Monotone Transformations of Continuous Variables

Why Monotone Transformations Matter

A continuous random variable is often observed or reported on a transformed scale. A measurement may be logged, exponentiated, rescaled, or converted using a reciprocal formula. To describe the transformed quantity correctly, we need its possible values and its probability density. For a strictly monotone transformation, each transformed value comes from exactly one original value, so the inverse function provides a direct route to the new density.

This tutorial develops that route for $Y=g(X)$ when $g$ is differentiable and strictly increasing or strictly decreasing over the interval where $X$ can occur. The result is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Suppose $X$ is a continuous random variable with density $f_X$ on an interval $I$. Let $g$ be strictly monotone and differentiable on the interior of $I$, with $g'(x)\ne 0$ there. Define $Y=g(X)$. The transformed support is the image of $I$ under $g$. If $g^{-1}$ is the inverse function, then, for $y$ in the interior of that image, $$ f_Y(y)=f_X\!\left(g^{-1}(y)\right)\left|\frac{d}{dy}g^{-1}(y)\right| =\frac{f_X\!\left(g^{-1}(y)\right)}{\left|g'\!\left(g^{-1}(y)\right)\right|}. $$ The density is zero outside the transformed support. Values at individual endpoints do not affect probabilities.

The assumptions matter. Strict monotonicity makes $g$ one-to-one, so an inverse exists on the transformed interval. The nonzero derivative lets us use the inverse derivative formula there. If a transformation is not one-to-one, a transformed value can have multiple preimages and this single-inverse formula does not apply directly; that case requires separate treatment.

Deriving the Formula from the CDF

Write $F_Y(y)=P(Y\le y)$. The effect of the inequality depends on whether $g$ increases or decreases. That distinction is essential when finding the cumulative distribution function (CDF).

Increasing transformation.
If $g$ is strictly increasing, then $g(X)\le y$ is equivalent to $X\le g^{-1}(y)$. Thus $F_Y(y)=F_X(g^{-1}(y))$ for $y$ in the transformed interval.

Decreasing transformation.
If $g$ is strictly decreasing, then $g(X)\le y$ is equivalent to $X\ge g^{-1}(y)$. For a continuous $X$, this gives $F_Y(y)=1-F_X(g^{-1}(y))$ in the transformed interval.

Differentiate.
For an increasing map, differentiating gives $f_Y(y)=f_X(g^{-1}(y))(g^{-1})'(y)$. For a decreasing map, differentiating $1-F_X(g^{-1}(y))$ introduces a minus sign, while $(g^{-1})'(y)<0$. In both cases the result is $f_X(g^{-1}(y))|(g^{-1})'(y)|$.

The absolute derivative is a scale adjustment. A small interval around $y$ corresponds approximately to an interval around $x=g^{-1}(y)$ whose width is $|(g^{-1})'(y)|$ times as large. The transformed density must account for that change in interval width. Equivalently, differentiating $g(g^{-1}(y))=y$ gives $g'(g^{-1}(y))(g^{-1})'(y)=1$, which yields the second form of the formula.

Applying the Inverse-Derivative Method

For each transformation, proceed in a fixed order: identify the interval on which $X$ has positive density, map that interval through $g$, solve $y=g(x)$ for $x$, and calculate the absolute derivative of the inverse. Substitute into the density formula only after those steps are clear.

Worked Example: Exponentiating a Standard Normal Variable

Let $X\sim N(0,1)$, and define $Y=e^X$. Find the support and density of $Y$, then calculate $P(1<Y<e)$. This is an increasing transformation.

Method and support. A standard normal variable can take any real value, and $e^x$ is strictly increasing. Its values are positive, so $Y$ has support $y>0$.

Inverse and derivative. Solving $y=e^x$ gives $x=\ln y$. Therefore $g^{-1}(y)=\ln y$ and $(g^{-1})'(y)=1/y$, which is positive for $y>0$.

The standard normal density is $f_X(x)=(1/\sqrt{2\pi})e^{-x^2/2}$. Substitution into the transformation formula gives

f_Y(y)=\frac{1}{\sqrt{2\pi}}\exp\!\left(-\frac{(\ln y)^2}{2}\right)\frac{1}{y},\qquad y>0.

Probability calculation. Since $Y=e^X$ is increasing, $1<e^X<e$ is equivalent to $0<X<1$. Thus

P(1<Y<e)=P(0<X<1)=\Phi(1)-\Phi(0)=0.8413-0.5000=0.3413.

The probability is $0.3413$ to four decimal places. As a check on the density, set $u=\ln y$, so $du=dy/y$. Integrating the displayed density over $y>0$ then gives the standard normal density integrated over all real $u$, which is 1.

Common mistake avoided: Using $y$ in place of $\ln y$ in the original density would evaluate $f_X$ at the transformed value rather than at its preimage. The original value corresponding to $y$ is $x=\ln y$.

Worked Example: Taking the Square Root of an Exponential Variable

Suppose $X$ has an exponential distribution with rate parameter 2, so $f_X(x)=2e^{-2x}$ for $x>0$, and define $Y=\sqrt{X}$. Find the density of $Y$ and calculate $P(Y\le 1)$.

