12 5lineartrans

# Linear Transformations on a Continuous Random Variable

.

## Recall

• The Expected Value, E(X) or $\mu$, of any random variable is a measure of the centre of the data.
• The Variance, Var(X) or $\sigma^2$ and Standard Deviation, SD(X) or $\sigma$, of any random variable are measures of the spread of the data
• The spread of the data refers to whether the data is tightly clustered around the mean, or widely spread.
• $SD(X) = \sqrt{ Var(X) }$

.

## Linear Transformations

We can apply the same linear transformations on E(X) and Var(X) on any random variables, whether they are Discrete or Continuous.
.

For any independent random variables, X, Y and any constants a, b, we have:

• $E\big(aX + b\big) = aE\big(X\big) + b$

.

• $E\big(X + Y\big) = E\big(X\big) + E\big(Y\big)$

.

• $\text{Var}\big(aX + b\big) = a^2 \text{Var}\big(X\big)$

.

• $\text{Var}\big(X +Y\big) = \text{Var} \big(X\big) + \text{Var} \big(Y\big)$

.

### Example 1

.

The continuous random variable Z has a mean of 5 and a variance of 2

Find

a) .. $E \big(3Z - 1\big)$

b) .. $\text{Var}\big(3Z - 1\big)$

c) .. $E \big( Z^2 \big)$

d) .. $E \big(3Z^2 – 1 \big)$
.

Solution:

a) .. Find $E\big(3Z - 1\big)$
.

… … … $E\big(3Z - 1\big) = 3E\big(Z\big) - 1$
.

… … … … … $= 3 \times 5 - 1$
.

… … … … … $= 14$

.

b) .. Find $\text{Var}\big(3Z - 1\big)$
.

… … … $\text{Var}\big(3Z - 1\big) = 3^2\text{Var}\big(Z\big)$
.

… … … … … $= 9 \times 2$
.

… … … … … $= 18$

.

c) .. Find $E \big(Z^2 \big)$
.

… … … Use $\text{Var} \big( X \big) = E \big( X^2 \big) - \Big[ E \big( X \big) \Big]^2$

.

… … … $\text{Var} \big( Z \big) = E \big( Z^2 \big) - \Big[ E \big( Z \big) \Big]^2$
.

… … … $2 = E \big( Z^2 \big) - \big( \, 5 \, \big)^2$
.

… … … $2 = E \big( Z^2 \big) - 25$
.

… … … $E \big( Z^2 \big) = 27$

.

d) .. Find $E \big( 3Z^2 – 1 \big)$
.

… … … $E\big(3Z^2 - 1\big) = 3E\big(Z^2\big) - 1$
.

… … … … … $= 3 \times 27 - 1$
.

… … … … … $= 80$
.

### Example 2

The maximum and minimum daily 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 for the daily minimum and 1.9 for the daily maximum.

A student is doing some lab work and needs to convert this to Kelvin.

a) .. State the expected values and variance for max and min temperature in Kelvin.

The same student decides to send his work to a friend in the USA. The students in USA need the values in Fahrenheit.

b) .. State the expected values and variance for max and min temperature in Fahrenheit.

Solution

a) .. State the expected values and variance for max and min temperature in Kelvin.

… … … … $0^\circ C = 273^\circ K$ This is just a straight linear shift.

… … To convert any temperature from Celsius to Kelvin, use $K = C + 273$
.

… … For E(X) use the rule $E\big(aX + b\big) = aE\big(X\big) + b$

… … … … $E \big(K \big) = E \big(C + 273 \big)$

… … … … $E \big(K \big) = E \big(C \big) + 273$
.

… … Daily Minimum: $E \big( K_{min} \big) = 9 + 273 = 282^\circ K$

… … Daily Maximum: $E \big( K_{max} \big ) = 15 + 273 = 288^\circ K$
.

… … Recall that $Var(X) = \sigma^2$

… … … … $Var \big( C_{min} \big) = 1.6^2 = 2.56$

… … … … $Var \big( C_{max} \big) = 1.9^2 = 3.61$
.

… … For Var(X) use the rule $\text{Var} \big( aX + b \big) = a^2 \text{Var} \big( X \big)$

… … … … $Var \big(K \big) = Var \big( C + 273 \big)$

… … … … $Var \big(K \big) = Var \big(C \big)$
.

… … Daily Minimum: $Var \big(K_{min} \big) = 2.56$

… … Daily Maximum: $Var \big( K_{max} \big) = 3.61$
.

b) .. State the expected values and variance for max and min temperature in Fahrenheit.

… … The conversion formula for Fahrenheit from Celsius is:

… … … … $F=\dfrac{5}{9}C+32$
.

… … For E(X) use the rule $E\big(aX + b\big) = aE\big(X\big) + b$

… … … … $E \big(F \big) = E \Big( \dfrac{5}{9}C+32 \Big)$

… … … … $E \big(F \big) = \dfrac{5}{9} E \big( C \big) +32$
.

… … Daily Minimum: $E \big( F_{min} \big ) = \dfrac{5}{9} \times 9 + 32 = 37^\circ F$

… … Daily Maximum: $E \big( F_{max} \big) = \dfrac{5}{9} \times 15 + 32 = 40.33^\circ F$
.

… … For Var(X) use the rule $\text{Var}\big(aX + b\big) = a^2 \text{Var} \big( X \big)$

… … … … $Var\big(F\big) = Var \Big( \dfrac{5}{9}C+32 \Big)$

… … … … $Var\big(F\big) = \Big( \dfrac{5}{9} \Big)^2 Var\big(C\big)$
.

… … Daily Minimum: $Var\big(F_{min}\big) = \Big( \dfrac{5}{9} \Big)^2 \times 2.56 = 0.79$

… … Daily Maximum: $Var\big( F_{max}\big) = \Big( \dfrac{5}{9} \Big)^2 \times 3.61 = 1.11$

.