CDF Derivation for Decreasing Transformations

Why Decreasing Transformations Need Care

A monotone transformation changes the way events involving a random variable are described. For an increasing function, values below a transformed threshold correspond to values below the original threshold. A decreasing function reverses that order: values below a transformed threshold correspond to values above an original threshold. That reversal changes the CDF calculation and makes atoms important.

The key distinction is between $P(X\ge a)$ and $P(X>a)$. They are equal when $P(X=a)=0$, but not in general. Since a CDF includes the probability at its endpoint, using $1-F_X(a)$ without checking for an atom can give the wrong transformed distribution.

Definition and CDF Formula

Definition: Let $X$ be a random variable with CDF $F_X$, and suppose $P(X\in J)=1$ for an interval $J$. Let $g:J\to\mathbb{R}$ be strictly decreasing and continuous, and define $Y=g(X)$. The image of $J$ under $g$ is denoted $g(J)$. For $y\in g(J)$, the inverse $g^{-1}(y)$ is the unique $x\in J$ satisfying $g(x)=y$.

For a real number $a$, write $F_X(a-)=\lim_{t\uparrow a}F_X(t)=P(X<a)$. This left limit differs from $F_X(a)=P(X\le a)$ by the mass at $a$: $F_X(a)-F_X(a-)=P(X=a)$.

Theorem: Under the conditions above, for every $y\in g(J)$, $$ F_Y(y)=P(Y\le y)=P\bigl(X\ge g^{-1}(y)\bigr) =1-F_X\bigl(g^{-1}(y)-\bigr). $$ If $P(X=g^{-1}(y))=0$, this simplifies to $F_Y(y)=1-F_X(g^{-1}(y))$. For $y$ outside $g(J)$, the CDF is determined by the range of $Y$; in particular, it is $0$ below $\inf g(J)$ and $1$ at or above $\sup g(J)$, when these bounds are finite.

The endpoint convention in the last sentence follows directly from $Y\in g(J)$ almost surely. At the lower bound $m=\inf g(J)$, $F_Y(m)=P(Y=m)$: it is zero if the lower bound is not attained, and otherwise equals the probability that $X$ maps to that endpoint. At the upper bound $M=\sup g(J)$, every possible value of $Y$ is at most $M$, so $F_Y(M)=1$, whether or not $M$ is attained.

Deriving the Formula

Fix $y\in g(J)$ and put $a=g^{-1}(y)$. Since $g$ is strictly decreasing, comparing $g(x)$ with $g(a)$ reverses the inequality: $g(x)\le g(a)$ exactly when $x\ge a$. Applying this pointwise equivalence to $X$ gives the event identity $\{Y\le y\}=\{X\ge a\}$. Therefore,

F_Y(y)=P(Y\le y)=P(X\ge a) =1-P(X<a)=1-F_X(a-).

The equality $P(X<a)=F_X(a-)$ follows from continuity of probability for increasing events. Choose any sequence $t_k<a$ with $t_k\uparrow a$. Then the events $\{X\le t_k\}$ increase to $\{X<a\}$, so $P(X<a)=\lim_{k\to\infty}P(X\le t_k)=\lim_{k\to\infty}F_X(t_k)=F_X(a-)$. Substituting $a=g^{-1}(y)$ proves the stated formula. If $P(X=a)=0$, then $F_X(a-)=F_X(a)$, giving the simplified version. $\square$

Continuity of $g$ ensures that its image of an interval is an interval, so values strictly between the range bounds are in $g(J)$. For values below the lower bound, $Y\le y$ is impossible; at or above the upper bound, it holds almost surely. The inverse formula is used only for values in the image, where the inverse is defined.

Worked Examples

Worked Example: A Continuous Uniform Variable

Let $X\sim \operatorname{Uniform}(0,1)$ and set $Y=2-X$. The transformation is strictly decreasing, and its image is $(1,2)$. Find $F_Y(y)$, including the range endpoints.

For $1<y<2$, the inverse threshold is $a=2-y$, which lies in $(0,1)$. A uniform variable has no atom at $a$, and $F_X(a)=a$. Thus

F_Y(y)=1-F_X(2-y)=1-(2-y)=y-1,\qquad 1<y<2.

If $y\le1$, then $Y>1$ almost surely, so $F_Y(y)=0$, including at $y=1$. If $y\ge2$, then $Y<2$ almost surely, so $F_Y(y)=1$, including at $y=2$. Therefore

F_Y(y)= \begin{cases} 0, & y\le 1,\\ y-1, & 1<y<2,\\ 1, & y\ge 2. \end{cases}

The values at the endpoints agree with the limits of the interior formula, but the endpoint checks are still needed: the range $(1,2)$ does not include either endpoint.

