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Distribution of a linear transformation of a discrete random variable

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Linear transformations of discrete distributions

For a nonzero multiplier a, the map x↦ax+b is one-to-one, so each transformed value keeps its paired probability. If a=0, all outcomes map to b and their probabilities combine to 1. The mean and spread identities hold in both cases

\[\begin{gathered}a\ne0:\quad\Pr(Y=ax+b)=\Pr(X=x)\\\operatorname{E}(Y)=a\operatorname{E}(X)+b\end{gathered}\]
  1. 1Transform each value Substitute each value of X into Y = aX + b. \(\displaystyle x=2,\quad Y=3(2)-1=5\)
  2. 2Keep paired probabilities Copy the probability from the original row to its transformed value \(\displaystyle \Pr(X=2)=\frac{1}{4}\Rightarrow\Pr(Y=5)=\frac{1}{4}\)
  3. 3Transform the mean Multiply the original mean by a and add b once \(\displaystyle \operatorname{E}(Y)=a\operatorname{E}(X)+b\)
  4. 4Transform the spread Square a for variance and use the absolute value of a for standard deviation \(\displaystyle \begin{gathered}\operatorname{Var}(Y)=a^2\operatorname{Var}(X)\\\operatorname{SD}(Y)=|a|\operatorname{SD}(X)\end{gathered}\)
Year 12, VCE Units 3 & 4, Chapter 10, 4 difficulty levels, No calculatorBuild a worksheet

Linear transformations of discrete distributions

For a nonzero multiplier a, the map x↦ax+b is one-to-one, so each transformed value keeps its paired probability. If a=0, all outcomes map to b and their probabilities combine to 1. The mean and spread identities hold in both cases

\[a\ne0:\quad\Pr(Y=ax+b)=\Pr(X=x),\qquad\operatorname{E}(Y)=a\operatorname{E}(X)+b\]
  1. Transform each value Substitute each value of X into Y = aX + b. \(\displaystyle x=2,\quad Y=3(2)-1=5\)
  2. Keep paired probabilities Copy the probability from the original row to its transformed value \(\displaystyle \Pr(X=2)=\frac{1}{4}\Rightarrow\Pr(Y=5)=\frac{1}{4}\)
  3. Transform the mean Multiply the original mean by a and add b once \(\displaystyle \operatorname{E}(Y)=a\operatorname{E}(X)+b\)
  4. Transform the spread Square a for variance and use the absolute value of a for standard deviation \(\displaystyle \operatorname{Var}(Y)=a^2\operatorname{Var}(X),\quad\operatorname{SD}(Y)=|a|\operatorname{SD}(X)\)

Worked example

A discrete random variable X has values -1, 2 and 4 with probabilities 1/4, 1/2 and 1/4. Define Y = -2X + 6. Construct the distribution of Y and find E(Y)

\[Y=-2X+6\]
  1. \(\displaystyle X=-1\Rightarrow Y=8,\quad X=2\Rightarrow Y=2,\quad X=4\Rightarrow Y=-2\)Apply the transformation to every value of X.
  2. \(\displaystyle \begin{array}{c|ccc}y&-2&2&8\\\Pr(Y=y)&\frac{1}{4}&\frac{1}{2}&\frac{1}{4}\end{array}\)Reorder the transformed values and keep each original probability with its mapped value
  3. \(\displaystyle \operatorname{E}(X)=(-1)\left(\frac{1}{4}\right)+2\left(\frac{1}{2}\right)+4\left(\frac{1}{4}\right)=\frac{7}{4}\)Find the original expected value from the probability-weighted sum
  4. \(\displaystyle \operatorname{E}(Y)=-2\operatorname{E}(X)+6=-2\left(\frac{7}{4}\right)+6=\frac{5}{2}\)Apply the linear expectation rule to the transformation

Common mistakes

Changing the probabilities
✕\(\displaystyle \Pr(Y=-2)=(-2)\times\frac{1}{4}=-\frac{1}{2}\)✓\(\displaystyle \Pr(Y=-2)=\Pr(X=4)=\frac{1}{4}\)
Probabilities are copied to transformed values and are not changed by a or b.
Leaving values unmapped
✕\(\displaystyle Y=3(2)=6\)✓\(\displaystyle Y=3(2)-1=5\)
The complete rule Y = aX + b must be applied, including its constant term
Using a signed standard deviation
✕\(\displaystyle \operatorname{SD}(-2X+6)=-2\operatorname{SD}(X)\)✓\(\displaystyle \operatorname{SD}(-2X+6)=|-2|\operatorname{SD}(X)=2\operatorname{SD}(X)\)
Standard deviation is non-negative, so the multiplier is used through its absolute value

Difficulty levels

introBuild a transformed table with a positive multiplier and calculate one moment
standardUse a negative constant in a payoff or cost transformation and calculate E(Y) and SD(Y)
challengeCalculate an event probability for Y and check the variance after transformation
examChoose the transformed-distribution table with no named probability or value error

Questions about this topic

How is a transformed distribution constructed?+

Calculate ax+b for every original value x and carry its probability with it. Combine probabilities if several original values map to the same new value. For a nonzero linear multiplier the map is one-to-one; for a=0 all mass lies at b.

Why do the probabilities stay unchanged?+

Each original outcome keeps its probability. A transformed value receives the sum of the probabilities of all outcomes that map to it

How does a linear transformation affect the mean and variance?+

For Y = aX + b, E(Y) = aE(X) + b and Var(Y) = a^2 Var(X); the constant shifts the mean but not the variance

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 10: Discrete random variables

In this chapter

Validate, complete and normalise a discrete probability distributionFinite events, probability assignments and complementsDistribution of a finite random variableRepresent a multi-stage finite sample spaceExpected value and linear expectation of a discrete random variableVariance, standard deviation and linear variance of a discrete random variableInterval, conditional and μ ± 2σ probabilities from a discrete distributionDistribution of a linear transformation of a discrete random variableConditional probability from exact representations