Distribution Calculations by Choosing a Convenient Scale

Why Choose a Convenient Scale?

A transformation can be easier to calculate after the input has been expressed on a convenient scale. For a uniform variable, the useful scale is often the unit interval. For an exponential variable, multiplying by its rate gives a unit-rate variable. For a normal variable, centering by its mean and dividing by its standard deviation gives a standard normal variable. These choices make the inverse transformation and the support easier to read.

The key is not merely to simplify an algebraic expression. A valid distribution calculation must identify the transformed variable’s support, calculate its cumulative distribution function (CDF) on that support, and obtain a density consistent with the CDF. We will use these checks for each distribution family.

Convention: A uniform distribution on an interval has constant density on that interval. An exponential distribution with rate \(\lambda>0\) has density \(f(x)=\lambda e^{-\lambda x}\) for \(x>0\). A normal distribution \(N(\mu,\sigma^2)\) has standard deviation \(\sigma>0\), so standardizing uses \((X-\mu)/\sigma\).

From a Unit Uniform Variable to an Exponential Variable

Suppose \(U\) is uniform on \((0,1)\). Its CDF is \(F_U(u)=u\) for \(0<u<1\), with value zero to the left of the interval and one to its right. The logarithm maps \((0,1)\) to \((-\infty,0)\). Negating the logarithm therefore maps the unit interval to the positive half-line, the support of an exponential variable. Dividing by \(\lambda\) sets the desired rate.

Theorem: If \(U\sim\operatorname{Unif}(0,1)\) and \(\lambda>0\), then \(T=-\log(U)/\lambda\) has an exponential distribution with rate \(\lambda\).

Proof. Since \(0<U<1\) with probability one, \(T>0\) with probability one. For \(t\leq 0\), \(F_T(t)=0\). For \(t>0\), reversing the inequality when multiplying by \(-\lambda\) gives

\begin{aligned} F_T(t) &=P\!\left(-\frac{\log U}{\lambda}\leq t\right)\\ &=P\!\left(U\geq e^{-\lambda t}\right)\\ &=1-e^{-\lambda t}. \end{aligned}

The last equality uses the uniform CDF: for \(0<e^{-\lambda t}<1\), the probability that \(U\) is at least \(e^{-\lambda t}\) is \(1-e^{-\lambda t}\). Differentiating the CDF for \(t>0\) gives \(f_T(t)=\lambda e^{-\lambda t}\). Thus the support, CDF, and density are those of an exponential distribution with rate \(\lambda\). \(\square\)

Worked Example: Turning a Unit Uniform Variable into an Exponential Variable

Let \(U\sim\operatorname{Unif}(0,1)\), and set \(T=-\log(U)/3\). Find the support, CDF, and density of \(T\).

Because \(U\) lies between zero and one with probability one, \(-\log U>0\), so \(T\) has support \((0,\infty)\). For \(t\leq0\), \(F_T(t)=0\). For \(t>0\),

\begin{aligned} F_T(t) &=P\!\left(-\frac{\log U}{3}\leq t\right)\\ &=P(U\geq e^{-3t})\\ &=1-e^{-3t}. \end{aligned}

Differentiating on the support yields \(f_T(t)=3e^{-3t}\) for \(t>0\), and the density is zero for \(t<0\). It integrates to one:

\int_0^\infty 3e^{-3t}\,dt =\left[-e^{-3t}\right]_0^\infty =1.

Therefore \(T\) is exponential with rate \(3\), not rate \(1/3\). The factor \(1/3\) in the transformation sets the rate to \(3\).

Normalizing an Exponential Scale

For an exponential variable, the rate determines the horizontal scale of the distribution. Multiplication by the rate removes that scale and produces a unit-rate exponential variable. This is useful when a calculation is simpler in terms of a single standard scale.

Theorem: If \(X\) has an exponential distribution with rate \(\lambda>0\), then \(W=\lambda X\) has an exponential distribution with rate \(1\).

Proof. Since \(X>0\) with probability one and \(\lambda>0\), \(W>0\) with probability one. For \(w\leq0\), \(F_W(w)=0\). For \(w>0\), the positive multiplier preserves the inequality, so

F_W(w) =P(\lambda X\leq w) =P(X\leq w/\lambda) =1-e^{-\lambda(w/\lambda)} =1-e^{-w}.

Differentiation gives \(f_W(w)=e^{-w}\) for \(w>0\), which is the unit-rate exponential density. \(\square\)

Worked Example: Mapping an Exponential Variable to a Uniform Variable

Let \(X\) have an exponential distribution with rate \(2\), and define \(R=e^{-2X}\). Find the support, CDF, and density of \(R\).

