Continuous random variables — PDF, CDF, percentiles: Question bank

Quick symbol reminder so nothing is used before it is named:

$f (x)$ = the probability density function (PDF) — "how thickly probability is spread at $x$ ." Its height is density, its area is probability.
$F (x) = P (X \leq x)$ = the cumulative distribution function (CDF) — "how much probability sits at or below $x$ ," the area of $f$ scooped from the far left up to $x$ .
$x_{p}$ = the == $p$ -th percentile==, the value solving $F (x_{p}) = p$ .
Support = the set of $x$ where $f (x) > 0$ ; outside it $f = 0$ .

True or false — justify

A PDF value

f (x)

can be larger than 1.

True.

f

is a density, not a probability; only the area under it must total 1. E.g.

f (x) = 2 x

[0, 1]

reaches

2

, and

f (x) = 5

[0, 0.2]

is fine since its area is

5 \times 0.2 = 1

For a continuous RV,

P (X = c) = 0

for every constant

c

True. A single point is an interval of width zero, so

\int_{c}^{c} f d t = 0

. Probability lives in areas, and a line segment has no area.

Because

P (X = c) = 0

, the event

X = c

is impossible.

False. Zero probability is not the same as impossible —

X

does land on some exact value every trial. "Impossible" (empty event) and "probability zero" coincide only for discrete variables, not continuous ones.

P (a \leq X \leq b) = P (a < X < b)

for a continuous RV.

True. The endpoints

a

and

b

each carry zero probability, so including or excluding them changes nothing. The

\leq

<

distinction is irrelevant here.

The CDF

F (x)

can be greater than 1 somewhere.

False.

F (x) = P (X \leq x)

is a probability, so

0 \leq F (x) \leq 1

always. It rises from

0

up to

1

and never overshoots — that's the total-area-equals-1 axiom seen through accumulation.

F

is always non-decreasing.

True. Moving

x

rightward adds more area under

f

, and since

f \geq 0

that added area can never be negative. So

F

can flatten (where

f = 0

) but never drop.

F

is flat on an interval, then

f = 0

there.

True. By the FTC

f (x) = F^{'} (x)

, and a flat stretch has slope zero, so the density is zero there — no probability is being accumulated on that interval.

Every non-negative function is a valid PDF.

False. Non-negativity is necessary but not sufficient; the total area must also equal exactly 1. E.g.

f (x) = 1

[0, 2]

is non-negative but has area

2 \neq = 1

The median always equals the mean for a continuous RV.

False. They coincide only when the density is symmetric about its centre. For a skewed density (like the exponential) the mean is pulled toward the long tail while the median stays where half the area sits.

If two RVs have the same median, they have the same distribution.

False. The median pins down one point where

F = 0.5

; infinitely many different shapes of

f

can share that single crossing while differing everywhere else.

f

must be continuous for

X

to be a continuous RV.

False. "Continuous RV" refers to

X

taking a continuum of values (so

F

is continuous with no jumps). The density

f

itself may have jumps — e.g. the uniform density is a flat block with sudden drops at its edges. See Uniform distribution.

Spot the error

"Since

f (0.7) = 1.4

for

f (x) = 2 x

, the probability that

X = 0.7

1.4

Error: reading probability off the height.

1.4

is a density, and it even exceeds 1, which a probability never can.

P (X = 0.7) = 0

; probability is the area, not the height.

"To find the 90th percentile I solve

f (x_{p}) = 0.9

Error: using $f$ instead of $F$ . A percentile is an accumulated quantity, so solve

F (x_{p}) = 0.9

. The density value at a point tells you nothing directly about how much probability lies below it.

F (x) = x^{2}

everywhere for

f (x) = 2 x

[0, 1]

Error: forgetting the support.

F (x) = x^{2}

only on

[0, 1]

; below the support

F = 0

and above it

F = 1

. A CDF must reach 0 to the far left and 1 to the far right.

P (1 \leq X \leq 3) = \int_{1}^{3} f = f (3) - f (1)

Error: subtracting the wrong function. Interval probability is

F (b) - F (a)

, i.e. difference of the CDF (the antiderivative), not of the density

f

. You subtract accumulated areas, not heights.

"For

f (x) = λ e^{- λ x}

x \geq 0

, the CDF is

F (x) = e^{- λ x}

Error: sign and boundary. Integrating gives

F (x) = [- e^{- λ t}]_{0}^{x} = 1 - e^{- λ x}

. The proposed answer decreases from 1 to 0, violating the non-decreasing rule — a clear signal of the flipped sign. See Exponential distribution.

