Linear Transformations on a Continuous Random Variable
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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.
- $SD(X) = \sqrt{ Var(X) }$
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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.
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For any independent random variables, X, Y and any constants a, b, we have:
- $E\big(aX + b\big) = aE\big(X\big) + b$
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- $E\big(X + Y\big) = E\big(X\big) + E\big(Y\big)$
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- $\text{Var}\big(aX + b\big) = a^2 \text{Var}\big(X\big)$
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- $\text{Var}\big(X +Y\big) = \text{Var} \big(X\big) + \text{Var} \big(Y\big)$
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Example 1
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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) .. $E \big(3Z^2 – 1 \big)$
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Solution:
… a) .. Find $E\big(3Z - 1\big)$
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… … … $E\big(3Z - 1\big) = 3E\big(Z\big) - 1$
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… … … … … $= 3 \times 5 - 1$
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… … … … … $= 14$
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… b) .. Find $\text{Var}\big(3Z - 1\big)$
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… … … $\text{Var}\big(3Z - 1\big) = 3^2\text{Var}\big(Z\big)$
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… … … … … $= 9 \times 2$
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… … … … … $= 18$
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… c) .. Find $E \big(Z^2 \big)$
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… … … Use $\text{Var} \big( X \big) = E \big( X^2 \big) - \Big[ E \big( X \big) \Big]^2$
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… … … $\text{Var} \big( Z \big) = E \big( Z^2 \big) - \Big[ E \big( Z \big) \Big]^2$
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… … … $2 = E \big( Z^2 \big) - \big( \, 5 \, \big)^2$
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… … … $2 = E \big( Z^2 \big) - 25$
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… … … $E \big( Z^2 \big) = 27$
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… d) .. Find $E \big( 3Z^2 – 1 \big)$
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… … … $E\big(3Z^2 - 1\big) = 3E\big(Z^2\big) - 1$
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… … … … … $= 3 \times 27 - 1$
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… … … … … $= 80$
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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 $K = C + 273$
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… … For E(X) use the rule $E\big(aX + b\big) = aE\big(X\big) + b$
… … … … $E \big(K \big) = E \big(C + 273 \big)$
… … … … $E \big(K \big) = E \big(C \big) + 273$
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… … Daily Minimum: $E \big( K_{min} \big) = 9 + 273 = 282^\circ K$
… … Daily Maximum: $E \big( K_{max} \big ) = 15 + 273 = 288^\circ K$
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… … Recall that $Var(X) = \sigma^2$
… … … … $Var \big( C_{min} \big) = 1.6^2 = 2.56$
… … … … $Var \big( C_{max} \big) = 1.9^2 = 3.61$
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… … For Var(X) use the rule $\text{Var} \big( aX + b \big) = a^2 \text{Var} \big( X \big)$
… … … … $Var \big(K \big) = Var \big( C + 273 \big)$
… … … … $Var \big(K \big) = Var \big(C \big)$
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… … Daily Minimum: $Var \big(K_{min} \big) = 2.56$
… … Daily Maximum: $Var \big( K_{max} \big) = 3.61$
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… b) .. State the expected values and variance for max and min temperature in Fahrenheit.
… … The conversion formula for Fahrenheit from Celsius is:
… … … … $F=\dfrac{5}{9}C+32$
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… … For E(X) use the rule $E\big(aX + b\big) = aE\big(X\big) + b$
… … … … $E \big(F \big) = E \Big( \dfrac{5}{9}C+32 \Big)$
… … … … $E \big(F \big) = \dfrac{5}{9} E \big( C \big) +32$
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… … Daily Minimum: $E \big( F_{min} \big ) = \dfrac{5}{9} \times 9 + 32 = 37^\circ F$
… … Daily Maximum: $E \big( F_{max} \big) = \dfrac{5}{9} \times 15 + 32 = 40.33^\circ F$
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… … For Var(X) use the rule $\text{Var}\big(aX + b\big) = a^2 \text{Var} \big( X \big)$
… … … … $Var\big(F\big) = Var \Big( \dfrac{5}{9}C+32 \Big)$
… … … … $Var\big(F\big) = \Big( \dfrac{5}{9} \Big)^2 Var\big(C\big)$
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… … Daily Minimum: $Var\big(F_{min}\big) = \Big( \dfrac{5}{9} \Big)^2 \times 2.56 = 0.79$
… … Daily Maximum: $Var\big( F_{max}\big) = \Big( \dfrac{5}{9} \Big)^2 \times 3.61 = 1.11$
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