Normal Location-Scale Transformations

Why Location-Scale Transformations Matter

A standard normal variate is a useful building block in simulation because a simple transformation changes its center and spread while preserving its normal shape. For example, a simulated normal claim amount might need a specified mean in dollars and a specified variance in dollars squared. The transformation must use the standard deviation—not the variance—as its scale.

This tutorial develops the transformation, establishes the distribution and moments it produces, and applies it to hypothetical simulation tasks. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let \(Z\sim N(0,1)\), and let \(\mu\) be a real number and \(\sigma>0\). The location-scale transformation \(X=\mu+\sigma Z\) has distribution \(N(\mu,\sigma^2)\). Here, \(\mu\) is the target mean and \(\sigma\) is the target standard deviation; the target variance is \(\sigma^2\).

Deriving the Distribution and Moments

The transformation has two parts. Multiplying \(Z\) by \(\sigma\) changes the spread, while adding \(\mu\) shifts the center. To verify the resulting distribution, take any real \(x\). Since \(\sigma>0\), dividing an inequality by \(\sigma\) does not reverse its direction:

\begin{aligned} F_X(x) &=P(X\le x)\\ &=P(\mu+\sigma Z\le x)\\ &=P\left(Z\le \frac{x-\mu}{\sigma}\right)\\ &=\Phi\left(\frac{x-\mu}{\sigma}\right), \end{aligned}

Here \(\Phi\) is the standard normal cumulative distribution function. This is the cumulative distribution function of \(N(\mu,\sigma^2)\). Differentiating with respect to \(x\) gives the corresponding density:

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

The moments follow directly from the properties \(E[Z]=0\) and \(\operatorname{Var}(Z)=1\). Linearity of expectation gives the mean; the variance rule for scaling gives the variance:

\begin{aligned} E[X]&=E[\mu+\sigma Z]=\mu+\sigma E[Z]=\mu,\\ \operatorname{Var}(X)&=\operatorname{Var}(\mu+\sigma Z) =\sigma^2\operatorname{Var}(Z)=\sigma^2. \end{aligned}

These calculations justify both the distributional result and its moments. The positive-scale condition is the usual convention because the target standard deviation is positive. If a nonzero negative multiplier were used instead, the distribution would still be normal, but its standard deviation would be the absolute value of that multiplier.

Key result: To generate \(X\sim N(\mu,v)\), where \(v>0\) is the target variance, generate \(Z\sim N(0,1)\) and set \(X=\mu+\sqrt{v}\,Z\). The square root is essential: the multiplier is the target standard deviation.

Worked Transformations

Worked Example: Simulating a Normal Amount With a Given Variance

A hypothetical payment amount \(X\), in dollars, is to be simulated from \(N(5{,}000,40{,}000)\). Three available standard normal variates are \(-1\), \(0\), and \(1\). Find the corresponding payment amounts and verify the target theoretical moments.

Method. The specified distribution is \(N(\mu,v)\) with \(\mu=5{,}000\) dollars and variance \(v=40{,}000\) dollars squared. Therefore the standard deviation is \( \sigma=\sqrt{40{,}000}=200\) dollars, and the transformation is \(X=5{,}000+200Z\).

For \(Z=-1\), \(X=5{,}000+200(-1)=4{,}800\). For \(Z=0\), \(X=5{,}000+200(0)=5{,}000\). For \(Z=1\), \(X=5{,}000+200(1)=5{,}200\). The generated amounts are therefore \(4{,}800\), \(5{,}000\), and \(5{,}200\) dollars.

The transformation result gives \(E[X]=5{,}000\) dollars and \(\operatorname{Var}(X)=200^2=40{,}000\) dollars squared, as required. The three values happen to have mean \(5{,}000\), but three simulated outcomes do not establish the theoretical moments; those follow from the transformation applied to a standard normal variate.

Check and trap avoided. The required scale is \(200\), since \(200^2=40{,}000\). Multiplying \(Z\) by \(40{,}000\) would incorrectly use the variance as the scale.

Worked Example: Transforming Variates and Checking Sample Moments

A hypothetical measurement \(T\), in hours, has target distribution \(N(12,9)\), where the second parameter is the variance. Transform the supplied standard normal variates \(-2\), \(-0.5\), and \(1.5\). Then calculate the mean and the variance of these three generated values using divisor \(3\), and compare them with the theoretical moments.

Method. The mean is \(\mu=12\) hours and the variance is \(9\) hours squared, so the standard deviation is \(\sigma=\sqrt{9}=3\) hours. Thus \(T=12+3Z\).

For \(Z=-2\), \(T=12+3(-2)=6\). For \(Z=-0.5\), \(T=12+3(-0.5)=10.5\). For \(Z=1.5\), \(T=12+3(1.5)=16.5\). The generated values are \(6\), \(10.5\), and \(16.5\) hours.

Their arithmetic mean is \( (6+10.5+16.5)/3=33/3=11\) hours. The deviations from that mean are \(-5\), \(-0.5\), and \(5.5\) hours. The variance calculated with divisor \(3\) is \( (25+0.25+30.25)/3=55.5/3=18.5\) hours squared.

