14.2Linearcombinations


 * Linear Combinations of Random Variables **

The expected value (E(X) or μ) of a random varible is a measure of the centre of the data (think mean or average in statistics). For revision (from methods) on how to calculate these for discrete and continuous variables click the links. The variance of a random variable is a measure of spread, that is, whether the data or probabilities are clustered together or spread over a wide range of values. For revision on how to calculate these for discrete and continuous variables see the links. Note: Variance ( Var(X) ) is equal to the standard deviation (σ) squared

For this section of the specialist course you will generally be given values for the expected value or variance and be asked to transform them or combine variables.


 * Expected value theorems **
 * E(aX) = aE(X)**, where a is a constant

.**E(aX + b) = aE(X) + b**, where a, b are constants

.**E(X + Y) = E(X) + E(Y)**, where X, Y are random variables and therefore **E(aX+bY) = aE(X)+bE(Y)**

Note that E(X 2 ) is not equal to E(X) 2.

math \text{because } \sigma^2=E(X- \mu)^2 math The constant a is squared after it comes out of the expectation theorem for E(X)
 * Variance theorems **
 * Var(aX) = a 2 Var(X)**

... . **Var(X + b) = Var(X)** (Shifting all values does not change the spread)

and **Var(aX + bY) = a 2 Var(X) + b 2 Var(Y)** (only if X and Y are independent)

The maximum and minimum temperature at Blackburn High School during June one year is found to be randomly distributed with a mean of 9 and 15 degrees Celsius respectively. The standard deviation is 1.6 and 1.9 respectively. __**Solution**__ 0 degrees Celsius is equal to 273 degrees Kelvin. This is just a straight linear shift. From our expectation theorems E(X+b) = E(X) +b Therefore the minimum and maximum expected values in Kelvin are 273+9 = 282K and 273+15 = 288K respectively.
 * Example **
 * a)** A student is doing some lab work and needs to convert this to Kelvin. State the expected values and **variance** for max and min temperature in Kelvin.

The variance is unchanged and hence is 2.56 and 3.61 respectively (variance is SD squared).


 * b)** The same student decides to send his work to a friend in the USA. They need to find these values in Fahrenheit (Silly USA). Find these vales.

__**Solution**__ The conversion formula for Fahrenheit from Celsius is: math F=\frac{5}{9}C+32 \\ math So there is a multiplicative and constant shift. From our theorems we get: math E( \frac{5}{9} X+32)= \frac{5}{9} E(X)+32 \\ E( \frac{5}{9} X_{\text{min}}+32)= 37 \\ \text{and } E( \frac{5}{9} X_{\text{max}}+32)=40.33 \\ math For our variance we get math Var(X_{\text{min}})=(\frac{5}{9})^2 (2.56) \\ Var(X_{\text{min}}) = 0.79 \\ \\ Var(X_{\text{max}})=(\frac{5}{9})^2(3.61) \\ Var(X_{\text{max}}) = 1.11 \\ math