Cumulative distribution functions: Question bank

True or false — justify

A CDF can decrease over some interval if the density dips low there

False — the CDF is a running total of non-negative probability, so it can flatten (add zero) but never fall; a decrease would mean negative probability was added.

For a continuous

X

F_{X} (x)

can be greater than

1

near a tall spike in the density

False — the CDF is a probability, always in

[0, 1]

; it is the PDF

f_{X}

that can exceed

1

because density is probability per unit length, not probability.

F_{X}

is flat (constant) on an interval

[a, b]

, then

X

never lands in

(a, b]

True — a flat stretch means

F_{X} (b) - F_{X} (a) = 0

, and since

P (a < X \leq b) = F_{X} (b) - F_{X} (a)

, no probability mass sits there.

Every CDF must be continuous everywhere

False — discrete distributions have jumps (the staircase for a die); CDFs are only required to be right-continuous, meaning the value at a jump is the top of the step.

lim_{x \to \infty} F_{X} (x) = 1

holds for every valid CDF

True — sweeping to

+ \infty

accumulates all the probability, and total probability is always exactly

1

For a discrete

X

P (X = x)

equals the height of the CDF at

x

False —

P (X = x)

is the size of the jump at

x

, namely

F_{X} (x) - F_{X} (x^{-})

, not the CDF's absolute value.

If two random variables have the same PDF shape but different scales, they have the same CDF

False — different scales stretch the density along

x

, so the accumulated area (the CDF) reaches any given level at a different

x

; only identical distributions share a CDF.

A function that goes from

0

1

and is non-decreasing is automatically a valid CDF

True (essentially) — non-decreasing, limits

0

and

1

, and right-continuity are exactly the defining conditions; any such function is the CDF of some random variable.

For continuous

X

P (X \leq x)

and

P (X < x)

are always equal

True — since

P (X = x) = 0

for continuous variables, including or excluding the single point

x

changes nothing.

For discrete

X

P (X \leq x)

and

P (X < x)

are always equal

False — at a value

x

that carries mass,

P (X \leq x)

includes the jump

P (X = x)

while

P (X < x)

excludes it, so they differ by that jump.

Spot the error

"To find

P (5 < X \leq 8)

I computed

F_{X} (5) - F_{X} (8) = - 0.2

." What went wrong?

The subtraction is backwards — you need

F_{X} (8) - F_{X} (5)

; because the CDF is non-decreasing, the larger argument always gives the larger value, so a negative result is the tell-tale sign of a flipped order.

X

is continuous, so

P (X = 3) = f_{X} (3)

, the density there." Where's the slip?

For continuous

X

P (X = 3) = 0

exactly;

f_{X} (3)

is a density (probability per unit length), and to get any real probability you must integrate it over an interval, not read it at a point.

"The median of

X

is wherever the PDF is tallest." What's confused?

That describes the mode (peak of the density); the median is where the CDF crosses

0.5

, i.e.

F_{X}^{- 1} (0.5)

, which splits area equally and can sit far from the peak in a skewed distribution.

"Since

F_{X}

is right-continuous, at a jump the CDF takes the value just below the step." Fix it.

Right-continuity means the CDF takes the value at the top of the step (the limit from the right equals the value); the point closes on the upper value, matching

F_{X} (x) = P (X \leq x)

"I inverted the CDF by writing

F_{X}^{- 1} (x) = 1/ F_{X} (x)

." What's the error?

The inverse CDF is a functional inverse (it undoes

F_{X}

, answering "which

x

gives this probability

p

?"), not a reciprocal

1/ F_{X}

; you solve

F_{X} (x) = p

for

x

P (X > x) = F_{X} (x)

because both describe the tail." Correct it.

F_{X} (x) = P (X \leq x)

is the left accumulation; the right tail is the complement

P (X > x) = 1 - F_{X} (x)

"The CDF of a fair die at

x = 3.5

is undefined because you can't roll

3.5

." Why is this wrong?

