CDF Derivation for Increasing Transformations

Why Increasing Transformations Simplify CDFs

When a random variable is transformed, its distribution can often be found by translating the event that defines its CDF back to the original variable. For a strictly increasing transformation, the order of values is preserved: comparing the transformed variable with a threshold is equivalent to comparing the original variable with the corresponding inverse threshold. This gives a direct route to the transformed CDF without first finding a density.

The essential point is to keep track of the domain and range. An inverse is defined only on the image of the transformation, and if the original variable is known to lie in the transformation's domain only almost surely, the event equivalence is also an almost-sure statement. These details matter even when they do not change the resulting probabilities.

The CDF Formula

Definition: Let $I$ be an interval, let $X$ be a random variable with $P(X\in I)=1$, and let $g:I\to\mathbb{R}$ be continuous and strictly increasing. Write $J=g(I)$. Define $Y=g(X)$ on $\{X\in I\}$, and assign any fixed value to $Y$ on the null event $\{X\notin I\}$. For $y\in J$, the inverse $g^{-1}(y)$ is defined, and $$ F_Y(y)=F_X\!\left(g^{-1}(y)\right). $$

Here $F_X(t)=P(X\le t)$ and $F_Y(y)=P(Y\le y)$. The arbitrary definition of $Y$ outside $\{X\in I\}$ makes it a random variable on the entire probability space; it cannot affect its distribution because that exceptional event has probability zero. Since $g$ is continuous and $X$ is measurable, this piecewise definition is measurable.

For $y$ in the image $J$, strict increase gives the order equivalence $g(x)\le y$ if and only if $x\le g^{-1}(y)$, for $x\in I$. The resulting equality of events holds on $\{X\in I\}$, which has probability one. Thus probabilities of the two events are equal, even if the events are not literally identical on the null complement.

Theorem: Under the conditions above, for every $y\in J$, $$ \{Y\le y\}=\{X\le g^{-1}(y)\}\quad\text{almost surely}, $$ and consequently $F_Y(y)=F_X(g^{-1}(y))$.

Proof. Fix $y\in J$, so that $g^{-1}(y)\in I$. On the event $\{X\in I\}$, the definition $Y=g(X)$ and strict increase of $g$ imply $$ Y\le y \quad\Longleftrightarrow\quad g(X)\le g\!\left(g^{-1}(y)\right) \quad\Longleftrightarrow\quad X\le g^{-1}(y). $$ Thus the two events can differ only when $X\notin I$, an event of probability zero. Taking probabilities yields $$ F_Y(y)=P(Y\le y)=P\!\left(X\le g^{-1}(y)\right) =F_X\!\left(g^{-1}(y)\right). $$ $\square$

The formula is stated for $y\in J$, where the inverse exists. If $J$ has a finite lower endpoint not included in $J$, then $Y$ is strictly above that endpoint almost surely, so the CDF there is zero. If $J$ has a finite upper endpoint not included in $J$, then $Y$ is below it almost surely, so the CDF there is one. When a finite endpoint belongs to $J$, use the inverse formula at that endpoint. Outside the range, the CDF is zero below the range and one above it.

Support Under an Increasing Transformation

A useful support statement is especially simple when the original variable has an interval as its support. The support of a distribution is the set of points for which every open neighborhood has positive probability. For a variable supported on a closed bounded interval, continuity and strict increase carry the interval's endpoints to the new support's endpoints.

Theorem: Suppose $X$ has support $[a,b]$, where $a<b$, and $g$ is continuous and strictly increasing on $[a,b]$. If $Y=g(X)$, then $Y$ has support $[g(a),g(b)]$.

Proof. Every value of $X$ lies in $[a,b]$ almost surely. Since $g$ is increasing, every value of $Y$ therefore lies in $[g(a),g(b)]$ almost surely. Any point outside this latter interval has a neighborhood disjoint from the possible values of $Y$, so it is not in the support.

Now take any $x\in[a,b]$ and let $y=g(x)$. For every open neighborhood $V$ of $y$, continuity of $g$ implies that the preimage of $V$ contains a relative neighborhood of $x$ in $[a,b]$. Since $x$ is in the support of $X$, that relative neighborhood has positive probability. Hence $P(Y\in V)>0$, so $y$ is in the support of $Y$. As $x$ ranges over $[a,b]$, $g(x)$ ranges over all of $[g(a),g(b)]$. This proves the support claim. $\square$

For unbounded or non-interval supports, the support requires more care than simply transforming two endpoints. The CDF formula remains valid on the image of the domain, but the support statement above is specifically for a variable whose support is the whole closed interval $[a,b]$.

Working With the Formula

Worked Example: A Nonlinear Transformation of a Uniform Variable

Let $X\sim\operatorname{Uniform}(0,1)$, and define $Y=X+X^2$. The function $g(x)=x+x^2$ is continuous on $[0,1]$, and $g'(x)=1+2x>0$ there, so it is strictly increasing. Its range is $[0,2]$. Solve $y=x+x^2$ for $x$: $$ x^2+x-y=0,\qquad g^{-1}(y)=\frac{-1+\sqrt{1+4y}}{2},\quad 0\le y\le 2. $$

The CDF of $X$ is $F_X(x)=x$ for $0\le x\le1$. Applying the increasing-transformation formula gives $$ F_Y(y)= \begin{cases} 0, & y<0,\\[3pt] \dfrac{\sqrt{1+4y}-1}{2}, & 0\le y\le2,\\[7pt] 1, & y>2. \end{cases} $$ At $y=0$, the middle expression is $0$; at $y=2$, it is $(3-1)/2=1$, as required at the support endpoints.