Since \(X>0\), the value of \(R\) is strictly between zero and one with probability one. For \(r\leq0\), \(F_R(r)=0\); for \(r\geq1\), \(F_R(r)=1\). When \(0<r<1\), the exponential function is increasing, but the map \(x\mapsto e^{-2x}\) is decreasing. Reversing the event inequality gives

\begin{aligned} F_R(r) &=P(e^{-2X}\leq r)\\ &=P\!\left(X\geq-\frac{\log r}{2}\right)\\ &=e^{-2(-\log r/2)}\\ &=r. \end{aligned}

Thus \(R\sim\operatorname{Unif}(0,1)\). Differentiating the CDF gives \(f_R(r)=1\) for \(0<r<1\), and zero outside that interval. The density integrates to one, \(\int_0^1 1\,dr=1\). This reverse transformation also shows why multiplying by the rate matters: \(e^{-2X}\), rather than \(e^{-X}\), cancels the rate \(2\) in the exponential tail probability.

Standardizing a Normal Input Before Transforming It

If \(X\sim N(\mu,\sigma^2)\), the centered and scaled variable \(Z=(X-\mu)/\sigma\) has the standard normal distribution. That scale makes the normal part of a calculation easy to express using the standard normal CDF, denoted by \(\Phi\), and density, denoted by \(\phi\). For a strictly increasing transformation such as the exponential function, it also makes the inverse particularly simple.

Worked Example: Exponentiating a Standardized Normal Variable

Let \(X\sim N(1,4)\), so \(\mu=1\) and \(\sigma=2\). Define \(Y=\exp((X-1)/2)\). Find the support, CDF, and density of \(Y\).

The standardized input \(Z=(X-1)/2\) has the standard normal distribution, and \(Y=e^Z\). Since an exponential is positive, \(Y\) has support \((0,\infty)\). For \(y\leq0\), \(F_Y(y)=0\). For \(y>0\), taking logarithms preserves the inequality:

\begin{aligned} F_Y(y) &=P(e^Z\leq y)\\ &=P(Z\leq\log y)\\ &=\Phi(\log y). \end{aligned}

For \(y>0\), differentiating gives

f_Y(y) =\phi(\log y)\frac{1}{y} =\frac{1}{y\sqrt{2\pi}} \exp\!\left(-\frac{(\log y)^2}{2}\right).

The density is zero for \(y<0\). Its integral over the support is one: using \(z=\log y\), so \(dy/y=dz\),

\int_0^\infty f_Y(y)\,dy =\int_{-\infty}^\infty \phi(z)\,dz =1.

The support and CDF limits agree as well: \(\Phi(\log y)\) tends to zero as \(y\) decreases to zero and to one as \(y\) increases without bound. The standardization has removed the original mean and variance from the calculation; they were chosen so that the input to the exponential is exactly standard normal.

How to Check the Scale and the Result

In each calculation, the transformation was chosen to simplify the input before identifying the output. The unit uniform variable makes logarithmic tails transparent; rate-scaling makes exponential variables comparable on a common scale; normal standardization lets the CDF be written directly with \(\Phi\). Regardless of the choice, check the result in a consistent order:

Map the support.
Determine the possible transformed values from the input support and the direction of the map. A decreasing map reverses event inequalities.

Calculate the CDF by events.
Express the event involving the transformed variable as an event involving the original variable. State the CDF separately outside the transformed support.

Differentiate and check.
Differentiate the CDF on the support, then verify that the density is nonnegative and integrates to one.

A frequent mistake is to confuse an exponential rate with a scale. If \(X\) has rate \(\lambda\), its density is \(\lambda e^{-\lambda x}\), so its typical horizontal scale is proportional to \(1/\lambda\). Consequently, \(\lambda X\) has rate \(1\). Dividing \(X\) by \(\lambda\) does not perform the same normalization: for \(X/\lambda\), the CDF at \(t>0\) is \(1-e^{-\lambda^2t}\), so the rate is \(\lambda^2\), not \(1\). Checking the CDF is a reliable way to catch this error.

Another common mistake is to use the correct inverse expression but forget that a decreasing map reverses the inequality. For \(R=e^{-\lambda X}\), the event \(R\leq r\) corresponds to \(X\geq-\log(r)/\lambda\), not \(X\leq-\log(r)/\lambda\). The latter would use the lower tail where the upper tail is required. Writing the event explicitly before substituting a CDF or survival probability prevents the error.

Key takeaway: Choose the scale that makes the transformation easiest to invert, but verify the conclusion independently: its support must match the map, its CDF must have the correct tail direction and endpoint limits, and its density must integrate to one.

Check Your Understanding

For each question, identify the useful scale or transformation and check the support before calculating.

If \(U\sim\operatorname{Unif}(0,1)\), what is the rate of \(-\log(U)/5\)? Give its support and CDF.
If \(X\) has an exponential distribution with rate \(4\), what distribution does \(4X\) have? Explain why \(X/4\) is not unit-rate exponential.
For \(X\) exponential with rate \(\lambda\), solve the event \(e^{-\lambda X}\leq r\) for \(X\), assuming \(0<r<1\). Why does the inequality reverse?
If \(Z\) is standard normal and \(Y=e^Z\), state the support and CDF of \(Y\).
What three checks should be made after deriving a transformed density?