Location and Scale Transformations of Sample Means

Why Transformations Matter

A sample mean describes the average of a random sample, so it is itself a random variable with a sampling distribution. Changing the measurement scale or adding a fixed amount to every observation changes that distribution in a predictable way. The same reasoning applies whether the observations represent measurements, hypothetical insurance payments, or values recorded in another unit.

The key is to transform every observation consistently, then express the new sample mean in terms of the original one. A shift changes the center but not the spread. A scale changes the spread, and a negative scale also reverses the ordering of possible values.

Definition: Let \(X_1,\ldots,X_n\) be an independent and identically distributed (IID) random sample from a population with \(E[X_i]=\mu\) and \(\operatorname{Var}(X_i)=\sigma^2<\infty\). Its sample mean is \(\bar{X}=(X_1+\cdots+X_n)/n\). For constants \(a\) and \(b\), transform each observation to \(Y_i=a+bX_i\). The transformed sample mean is \(\bar{Y}=(Y_1+\cdots+Y_n)/n\).

The Transformation Rules

The sample-mean identity follows directly from the definition:

\bar{Y} =\frac{1}{n}\sum_{i=1}^n(a+bX_i) =a+b\left(\frac{1}{n}\sum_{i=1}^nX_i\right) =a+b\bar{X}.

This identity is exact for every realized sample. It does not require normality or even independence. It immediately gives a rule for the sampling distribution: the distribution of \(\bar{Y}\) is the distribution of \(\bar{X}\) shifted by \(a\) and scaled by \(b\).

Theorem: Under the IID assumptions above, \(E[\bar{Y}]=a+b\mu\) and \(\operatorname{Var}(\bar{Y})=b^2\sigma^2/n\). If \(b\ne 0\), the distribution is the affine transformation \(a+b\bar{X}\) of the sampling distribution of \(\bar{X}\). If \(b=0\), \(\bar{Y}\) equals \(a\) with probability 1.

The expectation and variance formulas can be justified without knowing the specific distribution. Linearity of expectation gives \(E[\bar{X}]=\mu\). Independence gives \(\operatorname{Var}(\bar{X})=\sigma^2/n\), because the variance of a sum of independent variables is the sum of their variances. Applying the usual expectation and variance rules to \(a+b\bar{X}\) yields

E[\bar{Y}]=E[a+b\bar{X}]=a+b\mu, \qquad \operatorname{Var}(\bar{Y}) =\operatorname{Var}(a+b\bar{X}) =b^2\operatorname{Var}(\bar{X}) =\frac{b^2\sigma^2}{n}.

Consequently, the standard deviation of the transformed sample mean, also called its standard error, is \(|b|\sigma/\sqrt{n}\). The absolute value matters: a standard deviation cannot be negative. A shift \(a\) does not affect it, while a scale \(b\) multiplies it by \(|b|\). The variance uses \(b^2\), not \(b\).

How the Distribution Changes

For a positive scale, larger values remain larger. For a negative scale, the order reverses, so the cumulative distribution function needs special care. Write \(F_{\bar{X}}(t)=P(\bar{X}\le t)\). If \(b>0\), then

F_{\bar{Y}}(y) =P\left(\bar{X}\le\frac{y-a}{b}\right) =F_{\bar{X}}\left(\frac{y-a}{b}\right).

If \(b<0\), dividing an inequality by \(b\) reverses its direction. The exact formula, including when \(\bar{X}\) has point masses, is

F_{\bar{Y}}(y) =P\left(\bar{X}\ge\frac{y-a}{b}\right) =1-F_{\bar{X}}\left(\left(\frac{y-a}{b}\right)-\right).

Here \(F_{\bar{X}}(t-)=P(\bar{X}<t)\) is the left-hand limit of the distribution function at \(t\). It is needed because the event uses \(\ge\), not \(>\). For a continuous distribution, there is no point mass at the cutoff, and the formula simplifies to \(1-F_{\bar{X}}((y-a)/b)\). For a discrete distribution, omitting the left-hand limit can lose the probability at the cutoff.

