Sums of independent random variablesAQA A-Level Further Maths: Revision notes

Section 1

The mean of a sum

For any two random variables, discrete or continuous, $E(X+Y)=E(X)+E(Y).$ This is true whether or not $X$ and $Y$ are independent, and it works if one is discrete and the other continuous. If $E(X)=12$ and $E(Y)=7$ then $E(X+Y)=19$ .

Key termsexpectation of a sum

Section 2

The variance of a sum

If $X$ and $Y$ are independent, $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y).$ The variance of a sum adds because the two sources of variation combine. With $\text{Var}(X)=9$ and $\text{Var}(Y)=16$ , $\text{Var}(X+Y)=25$ , and the standard deviation is $\sqrt{25}=5$ .

Key termsindependent

Common mistake

Adding standard deviations. Here $3+4=7$ but the standard deviation of $X+Y$ is $5$ . Add variances, then take the square root.

Section 3

Independence matters

The result for the variance of a sum is only valid when $X$ and $Y$ are independent. State this assumption when you use it, using the context: for example, the preparation time for a dish does not affect its cooking time. Independence is not needed for the mean. If a question gives you the pdfs of two independent variables, find $E$ and $\text{Var}$ for each separately (using $\text{Var}=E(X^2)-[E(X)]^2$ ), then combine.

Section 4

Two independent observations are not the same as doubling

Let $X_1$ and $X_2$ be independent observations from the same distribution, with mean $\mu$ and variance $\sigma^2$ . $E(X_1+X_2)=2\mu,\qquad \text{Var}(X_1+X_2)=2\sigma^2.$ Doubling a single observation is different: $\text{Var}(2X)=4\sigma^2$ , because the same value is used twice, so the variation is not averaged out.

Common mistake

Writing $X_1+X_2=2X$ for two separate observations. They are different values, so the variance is $2\sigma^2$ , not $4\sigma^2$ .

Section 5

Worked example

$W$ has pdf $\frac16$ on $[4,10]$ and $B$ has pdf $\frac{b}{8}$ on $[0,4]$ , independent. $E(W)=7$ , $E(W^2)=52$ , so $\text{Var}(W)=3$ . $E(B)=\frac83$ , $E(B^2)=8$ , so $\text{Var}(B)=\frac89$ . For $T=W+B$ : $E(T)=7+\frac83=\frac{29}{3}$ and $\text{Var}(T)=3+\frac89=\frac{35}{9}$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Sums of independent random variables

$X$ and $Y$ are independent random variables with $E(X)=12$ , $\text{Var}(X)=9$ , $E(Y)=7$ and $\text{Var}(Y)=16$ .
A student claims that the standard deviation of $X+Y$ is $3+4=7$ . Show that this is wrong and state the correct value.2 marks
The times, in minutes, taken to prepare and to cook a dish are modelled by independent random variables $X$ and $Y$ , with $E(X)=8.5$ , $\text{Var}(X)=1.44$ , $E(Y)=12.3$ and $\text{Var}(Y)=2.56$ . The total time is $T=X+Y$ .
Find the standard deviation of $T$ , and state the assumption that makes it valid to add the variances.2 marks
The continuous random variable $X$ has probability density function $f(x)=\frac{x}{2}$ for $0\le x\le2$ , and $f(x)=0$ otherwise. The discrete random variable $Y$ is the score when a fair six-sided die is rolled, so $E(Y)=\frac72$ and $\text{Var}(Y)=\frac{35}{12}$ . $X$ and $Y$ are independent.
Find $E(X+Y)$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).