Linear Transformations of DRVs Revision Notes for Edexcel A-Level Further Mathematics

16.1.6 Linear Transformations of DRVs

Linear Transformation of Distribution (Coding)

A linear transformation of a distribution is performing a transformation to a variable of the form

$Y = mX + c$ .

lightbulbExample

Example Take the variable $X$ with the distribution:

$\begin{array}{c|c|c|c} x & 1 & 2 & 3 \\ P(X = x) & \frac{1}{3} & \frac{1}{2} & \frac{1}{6} \\ \end{array}$

$E(X) = 1 \times \frac{1}{3} + 2 \times \frac{1}{2} + 3 \times \frac{1}{6} = \frac{11}{6} = E(X)$

$E(X^2) = 1^2 \times \frac{1}{3} + 2^2 \times \frac{1}{2} + 3^2 \times \frac{1}{6} = \frac{23}{6} = E(X^2)$

$\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{23}{6} - \left(\frac{11}{6}\right)^2 = \frac{17}{36} = \text{Var}(X)$

Now take the distribution of $3X$ :

$\begin{array}{c|c|c|c} x & 3 & 6 & 9 \\ P(X = x) & \frac{1}{3} & \frac{1}{2} & \frac{1}{6} \\ \end{array}$

$E(3X) = 3 \times \frac{1}{3} + 6 \times \frac{1}{2} + 9 \times \frac{1}{6} = \frac{11}{2} = 3E(X)$

$E((3X)^2) = 3^2 \times \frac{1}{3} + 6^2 \times \frac{1}{2} + 9^2 \times \frac{1}{9} = \frac{69}{2}$

$\text{Var}(3X) = E(3X^2) - (E(3X))^2 = \frac{69}{2} - \left(\frac{11}{2}\right)^2 = \frac{17}{4} = 9 \text{Var}(X)$

Reason for relationship:

Consider the point:

The mean/middle point ( $1, 2, 3$ ). If we multiply by $3$ :

Everything is 3 x bigger and 3 x more spread out.
The middle is 3 times bigger $\Rightarrow E(3X) = 3E(X)$
Since the spread is bigger and the variance measures squares of spreads, the new variance will be multiplied by 9.

$\therefore \ E(kX) = kE(X)$

$\text{Var}(kX) = k^2 \text{Var}(X)$

lightbulbExample

Example Consider the distribution of $X+3$ :

$x$	4	5	6
$(P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$\frac{1}{6}$

$E(X+3) = \frac{29}{6} = E(X) + 3$

$E((X+3)^2) = \frac{143}{6}$

$\text{Var}(X+3) = \frac{17}{36} = \text{Var}(X)$

Moving all points the same distance moves the centre of the data set that distance but does nothing to the spread.

$\therefore E(X + k) = E(X) + k$

$\text{Var}(X + k) = \text{Var}(X)$

lightbulbExample

Example The random variable $Y$ has mean $2$ and variance $9$ .

Find:

$E(Y^2)$ $E(X) = 2 \quad \text{Var}(X) = 9$

$E(X^2) - 2^2 = 9 \quad \Rightarrow \quad E(X^2) = 13$

$\text{Var}(2X + 2)$

$\text{Var}(2X + 2) = 2^2 \times \text{Var}(X) = 4 \times 9 = 36$

Combinations of Random Variables

The Sum of Two Normal Distributions

e.g., Say $X \sim N(\mu_x$ , $\sigma_x^2)$ and $Y \sim N(\mu_y, \sigma_y^2)$ are independent. Then the sum of the two distributions is also normal:

$(X + Y) \sim N(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2)$

This is proved using the Probability Density Function (P.D.F.) of the normal distribution.

This also applies to the difference of two normal distributions:

$(X - Y) \sim N(\mu_x - \mu_y, \sigma_x^2 + \sigma_y^2)$

lightbulbExample

Example: If $X \sim N(30, 6)$ and $Y \sim N(35, 4)$ are independent, find $P(5Y > 3X + 66)$ .

Step 1: Rearrange to the form $P(aXe + bY > c) (or < c ).$ $P(5Y - 3X > 66)$

Step 2: State the distribution of the LHS (left-hand side) of the inequality, possibly performing a substitution for presentation purposes.

$Let W = 5Y - 3X .$

$3X \sim N(90, 54)$ $\left(3^2 \times 6\right)$
$5Y \sim N(175, 100)$ $\left(5^2 \times 4\right)$ Therefore,

$W \sim N(175 - 90, 100 + 54)$

$W \sim N(85, 154)$

Step 3: Calculate the probability:

$P(W > 66) = 0.937$

Sum / Difference of Two Poisson Distributions

The sum of two Poisson distributions leads to another Poisson distribution.

If $X \sim Po(\lambda_x)$ and $Y \sim Po(\lambda_y)$ are independent, then:

$(X + Y) \sim Po(\lambda_x + \lambda_y)$

(Table showing distributions for Binomial B(n, p) , Uniform, Geometric, and Poisson with their respective probability formulas, expectations, and variances.)

Notice the Poisson distribution has the same mean as the variance. This means that in the above definitions of $X$ and $Y$ : $\text{Var}(X) = \lambda_x \quad \text{and} \quad \text{Var}(Y) = \lambda_y$

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

Therefore, X + Y has the same mean as variance.

Now consider $X$ - $Y$ :

$E(X - Y) = \lambda_x - \lambda_y$

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

The two are different, meaning that the difference between the two Poisson distributions cannot be itself Poisson.

Linear Transformations of DRVs (Edexcel A-Level Further Mathematics): Revision Notes

16.1.6 Linear Transformations of DRVs

Linear Transformation of Distribution (Coding)

Reason for relationship:

Combinations of Random Variables

The Sum of Two Normal Distributions

Sum / Difference of Two Poisson Distributions

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