Worked Example: A Discrete Variable With Atoms

Suppose $X$ takes the values $0,1,3$ with probabilities $0.2,0.5,0.3$, respectively, and let $Y=5-2X$. Find the CDF of $Y$.

The transformed values are $5,3,-1$, with the same respective probabilities. Listing them in increasing order gives $Y=-1$ with probability $0.3$, $Y=3$ with probability $0.5$, and $Y=5$ with probability $0.2$. Hence

F_Y(y)= \begin{cases} 0, & y<-1,\\ 0.3, & -1\le y<3,\\ 0.8, & 3\le y<5,\\ 1, & y\ge5. \end{cases}

The atom at $y=3$ is a useful check on the formula. Here $g^{-1}(3)=1$, $F_X(1)=0.7$, and $F_X(1-)=0.2$. The correct calculation is $F_Y(3)=1-F_X(1-)=1-0.2=0.8$, which includes both the mass at $X=1$ and the mass at $X=3$. Using $1-F_X(1)=0.3$ would incorrectly omit the mass at $X=1$.

Worked Example: A Mixed Distribution and an Endpoint Atom

Let $X$ have probability $0.3$ at $0$, and with probability $0.7$ have a uniform distribution on $[0,2]$. More precisely, its CDF is $F_X(x)=0$ for $x<0$, $F_X(x)=0.3+0.35x$ for $0\le x<2$, and $F_X(x)=1$ for $x\ge2$. Set $Y=3-X$. Find its CDF.

The range of $Y$ is $[1,3]$. For $1<y<3$, the inverse threshold $a=3-y$ lies in $(0,2)$, where there is no atom. Consequently,

F_Y(y)=1-F_X(3-y) =1-\{0.3+0.35(3-y)\} =0.35(y-1).

At $y=1$, the inverse threshold is $2$. There is no atom at $X=2$, so $F_Y(1)=P(X\ge2)=0$, also given by $1-F_X(2-)=0$. At $y=3$, the inverse threshold is $0$, where $X$ has an atom. The left limit is $F_X(0-)=0$, so $F_Y(3)=1-F_X(0-)=1$. The resulting CDF is

F_Y(y)= \begin{cases} 0, & y<1,\\ 0.35(y-1), & 1\le y<3,\\ 1, & y\ge3. \end{cases}

The jump at $y=3$ has size $1-0.7=0.3$, exactly the probability that $X=0$. The continuous part contributes $0.7$ across the interval from $1$ to $3$; together the endpoint atom and that continuous probability sum to $1$.

Common Mistake: Using the CDF at the Threshold

A tempting calculation is $P(X\ge a)=1-F_X(a)$. But $F_X(a)=P(X\le a)$, so subtraction gives $P(X>a)$, not $P(X\ge a)$. The difference is $P(X=a)$. For a decreasing transformation, the inverse threshold itself must be included in the event, so the correct general expression is $1-F_X(a-)$.

The shorter expression $1-F_X(a)$ is valid only after establishing that $P(X=a)=0$. This is automatic for a continuous distribution at every threshold, but not for a discrete or mixed distribution. Another common error is to apply the inverse formula outside $g(J)$; there the inverse may not exist, and the CDF should instead be read from the range bounds.

Why the Distinction Matters

A CDF records cumulative probability with a closed inequality, $P(Y\le y)$. Under a decreasing map, that closed inequality becomes $P(X\ge a)$, which includes an endpoint on the opposite side of the original scale. The left limit in the formula is not a technical decoration: it preserves precisely that endpoint probability.

When using the result, a reliable sequence is to identify the transformed range, solve $g(a)=y$, reverse the inequality, and then decide whether the threshold has an atom. Only after that check should $F_X(a-)$ be replaced by $F_X(a)$.

Check Your Understanding

Use the event relation and endpoint conventions to answer these questions.

If $Y=7-X$, what event involving $X$ is equivalent to $Y\le y$? State the inverse threshold.
For a general random variable $X$, express $P(X\ge a)$ using $F_X(a-)$. Why might $1-F_X(a)$ be different?
Suppose $X$ is uniform on $(0,1)$ and $Y=1-X$. What are the lower and upper bounds of the range of $Y$, and what are $F_Y(0)$ and $F_Y(1)$?
In the discrete example, why is $F_Y(3)$ equal to $0.8$ rather than $0.3$?
If the lower bound of the range of $Y$ is not attained, what is the CDF at that lower bound? Explain using the definition of a CDF.