Discrete random variables — PMF, CDF: Question bank

A quick vocabulary refresher so every symbol here is earned:

True or false — justify

Every

p_{X}

value must lie between 0 and 1 inclusive.

True — each

p_{X} (x) = P (X = x)

is a probability, so

0 \leq p_{X} (x) \leq 1

; the upper bound follows because a single point's mass can never exceed the total mass of

1

If a function satisfies

g (x) \geq 0

and

\sum_{x} g (x) = 1

over a countable set, it is a valid PMF.

True — those two conditions are the entire rulebook; no extra structure (monotonicity, smoothness) is required for a legal PMF.

A PMF can equal

1

at a single value.

True — a degenerate RV that always takes one value

c

has

p_{X} (c) = 1

and

0

elsewhere; it obeys both axioms and is a perfectly legal (if boring) PMF.

For a discrete RV,

P (X = x)

can be positive for infinitely many values.

True — the value set may be countably infinite (e.g. Poisson on

0, 1, 2, \dots

); the masses just have to sum to

1

, which an infinite series can do.

A CDF can decrease over some interval if the PMF has a "negative bump."

False — PMF values are never negative, so the cumulative sum only ever adds non-negative mass; the CDF is non-decreasing everywhere, no exceptions.

F_{X} (x)

can exceed

1

for very large

x

False — the total mass is exactly

1

, so once every point is collected the running total saturates at

1

and stays there;

F_{X} (+ \infty) = 1

is a ceiling.

For a discrete RV,

F_{X}

is continuous.

False — it is a step function with genuine jumps of size

p_{X} (x_{i})

; it is only right-continuous, and a smooth curve would hide where the mass actually sits.

P (a < X \leq b)

and

P (a \leq X \leq b)

always give the same number.

False — they differ by exactly

p_{X} (a)

, the mass at the left endpoint; for a discrete RV that mass is often positive, so the bracket type matters.

F_{X} (3) = F_{X} (2.9)

, then

X

places no mass on any value in

(2.9, 3]

True — equal cumulative totals mean nothing new was scooped up between

2.9

and

3

, so

P (2.9 < X \leq 3) = 0

; in particular

p_{X} (3) = 0

The largest jump in the CDF sits at the most probable value.

True — the jump height at

x_{i}

p_{X} (x_{i})

, so the tallest riser of the staircase marks the mode (the value with greatest mass).

Spot the error

P (X = x)

is the area of the histogram bar, so a wider bar means more probability."

The error is importing continuous-density intuition; for a discrete RV probability is the bar's height (the mass itself), and bar width is a cosmetic drawing choice with no meaning.

P (2 < X \leq 5) = F_{X} (5) - F_{X} (5) = 0

Both terms should not be

F_{X} (5)

; the correct expression is

F_{X} (5) - F_{X} (2)

, subtracting the cumulative total at the lower endpoint so only the mass in

(2, 5]

survives.

"I recovered

p_{X} (4)

F_{X} (4^{-}) - F_{X} (4)

The subtraction is backwards; a jump goes up, so

p_{X} (4) = F_{X} (4) - F_{X} (4^{-})

— value-at-the-point minus value-just-to-the-left, which is non-negative.

"This function has

\sum_{x} g (x) = 1

, so it's a valid PMF" — but one value is

g (x_{0}) = - 0.2

Summing to

1

is necessary but not sufficient; the non-negativity condition

g (x) \geq 0

is violated, so a negative "mass" disqualifies it immediately.

"The CDF hits

1

at the largest value, so

F_{X}

must reach

1

at some finite

x

Only true when the value set is finite; for a countably infinite RV (e.g. geometric)

F_{X}

approaches

1

as a limit and never literally equals it at any finite value.

"Since

F_{X}

is right-continuous, it must also be left-continuous at each value."

A jump makes

F_{X}

discontinuous from the left:

F_{X} (x_{i}^{-}) < F_{X} (x_{i})

; right-continuity means the closed dot (the jump's landing) belongs to the right side, not that both sides match.

"For any RV,

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

because

b

is a single point with probability zero."

