Combining two random variables

9 minA Level6.2.1

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can find the mean and variance of combinations such as $X + Y$ , $X - Y$ and $2X + 3Y$ , explain why variances add even when you subtract, and say why the variance rule needs the variables to be independent.

Key to the symbols

Symbol	Meaning
$X$ , $Y$	two random variables
$a$ , $b$	constants, such as $2$ or $-1$
$aX + bY$	a linear combination: multiply each variable by a constant, then add
$E(X)$ , $\text{Var}(X)$ , $\text{sd}(X)$	mean, variance and standard deviation, as in the previous lesson

Note that $X - Y$ is the combination with $a = 1$ and $b = -1$ .

A journey in two parts

Mo walks to the bus stop and then rides the bus. The walk takes $X$ minutes and the ride takes $Y$ minutes, and both vary from day to day. The whole journey takes

$T = X + Y$

If the walk averages 8 minutes and the ride averages 25 minutes, the journey averages $8 + 25 = 33$ minutes. Means simply add.

Spread is less obvious. Some days both parts are slow, some days both are quick, and some days one cancels the other. The rule below tells you exactly how big the spread of the total is.

Definition

Independent random variables

$X$ and $Y$ are independent when knowing the value of one tells you nothing about the value of the other.

Mo's walking time and the bus time are reasonably independent: a slow walk does not make the bus slower.

Definition

Rules for aX + bY

For any random variables $X$ and $Y$ :

$E(aX + bY) = aE(X) + bE(Y)$

If $X$ and $Y$ are independent:

$\text{Var}(aX + bY) = a^2\,\text{Var}(X) + b^2\,\text{Var}(Y)$

The mean rule always works.
The variance rule needs independence.
Each multiplier is squared, exactly as in the one-variable rule.

Worked example: Example 1: the journey

Mo's walk $X$ has mean 8 minutes and variance 4. The bus ride $Y$ has mean 25 minutes and variance 9. $X$ and $Y$ are independent. Find the mean and standard deviation of the total time $T = X + Y$ .

Have a go first?

Why variances add even for a difference

It is tempting to think $\text{Var}(X - Y) = \text{Var}(X) - \text{Var}(Y)$ . Try a tiny example to see why that is wrong.

Let $X$ be $0$ or $2$ , each with probability $\frac{1}{2}$ . Then $E(X) = 1$ and every value is 1 away from the mean, so $\text{Var}(X) = 1$ . Let $Y$ be an independent copy with the same distribution, so $\text{Var}(Y) = 1$ .

The four equally likely pairs give these differences:

$X$	$Y$	$X - Y$
0	0	0
0	2	$-2$
2	0	2
2	2	0

$X - Y$ has mean 0, and its squared distances from 0 are $0, 4, 4, 0$ , with average $\frac{8}{4} = 2$ .

So $\text{Var}(X - Y) = 2 = 1 + 1$ . Subtracting has made the spread bigger, not zero. A high $X$ with a low $Y$ pushes the difference far one way, and a low $X$ with a high $Y$ pushes it far the other way.

The rule agrees: $X - Y$ has $b = -1$ , and $(-1)^2 = 1$ , so

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

Worked example: Example 2: a weighted total and a difference

$X$ and $Y$ are independent with $E(X) = 10$ , $\text{Var}(X) = 4$ , $E(Y) = 5$ and $\text{Var}(Y) = 9$ . Find the mean and standard deviation of (a) $T = 2X + 3Y$ and (b) $D = X - Y$ .

Have a go first?

Check 1

CalculateQuestion 1 · Recall

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