The CDF is defined for all real

x

;

F_{X} (3.5) = P (X \leq 3.5) = \frac{3}{6}

, counting outcomes

1, 2, 3

— the argument need not be an achievable value.

Why questions

Why is the CDF (not the PDF) the natural tool for asking "what fraction of predictions fall below a threshold?"

Because that question is literally

P (X \leq t)

— an accumulated probability — which is the CDF's definition; the PDF only gives local density and would need integrating to answer it.

Why must every CDF be right-continuous rather than left-continuous, given the

\leq

P (X \leq x)

The

\leq

includes the point

x

itself, so at a jump the mass at

x

is already counted at

x

; approaching from the right (values already past

x

) gives the same total, making the CDF continuous from the right.

Why does

E [X] = \int_{0}^{\infty} [1 - F_{X} (x)] d x

hold for non-negative

X

The quantity

1 - F_{X} (x) = P (X > x)

is the survival (tail) probability; summing how long

X

stays above each level, via integration by parts, rebuilds the average value.

Why is the inverse CDF the engine behind generating samples from any distribution?

U

is uniform on

[0, 1]

, then

X = F_{X}^{- 1} (U)

lands below any level

x

with probability exactly

F_{X} (x)

— see Inverse Transform Sampling — so feeding uniform noise through the quantile function reproduces the target distribution.

Why do the Kolmogorov–Smirnov test and calibration checks compare CDFs rather than PDFs?

CDFs are bounded in

[0, 1]

and defined everywhere without binning, so the maximum gap between an empirical CDF and a theoretical one is a clean, robust distance — see Kolmogorov-Smirnov Test — whereas PDF comparison depends on arbitrary bin widths.

Why can two distributions have the same mean and variance yet very different CDFs?

The mean and variance are only two summary numbers (see Expected Value and Variance); the CDF encodes the entire shape, so skew, tails, and multimodality can differ while the first two moments coincide.

Edge cases

What does the CDF look like for a constant random variable,

X = c

with certainty?

A single step:

F_{X} (x) = 0

for

x < c

and

F_{X} (x) = 1

for

x \geq c

; all probability piles up at one point, so the jump has full height

1

For a mixed variable (a continuous part plus a point mass at

x = 2

), how does the CDF behave at

2

It rises smoothly along the continuous part but shows a vertical jump of height

P (X = 2)

exactly at

x = 2

, combining staircase and ramp behaviour.

What is

F_{X} (x)

far to the left, at

x \to - \infty

, and does it ever start above

0

It must approach

0

; no probability has been accumulated yet, and starting above

0

would imply probability mass at

- \infty

, which is impossible.

If the density has an infinitely tall spike (like a very sharp peak), what happens to the CDF there?

The CDF stays bounded in

[0, 1]

and merely climbs steeply through that region; a tall-but-integrable spike adds finite area, so the running total rises fast but never exceeds

1

What is

F_{X}^{- 1} (p)

when the CDF is flat at level

p

over an interval (so many

x

satisfy

F_{X} (x) = p

The inverse is not unique there; by convention the quantile function takes the smallest such

x

F_{X}^{- 1} (p) = in f {x : F_{X} (x) \geq p}

, which keeps it well-defined even on flats and jumps.

What does

F_{X}^{- 1} (p)

return at the extremes

p = 0

p = 1

These can push to the support's boundaries —

- \infty

+ \infty

for unbounded distributions like the exponential — because no finite value has

0%

or a full

100%

strictly below it.

Recall Self-check: which single property catches the most traps?

Monotonic non-decreasing ::: Yes — it kills negative "probabilities", forces $F_{X} (b) \geq F_{X} (a)$ , forbids dips, and underwrites why we subtract in the correct order. Bounded in $[0, 1]$ ::: Also central — it's what distinguishes the CDF from the (possibly $> 1$ ) PDF and flags any impossible answer instantly.

See also: Probability Distributions · Random Variables · Survival Analysis · Copulas.