That reasoning holds for continuous RVs; for a discrete RV

P (X = b) = p_{X} (b)

may be positive, so

P (X < b) = F_{X} (b) - p_{X} (b)

, which is strictly less when mass sits on

b

Why questions

Why does the "add the point masses" recipe work for discrete RVs but not continuous ones?

Because discrete values are countable and the events

{X = x_{i}}

are disjoint, countable additivity lets us sum their masses; a continuous RV has uncountably many values each with mass

0

, so summing gives

0

and we must integrate a density instead.

Why must a valid PMF sum to exactly

1

and not just to something finite?

The events

{X = x_{i}}

are disjoint and together cover the whole sample space

Ω

, so their probabilities add to

P (Ω) = 1

by the axioms — anything less would mean some outcome carries no probability.

Why is the CDF defined with

\leq

(i.e.

P (X \leq x)

) rather than

<

The convention makes the jump belong to its value, giving right-continuity and the clean recovery rule

p_{X} (x_{i}) = F_{X} (x_{i}) - F_{X} (x_{i}^{-})

; using

<

would make the CDF left-continuous and shift every endpoint bookkeeping.

Why can we recover the entire PMF just from the CDF?

All the probability mass shows up as the jump heights of the staircase, and between jumps the CDF is flat; measuring each jump

F_{X} (x_{i}) - F_{X} (x_{i}^{-})

reads off

p_{X} (x_{i})

for every value.

Why does bracket type (

<

\leq

) matter so much more for discrete than continuous RVs?

For discrete RVs a single point can carry positive mass

p_{X} (a)

, so including or excluding

a

changes the answer by that amount; for continuous RVs each point has mass

0

, making the bracket cosmetic.

Why is the CDF non-decreasing even when the underlying PMF wiggles up and down?

The CDF is a cumulative total; moving right can only append more non-negative mass, never remove any, so the running sum can never drop even if individual masses rise and fall.

Edge cases

What is

F_{X} (x)

for

x

smaller than every possible value of

X

It is

0

— you have not yet passed any value carrying mass, so the running total (your bucket) is still empty; formally

F_{X} (- \infty) = 0

What does the CDF look like for a degenerate RV that always equals

c

A single step:

F_{X} (x) = 0

for

x < c

and

F_{X} (x) = 1

for

x \geq c

, one jump of height

1

c

because all the mass sits on that one value.

p_{X} (x) = 0

for some value

x

in the range, what happens to the CDF there?

Nothing visible — the staircase stays flat across that

x

with no jump, since a zero jump adds no mass; the value is simply not in the support.

Can a discrete RV take negative or non-integer values, and does that break the PMF/CDF machinery?

Yes it can (e.g. values

- 2, 0.5, π

) and nothing breaks; the machinery only needs the value set to be countable, not integer or positive.

For a countably infinite RV, is there a "last" jump that lands the CDF exactly on

1

No — the jumps get arbitrarily small and

F_{X} (x) \to 1

only in the limit

x \to + \infty

; there is no final value where it snaps to

1

, unlike the finite case.

At a jump value

x_{i}

, which value does the CDF actually take — the bottom or the top of the riser?

The top (closed dot on the right), because

F_{X} (x_{i}) = P (X \leq x_{i})

includes the mass at

x_{i}

; the bottom is the left-limit

F_{X} (x_{i}^{-})

, which the CDF has already left behind.

What is

P (X = x)

at a value where the CDF is flat (no jump)?

Zero — a flat stretch means no mass was added there, and

p_{X} (x) = F_{X} (x) - F_{X} (x^{-}) = 0

; that value carries no probability.

Connections

Probability Axioms — non-negativity and countable additivity underwrite every "true/false" verdict here.
Continuous random variables — PDF, CDF — the point-mass-vs-density contrast is the source of most bracket-type traps.
Expectation and Variance of Discrete RVs — built on the same PMF whose validity these questions probe.
Binomial Distribution, Poisson Distribution — the finite and countably-infinite examples behind the edge cases.
Conditional Probability — conditional PMFs inherit every property tested above.