CDF and Quantile Rules Under Location-Scale Changes

Why Distribution Rules Matter

A location-scale change expresses a random variable as $X=\mu+\sigma Z$, where $Z$ has a specified base distribution, $\mu\in\mathbb{R}$, and $\sigma>0$. The parameter $\mu$ shifts the values, while $\sigma$ stretches or contracts them without reversing their order. This representation is useful only if its probability statements are translated correctly: an event about $X$ must be put on the corresponding scale for $Z$.

The distribution function and quantiles summarize different aspects of a law. A CDF gives probabilities below a threshold; a quantile gives a threshold associated with a probability level. Under a positive location-scale change, both follow simple rules. Care is needed with survival probabilities and quantiles when the base law has atoms, because probabilities can accumulate at a single value and the CDF can jump.

CDF and Survival Rules

Definition: The cumulative distribution function of a real-valued random variable $W$ is $F_W(w)=P(W\leq w)$. Its strict survival function is $S_W(w)=P(W>w)=1-F_W(w)$. The inclusive upper-tail probability is $P(W\geq w)$, which need not equal $S_W(w)$ when $P(W=w)>0$.

Write $F_Z$ for the CDF of $Z$, and $F_X$ for the CDF of $X=\mu+\sigma Z$. Since $\sigma>0$, dividing an inequality by $\sigma$ preserves its direction. Thus the event $X\leq x$ is exactly the event $Z\leq (x-\mu)/\sigma$.

Theorem: If $X=\mu+\sigma Z$ with $\mu\in\mathbb{R}$ and $\sigma>0$, then for every $x\in\mathbb{R}$, $$ F_X(x)=F_Z\!\left(\frac{x-\mu}{\sigma}\right), \qquad S_X(x)=S_Z\!\left(\frac{x-\mu}{\sigma}\right). $$ The inclusive upper-tail rule is $$ P(X\geq x)=P\!\left(Z\geq\frac{x-\mu}{\sigma}\right). $$

The CDF identity follows directly from the event equivalence. The strict survival identity follows either by applying the same equivalence to $X>x$, or by taking complements of the CDF identity. The inclusive rule is stated separately because, at an atom, $P(W\geq w)$ includes the probability at $w$, while $P(W>w)$ does not.

For precision at jumps, define the left limit $F_W(w-)=\lim_{t\uparrow w}F_W(t)=P(W<w)$. Then $$ P(W\geq w)=1-F_W(w-), \qquad P(W=w)=F_W(w)-F_W(w-). $$ Because the transformation is strictly increasing, it preserves the probability at the corresponding point: $$ P(X=x)=P\!\left(Z=\frac{x-\mu}{\sigma}\right). $$ In particular, the jump in $F_X$ at $x$ is the jump in $F_Z$ at $(x-\mu)/\sigma$. For a continuous law there is no such jump, so strict and inclusive upper-tail probabilities agree at every threshold. For a discrete law they can differ.

Worked Example: A Normal Location-Scale Change

Let $Z\sim N(0,1)$, and set $X=10+2Z$. Find the CDF at $x=14$, the upper-tail probability at that threshold, and the 97.5th percentile.

First put the threshold on the base scale: $$ \frac{14-10}{2}=2. $$ Therefore $F_X(14)=\Phi(2)$, where $\Phi$ is the standard normal CDF. Numerically, $\Phi(2)\approx 0.97725$. Since the normal law is continuous, the strict and inclusive upper tails agree: $$ P(X>14)=P(X\geq14)=1-\Phi(2)\approx 0.02275. $$

The 97.5th percentile of the standard normal distribution is approximately $1.959964$. The corresponding percentile for $X$ is $$ 10+2(1.959964)=13.919928\approx13.92. $$ As a check, standardizing this value gives $(13.919928-10)/2=1.959964$, as required.

Worked Example: A Discrete Law With an Atom

Suppose $Z$ takes the value $-1$ with probability $0.4$ and the value $2$ with probability $0.6$. Let $X=3+2Z$. Find the distribution of $X$, the two upper-tail probabilities at its smaller value, and selected quantiles.

The transformation sends $-1$ to $3+2(-1)=1$, and $2$ to $3+2(2)=7$. Hence $P(X=1)=0.4$, $P(X=7)=0.6$, and the CDF is $$ F_X(x)= \begin{cases} 0, & x<1,\\ 0.4, & 1\leq x<7,\\ 1, & x\geq7. \end{cases} $$ At the atom $x=1$, $P(X>1)=0.6$, whereas $P(X\geq1)=1$. The difference is exactly the atom probability $P(X=1)=0.4$.

At probability level $p=0.4$, the lower quantile is $1$, because the CDF first reaches $0.4$ at $x=1$. At level $p=0.5$, it is $7$, because the CDF remains at $0.4$ until it jumps to $1$ at $7$. These conclusions will follow from the general lower-quantile definition below.

Lower Quantiles and Endpoint Conventions

Definition: For a real-valued random variable $W$ with CDF $F_W$, its lower quantile at $p\in(0,1)$ is $$ Q_W(p)=\inf\{w\in\mathbb{R}:F_W(w)\geq p\}. $$ Define the endpoint quantiles by $Q_W(0)=\operatorname{ess\,inf}W$ and $Q_W(1)=\operatorname{ess\,sup}W$, allowing the values $-\infty$ and $+\infty$. Here $\operatorname{ess\,inf}W$ is the greatest almost-sure lower bound, and $\operatorname{ess\,sup}W$ is the least almost-sure upper bound.