Worked Example: A Shifted and Rescaled Bernoulli Sample Mean

Suppose \(X_1,\ldots,X_4\) are IID Bernoulli variables with \(P(X_i=1)=0.25\) and \(P(X_i=0)=0.75\). For a hypothetical dollar-valued measure, transform each observation to \(Y_i=100+20X_i\). Find the sampling distribution of \(\bar{Y}\), its expectation and variance, and \(P(\bar{Y}\ge110)\). The method is to count the number of successes in the original Bernoulli sample before applying the affine transformation.

Let \(K=X_1+\cdots+X_4\). Then \(K\) has a binomial distribution with \(n=4\) and success probability \(0.25\), and \(\bar{X}=K/4\). The transformed mean is

\bar{Y}=100+20\bar{X}=100+5K.

The binomial probabilities, shown with denominator \(256\), are

\begin{aligned} P(K=0)&=(3/4)^4=81/256,\\ P(K=1)&=4(1/4)(3/4)^3=108/256,\\ P(K=2)&=6(1/4)^2(3/4)^2=54/256,\\ P(K=3)&=4(1/4)^3(3/4)=12/256,\\ P(K=4)&=(1/4)^4=1/256. \end{aligned}

Thus \(\bar{Y}\) takes values \(100,105,110,115,120\), with respective probabilities \(81/256,108/256,54/256,12/256,1/256\). These probabilities sum to \((81+108+54+12+1)/256=1\), as required.

A Bernoulli variable with success probability \(0.25\) has mean \(0.25\) and variance \(0.25(0.75)=0.1875\). Therefore

E[\bar{Y}]=100+20(0.25)=105, \qquad \operatorname{Var}(\bar{Y}) =\frac{20^2(0.1875)}{4} =18.75.

For the tail probability, \(\bar{Y}\ge110\) is equivalent to \(100+5K\ge110\), or \(K\ge2\). Hence

P(\bar{Y}\ge110) =\frac{54+12+1}{256} =\frac{67}{256} =0.2617.

The method avoids treating \(20\) as the variance multiplier: it is squared, so the variance is \(18.75\). As a check on the center, \(E[K]=4(0.25)=1\), giving \(E[\bar{Y}]=100+5(1)=105\), in agreement with the formula.

Worked Example: Transforming a Normal Sampling Distribution

Let \(X_1,\ldots,X_9\) be IID with \(X_i\sim N(50,100)\), where the second parameter is the variance. Define \(W_i=2X_i-30\). Find the distribution of \(\bar{W}\), its standard error, and \(P(\bar{W}>78)\). The method is to transform the normal population parameters and then use the exact sampling distribution of the mean of an IID normal sample.

Each transformed observation is normal, with mean \(2(50)-30=70\) and variance \(2^2(100)=400\). Since the observations remain independent, their mean is normal. Equivalently, \(\bar{W}=2\bar{X}-30\), with

\bar{W}\sim N\left(70,\frac{400}{9}\right).

The standard error is the square root of the variance:

\operatorname{SE}(\bar{W})=\sqrt{400/9}=\frac{20}{3}=6.6667

(to four decimal places). To compute the requested probability, express the cutoff in standard-error units:

\frac{78-70}{20/3} =\frac{8(3)}{20} =1.2.

Therefore \(P(\bar{W}>78)=P(Z>1.2)=1-\Phi(1.2)=0.1151\), where \(Z\sim N(0,1)\) and \(\Phi\) is its cumulative distribution function. The calculation avoids using the population standard deviation \(20\) as the standard error; averaging nine observations reduces the standard error to \(20/3\). The transformed mean is \(70\), and the variance \(400/9\) is consistent with a standard error of \(20/3\).

Worked Example: A Negative Scale and a Discrete Distribution

Suppose \(X_1,X_2\) are IID, with \(P(X_i=0)=P(X_i=8)=0.5\). Transform each observation to \(Z_i=12-1.5X_i\). Find the distribution, expectation, and variance of \(\bar{Z}\), and calculate \(P(\bar{Z}\le6)\). This example uses the distribution transformation directly; its main trap is that the negative scale reverses the ordering.