"Because

\int f = 1

, the maximum of

f

must be at most 1."

Error: confusing total area with peak height. A tall, narrow spike can have height

≫ 1

yet still enclose area 1. Nothing caps the height of a density; only the total area is fixed.

"The 100th percentile is the largest value

X

can take, so

F (x_{1.00}) = 1

gives it."

Error: percentiles need $0 < p < 1$ . For distributions with unbounded support (like the exponential),

F (x) = 1

has no finite solution — the "100th percentile" is

+ \infty

. Percentiles are defined strictly inside

(0, 1)

Why questions

Why do we use density instead of probability for continuous variables?

Because there are infinitely many exact values, each single value must carry probability 0; a density lets us describe how the total probability of 1 is spread out so intervals still get positive probability via area.

Why must a PDF integrate to exactly 1, not just to something finite?

The total area is the probability that

X

takes some value, which is a certainty — probability 1. A finite-but-not-1 area would mean the probabilities don't sum to certainty.

Why is

f (x) = F^{'} (x)

By the Fundamental Theorem of Calculus, the derivative of an accumulated area is the rate at which area is being added — and that rate is exactly the density. See Fundamental Theorem of Calculus.

Why is

F

invertible for a continuous RV with

f > 0

on its support?

Where

f > 0

F

is strictly increasing, so each probability level

p

is hit at exactly one

x

. That unique crossing is what makes

x_{p} = F^{- 1} (p)

well-defined. See Quantile function and inverse-transform sampling.

Why can't we just add up

P (X = x)

over all

x

like we do for discrete variables?

Each term is 0 and there are uncountably many of them; the sum is meaningless. Integration of a density replaces the sum, letting an uncountable collection of zero-probability points still build up positive interval probabilities. See Discrete random variables — PMF.

Why does the normal distribution use

Φ

for percentiles instead of a neat formula?

Its CDF

Φ (x) = \int_{- \infty}^{x} f

has no elementary closed form, so

Φ^{- 1}

(and z-scores) must be found by tables or numerics rather than algebra. See Normal distribution.

Why does the expectation use

\int x f (x) d x

rather than

\int x F (x) d x

Expectation weights each value

x

by its density of probability there, and

f

is that density;

F

is accumulated probability, the wrong weighting. See Expectation and Variance of continuous RVs.

Edge cases

What is

F (x)

far to the left of the support and far to the right?

F (- \infty) = 0

(no area scooped yet) and

F (+ \infty) = 1

(all the area collected). These limits are forced by the total-area-equals-1 axiom.

What happens to

P (a \leq X \leq b)

b \to a

It shrinks to

F (a) - F (a) = 0

. This is exactly why a single point (the width-zero interval) has probability zero.

Can a valid PDF be zero on part of its stated range?

Yes. The density may vanish on gaps within an interval; those gaps just contribute no probability. Only the total area over the whole line must be 1.

For the exponential

f (x) = λ e^{- λ x}

, what is

P (X < 0)

Exactly 0, because the support is

x \geq 0

and

f = 0

for negative

x

. Always check where the density is defined before integrating.

F

continuous at the edges of a uniform distribution even though

f

jumps there?

Yes —

F

ramps up smoothly (a straight line) with no jumps; the slope of

F

jumps (matching

f

's jumps), but

F

itself stays continuous because no single point carries mass.

What is the median of a distribution whose density is symmetric about

μ

Exactly

μ

: symmetry puts equal area on each side of the centre, so

F (μ) = 0.5

. Here median and mean coincide (as for the Normal distribution).

f (x) > 0

only on a single point, is that a valid PDF?

No. A single point has width zero, so its area is 0, not 1 — you cannot pack the required total probability onto a zero-width set with any finite density.

Recall One-line self-test

Cover every answer above and re-derive the justification, not just the verdict — if you can only say "true/false" you haven't learned the trap.

Connections

Continuous random variables — PDF, CDF, percentiles (the parent note)
Discrete random variables — PMF (where single points do carry probability)
Expectation and Variance of continuous RVs (why $\int x f d x$ , not $F$ )
Normal distribution (median = mean by symmetry; $Φ$ has no closed form)
Exponential distribution (skew makes median ≠ mean; sign trap in its CDF)
Uniform distribution ( $f$ jumps but $F$ stays continuous)
Fundamental Theorem of Calculus (why $f = F^{'}$ )
Quantile function and inverse-transform sampling (why $F$ is invertible)