Linear Transformations of Normal Variables

Why Transform a Normal Variable?

Measurements are often shifted or rescaled before they are reported. A temperature may be converted between scales, or a measurement may be adjusted by adding a fixed correction. When the original quantity follows a normal model, it is useful to know whether the transformed quantity is still normal and how its center and spread change. These answers follow directly from the normal density.

Here, a linear transformation means multiplying a random variable by a constant and adding a constant. This is a transformation of one random variable; it does not require a new assumption about how observations were collected. The result describes the probability model after the transformation.

Definition: If \(X\) is a random variable and \(a\) and \(b\) are fixed constants, then \(Y=aX+b\) is a linear transformation of \(X\). If \(X\sim N(\mu,\sigma^2)\) with \(\sigma>0\) and \(a\ne 0\), then \(Y\sim N(a\mu+b,a^2\sigma^2)\). Thus the mean of \(Y\) is \(a\mu+b\), its variance is \(a^2\sigma^2\), and its standard deviation is \(|a|\sigma\).

The formula covers adding a constant by setting \(a=1\), and multiplying by a constant by setting \(b=0\). A negative multiplier reverses the direction of the scale, but it cannot make a standard deviation negative. The absolute value in \(|a|\sigma\) accounts for that fact.

Transforming the Density

We derive the result from the density, rather than assuming that a transformed variable has a particular distribution. Suppose \(X\sim N(\mu,\sigma^2)\), where \(\sigma>0\), and set \(Y=aX+b\), with \(a\ne 0\). The normal density of \(X\) is

f_X(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).

The transformation can be inverted: \(x=(y-b)/a\). For a one-to-one differentiable transformation, the density of \(Y\) is the density of \(X\) at the inverse-transformed value, multiplied by the absolute derivative of that inverse. Here the derivative is \(1/a\), so its absolute value is \(1/|a|\). Therefore,

f_Y(y)=f_X\!\left(\frac{y-b}{a}\right)\frac{1}{|a|}.

Substituting the normal density gives

\begin{aligned} f_Y(y) &=\frac{1}{|a|\sigma\sqrt{2\pi}} \exp\!\left[-\frac{\left(\frac{y-b}{a}-\mu\right)^2}{2\sigma^2}\right]\\ &=\frac{1}{|a|\sigma\sqrt{2\pi}} \exp\!\left[-\frac{(y-(a\mu+b))^2}{2a^2\sigma^2}\right]. \end{aligned}

For the second equality, \(\frac{y-b}{a}-\mu=\frac{y-(a\mu+b)}{a}\), so squaring this difference divides its square by \(a^2\). The final density has the normal form, with center \(a\mu+b\) and variance \(a^2\sigma^2\). Since a standard deviation is the nonnegative square root of the variance, it is \(\sqrt{a^2\sigma^2}=|a|\sigma\). This establishes both that the transformed variable is normal and the values of its parameters.

Theorem: If \(X\sim N(\mu,\sigma^2)\), \(\sigma>0\), and \(a\ne 0\), then \(aX+b\sim N(a\mu+b,a^2\sigma^2)\). Equivalently, adding \(b\) shifts the mean by \(b\) without changing the standard deviation, while multiplying by \(a\) multiplies the mean by \(a\), the variance by \(a^2\), and the standard deviation by \(|a|\).

The restriction \(a\ne 0\) matters. If \(a=0\), then \(Y=b\) with probability 1: it is a constant, with variance zero, not a nondegenerate normal variable with a density of the form above. The change-of-variable derivation also requires \(a\ne0\), because only then can the transformation be inverted.

Applying the Transformation Rules

Worked Example: Adding a Fixed Amount

Suppose a hypothetical instrument reading \(X\), measured in millimeters, is modeled as \(X\sim N(50,8^2)\). A correction adds 12 millimeters to every reading, so \(Y=X+12\). Find the distribution of \(Y\).

The starting model is normal with \(\mu=50\) mm and \(\sigma=8\) mm. The transformation has \(a=1\) and \(b=12\), so the theorem applies. Its mean is

\mu_Y=a\mu+b=(1)(50)+12=62\text{ mm}.

Its standard deviation is

\sigma_Y=|a|\sigma=|1|(8)=8\text{ mm}.

Thus \(Y\sim N(62,8^2)\), with the mean and standard deviation stated in millimeters. Adding the same amount to every value shifts the distribution but does not change its spread.

Worked Example: Multiplying by a Negative Constant

Let a hypothetical variable \(X\), measured in degrees, satisfy \(X\sim N(10,3^2)\). Define \(W=100-2X\). Find the distribution of \(W\), and explain the role of the negative multiplier.

Write the transformation as \(W=aX+b\), with \(a=-2\) and \(b=100\). The input is normal and the multiplier is nonzero, so \(W\) is normal. Its mean is

\mu_W=a\mu+b=(-2)(10)+100=80.