The possible values of \(X_1+X_2\) are \(0,8,16\). Their probabilities are \(0.25,0.50,0.25\), respectively: both observations are 0, exactly one is 8, or both are 8. Consequently, \(\bar{X}\) takes values \(0,4,8\) with those same probabilities. Since \(\bar{Z}=12-1.5\bar{X}\), its distribution is

\begin{array}{c|ccc} \bar{Z} & 12 & 6 & 0\\ P(\bar{Z}) & 0.25 & 0.50 & 0.25 \end{array}

For one observation, \(E[X_i]=0(0.5)+8(0.5)=4\), and \(E[X_i^2]=0^2(0.5)+8^2(0.5)=32\). Thus \(\operatorname{Var}(X_i)=32-4^2=16\). The formulas give

E[\bar{Z}]=12-1.5(4)=6, \qquad \operatorname{Var}(\bar{Z}) =\frac{(-1.5)^2(16)}{2} =18.

From the displayed distribution, \(P(\bar{Z}\le6)=P(\bar{Z}=6)+P(\bar{Z}=0)=0.50+0.25=0.75\). As a check, its mean is \(12(0.25)+6(0.50)+0(0.25)=6\), and its variance is \((12-6)^2(0.25)+(6-6)^2(0.50)+(0-6)^2(0.25)=18\). The sign of the scale affects the order of the values, but its square determines the variance.

Assumptions, Uses, and Exam Traps

The identity \(\bar{Y}=a+b\bar{X}\) holds for any observations and constants. The stated moment formulas require an IID sample with a finite population variance. If observations are independent but not identically distributed, the transformed mean still has expectation \(a+b\sum_i E[X_i]/n\), while its variance is \(b^2\sum_i\operatorname{Var}(X_i)/n^2\). If observations are dependent, covariance terms must also be included; the IID variance formula need not apply.

A linear transformation preserves the general shape of a distribution, but it does not make an arbitrary sampling distribution normal. In particular, the exact normal result in the second example depends on the original population being normal. For a nonnormal population, the shift-and-scale identities for the mean and variance still hold, but a normal sampling distribution does not follow from those identities alone.

Do not forget to square the scale for variance. Multiplying all observations by \(b\) multiplies standard deviation by \(|b|\) and variance by \(b^2\).
Do not scale the shift by sample size. Adding \(a\) to each observation adds \(a\), not \(na\), to the sample mean.
Do not ignore a negative scale. It reverses inequalities and the ordering of support values, although variance remains nonnegative.
Keep units consistent. If every observation is converted by a factor of 100, the sample mean is also multiplied by 100, while its variance is multiplied by \(100^2\).
Distinguish a point probability from a tail probability. With discrete data, the probability at the cutoff may matter when translating a CDF or a strict versus non-strict inequality.

These rules are useful when comparing measurements recorded in different units, shifting a variable by a fixed baseline, or translating a known sampling distribution into a new scale. They let you update the center, spread, and probability statements without recomputing the full distribution from the observations.

Check Your Understanding

For each problem, identify the transformation of the sample mean before calculating.

Let \(X_1,\ldots,X_5\) be IID with mean 12 and variance 25. Set \(T_i=3+2X_i\). Find \(E[\bar{T}]\) and \(\operatorname{Var}(\bar{T})\). (Answer: \(27\) and \(20\), respectively.)
Let \(X_1,\ldots,X_5\) be IID Bernoulli with success probability \(0.4\), and set \(V_i=10-5X_i\). Find the expectation and variance of \(\bar{V}\). (Answer: \(E[\bar{V}]=8\), \(\operatorname{Var}(\bar{V})=1.2\).)
Let \(X_1,\ldots,X_4\) be IID with \(X_i\sim N(20,16)\), and define \(U_i=7+0.5X_i\). Find \(P(\bar{U}>19)\). (Answer: \(0.0228\).)
Let \(X_1,X_2\) be IID and equally likely to be 2 or 6, and let \(Z_i=5-2X_i\). Find \(E[\bar{Z}]\), \(\operatorname{Var}(\bar{Z})\), and \(P(\bar{Z}<0)\). (Answer: \(-3\), \(8\), and \(0.75\).)
A sample mean \(\bar{X}\) has values \(-1,0,2\) with probabilities \(0.2,0.5,0.3\). Let \(T=4-3\bar{X}\). Give the distribution of \(T\), its expectation, and its variance. (Answer: \(T=7,4,-2\) with probabilities \(0.2,0.5,0.3\); \(E[T]=2.8\), \(\operatorname{Var}(T)=11.16\).)

What You'll Learn