(X \leq 3) = 1.0$P(X≤3)=1.0. So the median is $m = 2$m=2.

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The Cumulative Distribution Function (Discrete)

For a discrete random variable, the cumulative distribution function (CDF) is:

$F(x) = P(X \leq x) = \sum_{x_i \leq x} P(X = x_i)$F(x)=P(X≤x)=∑xi​≤x​P(X=xi​)

The CDF is a step function that increases from 0 to 1. To recover the probability mass function (PMF) from the CDF:

$P(X = x) = F(x) - F(x^{-})$P(X=x)=F(x)−F(x−)

where $F(x^{-})$F(x−) is the limit of $F$F from the left.

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Covariance and Independence (Introduction)

For two discrete random variables $X$X and $Y$Y:

$E(X + Y) = E(X) + E(Y) \quad \text{(always)}$E(X+Y)=E(X)+E(Y)(always)

If $X$X and $Y$Y are independent:

$E(XY) = E(X)E(Y)$E(XY)=E(X)E(Y) $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y)$Var(X+Y)=Var(X)+Var(Y)

If they are not independent, then:

$\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X, Y)$Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)

where $\text{Cov}(X, Y) = E(XY) - E(X)E(Y)$Cov(X,Y)=E(XY)−E(X)E(Y).

Exam Tip: In exam questions, always check whether the variables are stated to be independent. If they are, you can add variances directly. If not, you need the covariance term.

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Summary

• $E(X) = \sum x \cdot P(X = x)$E(X)=∑x⋅P(X=x) — the weighted average of all possible values.
• $E(g(X)) = \sum g(x) \cdot P(X = x)$E(g(X))=∑g(x)⋅P(X=x) — apply the function first, then weight by probabilities.
• $E(aX + b) = aE(X) + b$E(aX+b)=aE(X)+b — expectation is linear.
• $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2 — the computational formula for variance.
• $\text{Var}(aX + b) = a^2 \text{Var}(X)$Var(aX+b)=a2Var(X) — variance scales by the square of the constant.
• For independent random variables: $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y)$Var(X+Y)=Var(X)+Var(Y).

Exam Tip: When finding $\text{Var}(X)$Var(X), always compute $E(X^2)$E(X2) and $E(X)$E(X) in a clear table. Show your working for each step — this method is less error-prone than using the definition directly.