Linear transformations of moments
A linear transformation changes the centre by multiplying and adding. It changes spread through the multiplier only, while independent totals add their means and variances
- Transform the mean linearly Multiply the mean by the coefficient of X, then add the constant \(\displaystyle E(aX+b)=aE(X)+b\)
- Square the coefficient for variance The constant changes location but not spread, so it is absent from the variance rule \(\displaystyle \operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)\)
- Use the absolute coefficient for standard deviation Standard deviation is non-negative, so a negative coefficient is changed to its absolute value \(\displaystyle \operatorname{SD}(aX+b)=|a|\operatorname{SD}(X)\)
- Add moments for independent totals For independent identically distributed variables, means and variances each add across the copies \(\displaystyle E(S_n)=nE(X),\quad\operatorname{Var}(S_n)=n\operatorname{Var}(X)\)
- Recover the second moment Add the variance to the square of the mean \(\displaystyle E(X^2)=\operatorname{Var}(X)+[E(X)]^2\)
Worked example
Use the displayed mean and variance of the continuous random variable to find its second moment and the mean, variance and standard deviation of its linear transformation. For a second example, take nine independent, identically distributed copies of the transformed variable and find the mean, variance and standard deviation of their total
- \(\displaystyle E(Y)=(-2)(4)+7=-1\)Multiply the mean by -2 and then add 7.
- \(\displaystyle \operatorname{Var}(Y)=(-2)^2\left(\frac{9}{4}\right)=9\)Square the multiplier and leave the added constant out of the variance
- \(\displaystyle \operatorname{SD}(X)=\sqrt{\frac{9}{4}}=\frac{3}{2}\)Take the principal square root of the original variance
- \(\displaystyle \operatorname{SD}(Y)=|-2|\left(\frac{3}{2}\right)=3\)Multiply the original standard deviation by the absolute value of the coefficient
- \(\displaystyle E(X^2)=\operatorname{Var}(X)+[E(X)]^2\)Recover the second moment of the original variable by adding its variance to its squared mean
- \(\displaystyle E(X^2)=\frac94+4^2=\frac94+16=\frac{73}{4}\)Substitute the displayed original moments and use a common denominator
- \(\displaystyle E(S)=\sum_{i=1}^{9}E(Y_i)=9E(Y)\)Linearity adds the means and identical distributions give the same mean for every copy
- \(\displaystyle E(S)=9\times(-1)=-9\)Multiply the transformed mean by the number of copies
- \(\displaystyle \operatorname{Var}(S)=\sum_{i=1}^{9}\operatorname{Var}(Y_i)=9\operatorname{Var}(Y)\)Independence allows the variances to add and identical distributions give the same variance for every copy
- \(\displaystyle \operatorname{Var}(S)=9\times(-2)^2\times\frac94=81\)Multiply the transformed variance by the number of independent copies, not its square
- \(\displaystyle \operatorname{SD}(S)=\sqrt{\operatorname{Var}(S)}=\sqrt{81}=9\)Take the non-negative principal square root after adding the variances
Common mistakes
Difficulty levels
Questions about this topic
Why does the added constant not affect variance?+
Adding a constant moves every value by the same amount, so each distance from the mean is unchanged. The spread therefore depends on the multiplier only
Why is the absolute value used for standard deviation?+
Standard deviation cannot be negative. A negative multiplier reverses the order of values but changes their spread by its positive magnitude
How do moments behave for a total of independent copies?+
The means add and the variances add. The standard deviation is the principal square root of the total variance, not the sum of individual standard deviations
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 12: Continuous probability distributions