The standard deviation is

\sigma_W=|a|\sigma=|-2|(3)=6.

Therefore \(W\sim N(80,6^2)\). Multiplication by \(-2\) reverses the order of values and doubles their distances from the center. The added 100 then shifts the center. The standard deviation is 6, not \(-6\), because spread is nonnegative.

Worked Example: Rescaling and Then Shifting a Measurement

Suppose a hypothetical length \(X\) in centimeters follows \(X\sim N(120,15^2)\). A conversion and offset define \(C=0.8X+5\), where \(C\) is reported in a new measurement scale. Find its normal model and describe how its spread compares with that of \(X\).

Here \(a=0.8\), \(b=5\), \(\mu=120\) cm, and \(\sigma=15\) cm. The starting variable is normal, and \(a\ne0\), so the transformed variable is normal. Its mean, in the new scale’s units, is

\mu_C=a\mu+b=(0.8)(120)+5=96+5=101.

Its standard deviation is

\sigma_C=|a|\sigma=|0.8|(15)=12.

Thus \(C\sim N(101,12^2)\), with both parameters expressed in the new scale’s units. The multiplier reduces the standard deviation from 15 to 12, while adding 5 changes the mean but leaves that spread unchanged.

Why the Mean and Spread Change Differently

The parameter formulas have a simple interpretation. Adding \(b\) moves every possible value by the same amount, so it moves the center but leaves the distance between any two transformed values unchanged. Multiplying by \(a\) scales those distances by \(|a|\). If \(|a|>1\), distances from the center increase; if \(0<|a|<1\), they decrease. If \(a<0\), the order is also reversed, but the size of each distance is still scaled by \(|a|\).

The variance uses \(a^2\) because variance is based on squared distances from the mean. In particular, for \(Y=aX+b\), subtracting its mean gives \(Y-(a\mu+b)=a(X-\mu)\). Squaring multiplies the squared deviation by \(a^2\); adding \(b\) has no effect on that deviation. This agrees with the variance in the density derivation.

These results require no assumption that the constants are random or that the transformation is estimated from data: \(a\) and \(b\) are fixed. For a single transformed variable, no independence assumption is needed. The key distributional condition is that the original \(X\) is normal. If that condition is not met, the transformed variable need not be normal.

Write the transformation in a consistent form.
Express it as \(Y=aX+b\), identifying the multiplier and added constant, including their signs.

Check the starting model and multiplier.
Confirm that \(X\sim N(\mu,\sigma^2)\) with \(\sigma>0\). The nondegenerate normal result applies when \(a\ne0\).

Transform the parameters.
Calculate \(\mu_Y=a\mu+b\) and \(\sigma_Y=|a|\sigma\). If variance is needed, calculate \(\sigma_Y^2=a^2\sigma^2\).

Report the model in context.
State that \(Y\) is normal, give its parameters, and express the mean and standard deviation in the units of \(Y\).

Common Mistakes

Keeping the original mean after a shift. If \(Y=X+b\), its mean is \(\mu+b\), not \(\mu\). The spread stays the same, but the center moves.
Using \(a\sigma\) as the standard deviation when \(a<0\). The standard deviation is \(|a|\sigma\), which cannot be negative. The variance is \(a^2\sigma^2\).
Adding \(b\) to the standard deviation. A fixed shift moves every value equally and does not change distances from the mean. It therefore changes the mean, not the standard deviation.
Claiming the result is a nondegenerate normal distribution when \(a=0\). In that case \(Y=b\) always, so its spread is zero and it is a constant rather than a normal variable with a positive standard deviation.
Assuming any transformation of any variable produces a normal variable. The theorem applies because the starting variable is normal and the transformation is linear. It does not say that transforming a nonnormal variable this way must make it normal.

Key takeaway: For \(X\sim N(\mu,\sigma^2)\) and fixed \(a\ne0\), the transformed variable \(aX+b\) is normal with mean \(a\mu+b\) and standard deviation \(|a|\sigma\). A shift changes location; a multiplier changes scale, and a negative multiplier also reverses direction.

Check Your Understanding

For each question, identify the transformed mean and standard deviation where requested. Assume any stated normal model has positive standard deviation.

If \(X\sim N(20,4^2)\) and \(Y=X-3\), what are the mean and standard deviation of \(Y\)?
If \(X\sim N(5,2^2)\) and \(V=-3X\), what are the mean and standard deviation of \(V\)? Why is the standard deviation not negative?
If \(X\sim N(\mu,\sigma^2)\) and \(W=2X+7\), write the distribution of \(W\) using \(\mu\) and \(\sigma\).
What happens to the distribution when \(a=0\) in \(Y=aX+b\)? Why does the density derivation not apply?
Explain why adding a fixed constant changes the mean but not the standard deviation.