The theoretical mean and variance remain \(12\) hours and \(9\) hours squared, respectively. The particular three variates need not have mean zero or variance one, so their transformed sample moments need not equal the target moments.

Check and trap avoided. The transformation uses the standard deviation \(3\), not the variance \(9\). The sample calculation is also explicitly a variance with divisor \(3\); it is not a replacement for the theoretical variance of the distribution.

Generating Multiple Normal Quantities

An affine transformation can produce a normal marginal distribution for each quantity, but that fact alone does not guarantee that several generated quantities are independent. If the same standard normal variate is used repeatedly, the resulting quantities can be strongly dependent. Independence requires independent input variates when independent outputs are intended.

Worked Example: Same Marginals, Dependence From a Shared Variate

A hypothetical simulation requires paired quantities \(X\) and \(Y\), with marginal distributions \(X\sim N(80,100)\) and \(Y\sim N(20,9)\). An implementation uses the same \(Z\sim N(0,1)\) for both transformations. Write the transformations, find the pair when \(Z=-1\) and when \(Z=0.5\), and calculate the theoretical covariance and correlation.

Method. The target standard deviations are \(\sqrt{100}=10\) and \(\sqrt{9}=3\). Using the same \(Z\) gives \(X=80+10Z\) and \(Y=20+3Z\).

When \(Z=-1\), \(X=80-10=70\) and \(Y=20-3=17\). When \(Z=0.5\), \(X=80+5=85\) and \(Y=20+1.5=21.5\). Each marginal has the specified normal distribution by the location-scale result.

For the covariance, constants do not affect covariance, and scaling multiplies it by the product of the scales: \( \operatorname{Cov}(X,Y)=10(3)\operatorname{Var}(Z)=30\). The marginal standard deviations are \(10\) and \(3\), so the correlation is \(30/(10\cdot3)=1\). Indeed, \(Y=20+3(X-80)/10\), an exact increasing linear relationship.

Check and trap avoided. Correct marginal distributions do not imply independent paired values. If the required \(X\) and \(Y\) are independent, generate them using independent standard normal variates \(Z_1\) and \(Z_2\), respectively; then their covariance is zero.

Practical Checks and Common Mistakes

Before using a location-scale transformation, identify the target mean and variance from the distribution specification, then take the square root of the variance to obtain the multiplier. Keep units in view: if \(X\) is measured in dollars, its standard deviation is in dollars and its variance is in dollars squared. A transformation should add quantities with compatible units.

A reliable check is to substitute the chosen scale into the moment formulas. If the proposed transformation is \(X=\mu+cZ\), then its mean is \(\mu\) and its variance is \(c^2\), because \(\operatorname{Var}(Z)=1\). Thus a target variance \(v\) requires \(c^2=v\), or \(c=\sqrt v\) under the positive-scale convention.

Common exam trap: For \(X\sim N(50,64)\), the target standard deviation is \(\sqrt{64}=8\). With \(Z=1.5\), the correct simulated value is \(50+8(1.5)=62\). The incorrect calculation \(50+64(1.5)=146\) treats the variance as the standard deviation and produces the wrong scale.

A second important check concerns what is being verified. The theoretical statements \(E[X]=\mu\) and \(\operatorname{Var}(X)=\sigma^2\) describe the distribution over repeated generation. A finite list of simulated values can have a different sample mean and variance. That difference is not, by itself, evidence that the transformation is wrong. Conversely, computing sample moments from a short list does not prove the intended distribution was generated.

For a single target normal variable, the transformation is direct. For a collection intended to be independent, the inputs must also be independent. Reusing an input can preserve each output's marginal distribution while changing the joint behavior, as the shared-variate example demonstrates.

Check Your Understanding

For each problem, identify the scale carefully and distinguish theoretical moments from moments of a finite simulated sample.

A hypothetical loss \(L\) has distribution \(N(1{,}200,10{,}000)\), in dollars. What transformation uses \(Z\sim N(0,1)\), and what value results for \(Z=0.6\)? (Answer: \(L=1{,}200+100Z\); the value is \(1{,}260\) dollars.)
A simulated quantity is defined as \(W=7+4Z\), where \(Z\sim N(0,1)\). State the distribution, theoretical mean, and theoretical variance of \(W\). (Answer: \(W\sim N(7,16)\), with mean \(7\) and variance \(16\).)
A target variable has mean \(30\) and variance \(25\). For supplied variates \(-1\) and \(2\), find both transformed values and their arithmetic mean. (Answer: the values are \(25\) and \(40\), and their arithmetic mean is \(32.5\).)
Two variables are generated as \(A=10+2Z\) and \(B=4-5Z\) using the same \(Z\sim N(0,1)\). Find their marginal distributions and covariance. (Answer: \(A\sim N(10,4)\), \(B\sim N(4,25)\), and \(\operatorname{Cov}(A,B)=-10\).)
A hypothetical normal amount has mean \(900\) dollars and variance \(14{,}400\) dollars squared. An implementation transforms \(Z=-0.25\) using the variance as its multiplier. Give the correct value and identify the implementation error. (Answer: the correct value is \(900+120(-0.25)=870\) dollars; the implementation must use the standard deviation \(120\), not the variance \(14{,}400\).)

What You'll Learn