Uniform, Exponential, and Beta distributions: Question bank

This is a trap-spotting page for Uniform, Exponential, and Beta distributions. Every line below is a question ::: answer reveal. Read the question, commit to an answer in your head, THEN reveal. If your gut answer disagrees with the reasoning, you found a misconception worth fixing.

Prerequisites worth having open: Continuous Random Variables and PDFs, Poisson Process, Bayesian Inference and Conjugate Priors.

True or false — justify

A density value bigger than 1 is impossible

False — a density is not a probability. For

U (0, 0.2)

the height is

\frac{1}{0.2} = 5

; only the area under the curve must be

\leq 1

, and here width

0.2 \times

height

5 = 1

For

U (a, b)

, the probability of landing exactly at the midpoint is the highest

False — every single point has probability exactly

0

for a continuous variable. The midpoint is only special as the mean, not as a more-likely outcome.

The exponential distribution is the continuous cousin of the geometric distribution

True — both count "waiting until the first success/event" and both are the only memoryless distributions in their world (continuous vs. discrete).

If waiting times between bus arrivals are

Exp (λ)

, then after waiting 10 minutes you are "due" for a bus soon

False — memorylessness means the remaining wait has the same distribution as a fresh start. The process does not remember that you waited.

Beta (1, 1)

is a completely different animal from the uniform distribution

False — plug

α = β = 1

: the PDF becomes

\frac{1}{B ( 1 , 1 )} x^{0} (1 - x)^{0} = 1

[0, 1]

, which is exactly

U (0, 1)

Larger

λ

Exp (λ)

means longer waiting times

False —

λ

is the rate; mean wait is

1/ λ

. Bigger rate → events pack closer → shorter waits.

Increasing both

α

and

β

(keeping their ratio fixed) sharpens a Beta's peak

True — the mean

\frac{α}{α + β}

stays put, but larger pseudo-counts mean more "evidence", so the distribution narrows around that mean (higher confidence).

The exponential PDF is a valid probability density even though

λ e^{- λ t}

can exceed 1

True — for

λ = 5

f (0) = 5

. Again density

\neq =

probability;

\int_{0}^{\infty} λ e^{- λ t} d t = 1

is what matters.

A Beta distribution can be U-shaped

True — when

α < 1

and

β < 1

the exponents

α - 1, β - 1

are negative, so the density blows up toward both

0

and

1

, giving a bathtub shape (belief concentrated at the extremes).

The variance of

U (a, b)

depends on where the interval sits on the number line

False —

Var = \frac{( b - a ) ^{2}}{12}

depends only on the width

b - a

; shifting

[a, b]

left or right changes the mean, not the spread.

Spot the error

"Mean of

Exp (λ = 2)

2

because the

2

is right there in the formula."

The error is reading

λ

as the mean. Mean is

1/ λ = 0.5

. The visible number is the rate, not the time.

"To find the 95th percentile of

Exp (λ)

, I set

e^{- λ t} = 0.95

and solve."

Wrong tail. The CDF is

F (t) = 1 - e^{- λ t}

; the 95th percentile solves

1 - e^{- λ t} = 0.95

, i.e.

e^{- λ t} = 0.05

. You solved for the 5th percentile.

"Beta has support on all real numbers, like the normal distribution."

False support — Beta lives strictly on

[0, 1]

, which is exactly why it models probabilities and proportions.

"Since

U (a, b)

has constant density, its CDF is also constant."

The PDF is constant, but the CDF

F (x) = \frac{x - a}{b - a}

rises linearly from

0

1

— a ramp, not a flat line.

"For

Beta (α, β)

, the mode equals the mean

\frac{α}{α + β}

Only when

α = β

(symmetric). In general the mode is

\frac{α - 1}{α + β - 2}

(for

α, β > 1

); mean and mode differ for skewed Betas.

Γ (α) = α!

Γ (3) = 6

Off by one shift —

Γ (n) = (n - 1)!

, so

Γ (3) = 2! = 2

, not

6

. This shift is why

Beta (α, β)

uses exponents

α - 1, β - 1

"The exponential can model a component whose failure rate rises with age."

