Linear combinations of random variables

Scaling and shifting one variable Paper 6

If $X$ is a random variable and $a$ and $b$ are constants, then
$$\mathrm{E}(aX+b)=a\mathrm{E}(X)+b\text{and}\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\mathrm{.}$$
These hold for any distribution, discrete or continuous. Proofs are not required, but the reasons are worth knowing, because they stop the common errors:

• adding $b$ moves every value by the same amount, so the mean moves by $b$ but the spread does not change: $b$ never appears in the variance;
• multiplying by $a$ stretches every distance from the mean by a factor $a$, so the standard deviation is multiplied by $\left|a\right|$ and the variance, which is measured in squared units, by $a^2$.

So $\mathrm{Var}(5-2X)=(-2{)}^2\mathrm{Var}(X)=4\mathrm{Var}(X)$: a negative multiplier still increases the variance, and a variance can never be negative.

A familiar use is a change of units. If a temperature $C$ in degrees Celsius has mean 20 and variance 9, then $F=1.8C+32$ has mean $1.8\times 20+32=68$ and variance ${1.8}^2\times 9=29.16$.

Combining two variables Paper 6

For any two random variables $X$ and $Y$,
$$\mathrm{E}(aX+bY)=a\mathrm{E}(X)+b\mathrm{E}(Y)\mathrm{.}$$
If $X$ and $Y$ are independent, the variances also combine:
$$\mathrm{Var}(aX+bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)\mathrm{.}$$
The most important special case is the difference. With $a=1$ and $b=-1$,
$$\mathrm{Var}(X-Y)=\mathrm{Var}(X)+(-1{)}^2\mathrm{Var}(Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)\mathrm{.}$$
Variances add even when the variables are subtracted. Think of the uncertainty: if each of two parcel masses is uncertain, their difference is uncertain because of both of them, so it is more variable than either, never less.

Independence matters for the variance result only. For example, the height of a person and the length of their arm are not independent, and the variance of their sum is not the sum of the variances. In context questions, "state an assumption" usually wants independence, said in context.

Adding copies or multiplying by a number Paper 6

There is a real difference between $X_1+X_2+\dots +X_n$ and $nX$:

• $X_1+X_2+\dots +X_n$ is the total of $n$ independent observations, such as the total mass of 12 different packets. Its mean is $n\mu$ and its variance is $n{\sigma }^2$, because the $n$ variances add.
• $nX$ is one observation multiplied by $n$, such as 12 times the mass of a single packet. Its mean is also $n\mu$, but its variance is $n^2{\sigma }^2$.

The total of 12 different packets varies less than 12 times one packet, because in a total the heavy and light packets partly cancel out. Before writing a variance, decide which situation the question describes: "the total mass of 12 packets" is a sum; "a recipe needs 3 times the mass of one scoop" is a multiple.

Normal and Poisson variables Paper 6

The rules above give the mean and variance of a combination. Two further facts give its distribution:

• if $X$ is normal, then $aX+b$ is normal;
• if $X$ and $Y$ are independent and normal, then $aX+bY$ is normal.

So to find a probability about a combination of normal variables, find its mean and variance, then standardise as in Paper 5. A comparison such as "$X$ is greater than $Y$" is turned into a single variable: $\mathrm{P}(X>Y)=\mathrm{P}(X-Y>0)$.

For Poisson variables: if $X\sim \mathrm{Po}(\lambda )$ and $Y\sim \mathrm{Po}(\mu )$ are independent, then
$$X+Y\sim \mathrm{Po}(\lambda +\mu )\mathrm{.}$$
This is why the rate of a Poisson process can be scaled to a longer interval. Only the sum is Poisson. $2X$ has mean $2\lambda$ but variance $4\lambda$, so it cannot be Poisson; and $X-Y$ can be negative, so it cannot be Poisson either.