Method and support. The support of $X$ is $x>0$. The square-root function is strictly increasing there and maps positive values to positive values, so $Y$ has support $y>0$.

Inverse and derivative. From $y=\sqrt{x}$, we obtain $x=y^2$. The inverse derivative is $dx/dy=2y$, which is positive on the support. Therefore,

f_Y(y)=f_X(y^2)\,2y =(2e^{-2y^2})(2y) =4y e^{-2y^2},\qquad y>0.

Probability calculation. Because the transformation is increasing, $Y\le1$ is equivalent to $X\le1$. Using the exponential CDF,

P(Y\le1)=P(X\le1)=1-e^{-2(1)}=1-e^{-2}=0.8647.

The probability is $0.8647$ to four decimal places. The density also checks: its integral over the support is

\int_0^\infty 4y e^{-2y^2}\,dy =\int_0^\infty e^{-u}\,du =1,\qquad u=2y^2,\quad du=4y\,dy.

Common mistake avoided: The density uses $dx/dy=2y$, not $dy/dx=1/(2\sqrt{x})$. The formula multiplies by the absolute derivative of the inverse transformation.

Worked Example: A Decreasing Reciprocal Transformation

Let $X\sim\operatorname{Uniform}(1,4)$, with density $f_X(x)=1/3$ for $1<x<4$. Define $Y=6/X$. Find the support and density of $Y$, and calculate $P(Y>3)$.

Method and support. The function $6/x$ is strictly decreasing for $x>0$. It maps the endpoints 1 and 4 to 6 and $3/2$, respectively, so the transformed interval is $3/2<y<6$.

Inverse and derivative. Solving $y=6/x$ gives $x=6/y$, and

\left|\frac{dx}{dy}\right| =\left|-\frac{6}{y^2}\right| =\frac{6}{y^2}.

It follows that

f_Y(y)=\frac{1}{3}\frac{6}{y^2} =\frac{2}{y^2},\qquad \frac{3}{2}<y<6.

Probability calculation. Since $6/x$ is decreasing, $6/X>3$ is equivalent to $X<2$, not $X>2$. The uniform density gives

P(Y>3)=P(X<2)=\frac{2-1}{4-1}=\frac{1}{3}=0.3333.

As a check, integrate the transformed density over the values above 3:

\int_3^6 \frac{2}{y^2}\,dy =\left[-\frac{2}{y}\right]_3^6 =-\frac{1}{3}+\frac{2}{3} =\frac{1}{3}.

The two probability calculations agree. The common trap is to keep the inequality direction unchanged for a decreasing function; reversing it gives the wrong event.

What to Check Before Giving a Transformed Density

A correct-looking formula can still describe the wrong random variable if the support or inverse has been mishandled. Before finalizing a result, check the following:

Monotonicity: confirm that $g$ is one-to-one on the interval where $X$ can occur.
Support: map the possible values of $X$ through $g$. For a decreasing map, the order of the endpoint values reverses.
Inverse: solve $y=g(x)$ for $x$, then evaluate $f_X$ at that inverse value.
Derivative: use the absolute derivative of the inverse, not the derivative of $g$ itself.
Probability inequalities: when using the CDF method, reverse the inequality for a decreasing transformation.
Density check: where practical, integrate the proposed density over its support and confirm that the result is 1.

The formula does not say that the transformed density is simply the original density with a new variable name. The inverse identifies which original value produced $y$, and the derivative accounts for how the transformation stretches or compresses intervals. Both pieces are necessary.

Key takeaway: For a differentiable, strictly monotone transformation $Y=g(X)$, map the support through $g$, substitute $g^{-1}(y)$ into $f_X$, and multiply by $\left|(g^{-1})'(y)\right|$. The absolute value handles either direction of monotonicity.

Check Your Understanding

For each problem, identify the transformed support and use the inverse derivative to determine the requested density or probability.

Let $X\sim\operatorname{Uniform}(0,5)$ and $Y=2X+4$. Find the support and density of $Y$. (Answer: $4<y<14$, $f_Y(y)=1/10$ on that interval.)
Let $X$ be exponential with rate 3 and $Y=X/2$. Find the density of $Y$. (Answer: $y>0$, $f_Y(y)=6e^{-6y}$.)
Let $X\sim\operatorname{Uniform}(0,\pi/2)$ and $Y=\cos X$. Find the support and density of $Y$. (Answer: $0<y<1$, $f_Y(y)=2/(\pi\sqrt{1-y^2})$.)
Let $X\sim N(2,9)$ and $Y=4-X$. Find the density of $Y$ and $P(Y>2)$. (Answer: $f_Y(y)=\frac{1}{3\sqrt{2\pi}}\exp(-(y-2)^2/18)$ for real $y$, and $P(Y>2)=0.5000$.)
Let $X\sim\operatorname{Uniform}(2,5)$ and $Y=10/X$. Find the support and density of $Y$, then calculate $P(Y>10/3)$. (Answer: $2<y<5$, $f_Y(y)=10/(3y^2)$, and $P(Y>10/3)=1/3=0.3333$.)

What You'll Learn