For example, $P(Y>3/4)=1-F_Y(3/4)$. Since $\sqrt{1+4(3/4)}=\sqrt4=2$, $$ F_Y(3/4)=\frac{2-1}{2}=\frac12, \qquad P(Y>3/4)=1-\frac12=\frac12. $$ This agrees with the original-variable calculation: $g^{-1}(3/4)=1/2$, so $Y>3/4$ is equivalent, up to a null event, to $X>1/2$, which has probability $1/2$.

Worked Example: A Rational Transformation on a Bounded Interval

Let $X\sim\operatorname{Uniform}(0,2)$ and set $Y=3X/(1+X)$. On $[0,2]$, $g(x)=3x/(1+x)$ is continuous and has derivative $g'(x)=3/(1+x)^2>0$. Its range is $[0,2]$. Solving $y=3x/(1+x)$ gives $y(1+x)=3x$, hence $y=x(3-y)$ and $$ g^{-1}(y)=\frac{y}{3-y},\quad 0\le y\le2. $$

Since $F_X(x)=x/2$ on $[0,2]$, the transformed CDF is $$ F_Y(y)= \begin{cases} 0, & y<0,\\[3pt] \dfrac{y}{2(3-y)}, & 0\le y\le2,\\[7pt] 1, & y>2. \end{cases} $$ The denominator is positive throughout the stated interval. At its upper endpoint, $F_Y(2)=2/[2(1)]=1$.

To calculate $P(Y\le1)$, first invert the threshold: $g^{-1}(1)=1/(3-1)=1/2$. Therefore $$ P(Y\le1)=F_X(1/2)=\frac{1/2}{2}=\frac14. $$ Equivalently, the transformed CDF gives $F_Y(1)=1/[2(2)]=1/4$. Consequently, $P(Y>1)=1-1/4=3/4$. The two calculations agree.

Worked Example: Transforming an Exponential Variable

Let $X$ have an exponential distribution with rate $2$, so $F_X(x)=1-e^{-2x}$ for $x\ge0$ and $F_X(x)=0$ for $x<0$. Define $Y=X/(1+X)$. The function is continuous and strictly increasing on $[0,\infty)$, since $g'(x)=1/(1+x)^2>0$. It maps $[0,\infty)$ onto $[0,1)$, and solving $y=x/(1+x)$ gives $x=y/(1-y)$.

For $0\le y<1$, the CDF is therefore $$ F_Y(y)=F_X\!\left(\frac{y}{1-y}\right) =1-\exp\!\left(-\frac{2y}{1-y}\right). $$ Including the values outside the range, $$ F_Y(y)= \begin{cases} 0, & y<0,\\[3pt] 1-\exp\!\left(-\dfrac{2y}{1-y}\right), & 0\le y<1,\\[7pt] 1, & y\ge1. \end{cases} $$ The support is $[0,1]$: although $Y$ never equals $1$, values of $Y$ can be arbitrarily close to $1$ with positive probability in every neighborhood below it.

For $P(Y\le1/2)$, the inverse threshold is $(1/2)/(1-1/2)=1$, so $$ P(Y\le1/2)=F_X(1)=1-e^{-2}. $$ Numerically, $e^{-2}$ is approximately $0.1353$, giving $P(Y\le1/2)\approx0.8647$. This also follows directly because $X/(1+X)\le1/2$ is equivalent to $X\le1$ for $X\ge0$.

Using the CDF and Avoiding Common Errors

The formula is useful because it works whether $X$ is continuous, discrete, or has a mixture of probability mass and continuous variation. No density for $X$ is required. Once $F_Y$ is known, probabilities of intervals can be calculated using CDF differences; for instance, $P(c<Y\le d)=F_Y(d)-F_Y(c)$. If $X$ is continuous, individual threshold equalities have probability zero, so changing a strict endpoint to a non-strict one does not change such a probability. Without continuity, endpoint conventions must be respected.

A common mistake is to write $\{Y\le y\}=\{X\le g^{-1}(y)\}$ as an unconditional literal identity when $g$ is defined only on $I$ and $X\in I$ is known only almost surely. On $\{X\notin I\}$, $g(X)$ may not even be defined. The correction is to define $Y$ on that null event (or assume $X\in I$ everywhere), state that the events agree almost surely, and then take probabilities. Another frequent error is using $g^{-1}(y)$ when $y$ is outside $g(I)$; there the inverse is not defined, and the CDF must instead be determined from the range.

Key takeaway: For a continuous, strictly increasing $g$, translate the threshold $y$ back to $g^{-1}(y)$, but only when $y$ lies in the image of the domain. State the transformed range, check endpoint values, and make clear whether event equalities hold everywhere or almost surely.

Check Your Understanding

Use the inverse-threshold argument and state any range restrictions needed.

If $X$ has support $[0,3]$ and $g(x)=2x+x^2$, what is the support of $Y=g(X)$? Identify the inverse threshold formula without needing to specify $F_X$.
Suppose $X\sim\operatorname{Uniform}(0,1)$ and $Y=2X+X^2$. What equation must be solved to evaluate $F_Y(y)$ for $y$ in the transformed range?
For the exponential example, explain why $F_Y(1)=1$ even though $1$ is not in the image of $[0,\infty)$ under $g(x)=x/(1+x)$.
Why does strict increase, rather than just the existence of a formula resembling an inverse, establish the needed order equivalence?
If $P(X\in I)=1$ but not every outcome has $X\in I$, in what sense do the transformed events agree, and why is that sufficient for the CDF calculation?