No — exponential encodes a constant hazard rate (memorylessness). Aging (rising hazard) needs a Weibull or gamma, not exponential.

Why questions

Why must the uniform density equal exactly

\frac{1}{b - a}

and no other constant?

Because a constant

c

over width

(b - a)

must integrate to

1

c (b - a) = 1

. Any other value would make the total probability wrong.

Why does the exponential come out of a Poisson process specifically?

The chance of zero events in time

t

e^{- λ t}

; "wait exceeds

t

" is the same event, so

P (T > t) = e^{- λ t}

and differentiating

1 - e^{- λ t}

gives the exponential PDF.

Why is Beta called the "conjugate prior" for the Bernoulli/Binomial?

Because a Beta prior times a Binomial likelihood is proportional to another Beta — you just add successes to

α

and failures to

β

. The belief stays in the same family, so updating is bookkeeping. See Bayesian Inference and Conjugate Priors.

Why do we interpret

α

and

β

as "pseudo-counts"?

Because posterior

Beta (α + h, β + t)

shows heads

h

and tails

t

literally added on — so

α - 1, β - 1

act like prior imaginary successes and failures baked into your belief before any real data.

Why does memorylessness make exponential the natural model for "time between server requests"?

Requests arrive independently at a steady average rate; how long you've already waited gives no clue about the next one — exactly the "process forgets the past" behaviour exponential enforces.

Why is

U (- 1/ n, 1/ n)

the right initializer scale in Xavier and He Initialization?

Its variance

\frac{( 2/ n ) ^{2}}{12} = \frac{1}{3 n}

scales like

1/ n

, keeping signal variance roughly constant across layers so gradients neither vanish nor explode.

Why can the uniform distribution represent "total ignorance"?

Because it assigns equal density everywhere in its range — no point is preferred, which is the mathematical statement of having no reason to favour any outcome. It's also the maximum-entropy distribution on a bounded interval.

Why does Beta's mean

\frac{α}{α + β}

read like a success fraction?

Because after seeing

α - 1

successes and

β - 1

failures, your best estimate of the true probability is essentially (successes)/(total), smoothed by the prior — a Laplace-style estimate.

Edge cases

What happens to

U (a, b)

b \to a

The width shrinks to

0

, the density

\frac{1}{b - a} \to \infty

, and the distribution collapses to a point mass at

a

— variance

\frac{( b - a ) ^{2}}{12} \to 0

What is the exponential PDF at

t = 0

, and is that a problem?

f (0) = λ

, a finite positive value — the density is largest right at the start and decays from there. Not a problem: it's still a valid density since the area is

1

What does

Beta (α, β)

do at the boundaries

x = 0

and

x = 1

when

α < 1

β < 1

The density diverges (goes to

+ \infty

) at that boundary, yet the integral stays finite — the spike is integrable, modelling strong belief that the true probability sits near an extreme.

Is the exponential defined for

λ \leq 0

No —

λ

must be

> 0

. A zero or negative rate makes

λ e^{- λ t}

non-normalizable or negative, so it cannot be a density.

What distribution is

Beta (α, β)

when

α

and

β

grow huge with equal ratio?

It concentrates into an ever-narrower spike around its mean — in the limit it behaves like a point mass, i.e. near-certainty about the probability's value.

For

Exp (λ)

, what is

P (T = t)

for any specific

t

Exactly

0

— like all continuous variables, only intervals carry positive probability; asking for an exact instant always gives zero.

What happens to the exponential's mean and variance as

λ \to \infty

Mean

1/ λ \to 0

and variance

1/ λ^{2} \to 0

— events arrive so fast the waiting time collapses toward zero with vanishing spread.

Recall Quick self-test

$λ$ is rate or time? ::: Rate — mean time is $1/ λ$ . $Beta (1, 1)$ equals which distribution? ::: $U (0, 1)$ . Which continuous distribution alone is memoryless? ::: The exponential. Does a density above 1 break anything? ::: No — only the total area must be 1.