A LevelMathsCambridge (CIE)Probability & Statistics 2FlashcardsStatistical DistributionsLinear Combinations of Random Variables

Linear Combinations of Random Variables (Cambridge (CIE) A Level Maths: Probability & Statistics 2): Flashcards

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FrontLinear Combinations of Random Variables
For constants $a$ and $b$ , fill in the two results:
$E (a X + b) =______$
$Var (a X + b) =______$

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Cards in this collection (10)

For constants $a$ and $b$ , fill in the two results:
$E (a X + b) =______$
$Var (a X + b) =______$
The completed results are:
$E (a X + b) = a E (X) + b$
$Var (a X + b) = a^{2} Var (X)$
The mean is affected by both the multiplication and the addition, but the variance is affected only by the multiplication.
The $a$ is squared in the variance, because variance is built from squared deviations.
Why does adding a constant leave the variance unchanged?
Adding a constant slides every value along by the same amount, so the whole distribution shifts without being stretched.
Variance measures spread about the mean, and shifting the values and the mean together leaves every deviation exactly as it was.
So $Var (X + 7)$ and $Var (X)$ are the same number.
If $X$ is normal, is $a X + b$ normal? What if $X$ is Poisson?
If $X$ is normal then $a X + b$ is also normal, specifically $N (a μ + b, a^{2} σ^{2})$ .
A binomial, geometric or Poisson variable does not keep its type: doubling a Poisson count gives only even numbers, which no Poisson distribution does.
That closure under linear change is one of the things that makes the normal distribution so convenient.
For independent $X$ and $Y$ , fill in the two results:
$E (a X + b Y) =______$
$Var (a X + b Y) =______$
The completed results are:
$E (a X + b Y) = a E (X) + b E (Y)$
$Var (a X + b Y) = a^{2} Var (X) + b^{2} Var (Y)$
The result for the expectation holds for any two random variables, but the one for the variance needs them to be independent.
Both constants are squared in the variance, exactly as in the single-variable case.
True or False?
$Var (X - Y)$ is the same as $Var (X + Y)$ for independent $X$ and $Y$ .
True.
Subtracting $Y$ is adding $- Y$ , and the coefficient is squared in the variance, so $(- 1)^{2} = 1$ and nothing changes.
Variances therefore add for a difference just as they do for a sum, which is the single most commonly lost mark in this topic.
If $X$ and $Y$ are independent normal variables, what is the distribution of $a X + b Y$ ?
It is normal as well, with mean $a μ_{1} + b μ_{2}$ and variance $a^{2} σ_{1}^{2} + b^{2} σ_{2}^{2}$ .
This is what makes questions about differences and totals of normal variables tractable at all.
It holds for a difference too, with the means subtracting while the variances still add.
If $X$ and $Y$ are independent Poisson variables, when is a combination of them Poisson?
Only for the plain sum: if $X ~ Po (λ)$ and $Y ~ Po (μ)$ then $X + Y ~ Po (λ + μ)$ .
Once constants are involved it fails: $a X + b Y$ is not Poisson for values of $a$ and $b$ other than 0 and 1.
The normal is closed under any linear combination, but the Poisson is closed only under addition.
$X$ and $Y$ are independent with $Var (X) = 3$ and $Var (Y) = 4$ . What is $Var (4 X - Y)$ ?
It is $4^{2} \times 3 + (- 1)^{2} \times 4 = 48 + 4 = 52$ .
Both coefficients are squared and the two terms are then added, however the original expression was signed.
Writing the squares in explicitly, rather than doing them in your head, is what stops the minus sign getting through.
What is the difference between $2 X$ and $X_{1} + X_{2}$ ?
$2 X$ means one observation of $X$ , doubled; $X_{1} + X_{2}$ means two separate observations added together.
They have the same expected value, $2 E (X)$ , but different variances: $Var (2 X) = 4 Var (X)$ while $Var (X_{1} + X_{2}) = 2 Var (X)$ .
If $X$ could be 0 or 1 then $2 X$ can only be 0 or 2, whereas $X_{1} + X_{2}$ can also be 1, which shows they are genuinely different variables.
A carton holds six eggs. Why is the total mass not modelled by $C + 6 E$ ?
Because the six eggs have six different masses, so the model must be $C + E_{1} + E_{2} + E_{3} + E_{4} + E_{5} + E_{6}$ .
Writing $6 E$ would mean weighing one egg and multiplying by six, which would give a larger variance than the real situation has.
Asking whether the question means one thing repeated or several separate things is what decides between the two.

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