For $0<p<1$, the set in the definition is nonempty and bounded below: the CDF tends to $1$ as $w$ tends to $+\infty$, and to $0$ as $w$ tends to $-\infty$. Thus the lower quantile is finite. This definition also works for discrete laws. It chooses the first point at which the CDF reaches or exceeds $p$, so a quantile can remain constant across a whole range of probability levels.

The endpoint convention records the extreme values of the distribution up to probability-zero events. For example, a bounded continuous distribution on $[0,1]$ has endpoint quantiles $0$ and $1$, even though the probability of either endpoint may be zero. An unbounded distribution may have an infinite endpoint quantile.

Quantiles Under Location-Scale Changes

Theorem: If $X=\mu+\sigma Z$ with $\mu\in\mathbb{R}$ and $\sigma>0$, then $$ Q_X(p)=\mu+\sigma Q_Z(p),\qquad 0<p<1. $$ The same relation holds at the endpoints $p=0$ and $p=1$, using the extended-real endpoint quantiles defined above.

To prove the interior rule, fix $p\in(0,1)$ and consider the set where the CDF reaches at least $p$. By the CDF identity, $$ \{x:F_X(x)\geq p\} = \left\{\mu+\sigma z:F_Z(z)\geq p\right\}. $$ Since the map $z\mapsto\mu+\sigma z$ is strictly increasing and continuous, it carries the infimum of the set on the right to the infimum of the set on the left. Therefore $$ \inf\{x:F_X(x)\geq p\} = \mu+\sigma\inf\{z:F_Z(z)\geq p\}, $$ which is exactly $Q_X(p)=\mu+\sigma Q_Z(p)$.

For the endpoints, a number $a$ is an almost-sure lower bound for $Z$ precisely when $\mu+\sigma a$ is an almost-sure lower bound for $X$. Since $\sigma>0$, the greatest such lower bound transforms to $\mu+\sigma\operatorname{ess\,inf}Z$. The same argument with almost-sure upper bounds gives $\operatorname{ess\,sup}X=\mu+\sigma\operatorname{ess\,sup}Z$. These identities include infinite endpoints because a positive affine map preserves the directions of unboundedness. This proves the endpoint cases as well. $\square$

Worked Example: Quantiles of a Discrete Distribution

Continue with the two-point variable $Z$ from the previous example. Its CDF is $0$ below $-1$, $0.4$ from $-1$ up to but not including $2$, and $1$ at and above $2$. Therefore $$ Q_Z(p)= \begin{cases} -1, & 0<p\leq0.4,\\ 2, & 0.4<p<1. \end{cases} $$ For $X=3+2Z$, the quantile theorem gives $$ Q_X(p)= \begin{cases} 1, & 0<p\leq0.4,\\ 7, & 0.4<p<1. \end{cases} $$ In particular, $Q_X(0.4)=1$ and $Q_X(0.5)=7$. The endpoint quantiles are $Q_X(0)=1$ and $Q_X(1)=7$, matching the smallest and largest values that $X$ takes with positive probability.

Worked Example: A Uniform Base Law

Let $Z$ be uniform on $[0,1]$, so its CDF is $0$ for $z<0$, $z$ for $0\leq z\leq1$, and $1$ for $z>1$. Set $X=-2+5Z$. At $x=0$, the standardized threshold is $$ \frac{0-(-2)}{5}=0.4. $$ Thus $F_X(0)=0.4$ and $P(X>0)=P(X\geq0)=0.6$; equality of the two tails holds because this law has no atom at zero.

For $0<p<1$, $Q_Z(p)=p$, so $Q_X(p)=-2+5p$. In particular, $Q_X(0.2)=-2+5(0.2)=-1$, and $Q_X(0.8)=-2+5(0.8)=2$. The endpoint quantiles are $-2$ and $3$, the essential lower and upper endpoints of the transformed uniform law.

Common Pitfall: Confusing Strict and Inclusive Tails

A frequent error is to use $P(W\geq w)=1-F_W(w)$. This is incorrect whenever $W$ can equal $w$ with positive probability. Since $F_W(w)=P(W\leq w)$, its complement is $P(W>w)$, not $P(W\geq w)$. The correct inclusive formula is $P(W\geq w)=1-F_W(w-)$. In the two-point example, $1-F_X(1)=0.6$ gives $P(X>1)$; it does not give $P(X\geq1)$, which is $1-F_X(1-)=1$.

A related mistake is to treat a quantile as an ordinary inverse of the CDF at every probability level. A discrete CDF has flat portions and jumps, so it need not have an ordinary inverse. The lower-quantile definition using the infimum and the condition $F_W(w)\geq p$ resolves this ambiguity and makes the affine transformation rule valid without assuming continuity.

Key takeaway: For $\sigma>0$, standardize a threshold with $(x-\mu)/\sigma$ to obtain CDF and tail probabilities, and transform a base quantile $q$ to $\mu+\sigma q$. At atoms, distinguish $P(X>x)$ from $P(X\geq x)$; their difference is $P(X=x)$.

Check Your Understanding

Use the definitions and transformation rules to answer these questions.

If $X=4+3Z$, express $F_X(x)$ and $P(X>x)$ in terms of the distribution of $Z$.
Why can $P(X\geq x)$ differ from $1-F_X(x)$? Write the correct formula for the inclusive upper tail.
For a discrete CDF, why is the lower-quantile definition based on an infimum preferable to an ordinary inverse?
If $Q_Z(0.9)=2$ and $X=-1+5Z$, find $Q_X(0.9)$.
For $Z$ uniform on $[0,1]$ and $X=2+4Z$, find $F_X(4)$, $P(X>4)$, and $Q_X(0.25)$.