1.3.5 · D5Probability & Statistics
Question bank — Random variables (discrete and continuous)
Every item below targets a specific misconception. The reasoning is the point — a bare "true" or "false" earns nothing.
True or false — justify
A random variable is a specific number like "3".
False. It is a function mapping outcomes to numbers; "3" is a value the function can output, not the variable itself.
For a discrete RV, can be greater than 1.
False. Each is a probability, so it lives in ; the whole set must also sum to exactly 1.
For a continuous RV, the density can be greater than 1.
True. is a density (probability per unit length), not a probability — e.g. has on that interval.
For a continuous RV, means the value is impossible.
False. It means that single point carries zero area; you can still observe values arbitrarily close to . Impossibility and zero probability differ for continuous RVs.
must be one of the values can actually take.
False. A fair die has , which is never rolled. The mean is a balance point, not an attainable outcome.
Two different random variables can share the same PMF.
True. The PMF describes the distribution of values, not the underlying experiment; "dots on a die" and "1..6 drawn from a bag" have identical PMFs.
For any valid PDF, guarantees everywhere.
False. Normalization constrains total area, not height; a narrow tall spike can exceed 1 while still integrating to 1.
The CDF is always non-decreasing.
True. As grows you include more outcomes, so accumulated probability can only stay flat or rise, never drop.
A Binomial RV with is exactly a Bernoulli RV.
True. One trial, count of successes — the formula collapses to the Bernoulli PMF.
Spot the error
" for a continuous RV."
Wrong: you must integrate, not sum. ; a continuous RV has no mass at individual points to sum.
"Since for continuous RVs, we can never compute any probability from them."
Wrong: probabilities of intervals are perfectly well defined via area under . Only single-point probabilities vanish.
" — just add up the probabilities."
Wrong: each value must be weighted by its probability. The correct expression is ; a bare sum of probabilities is always 1.
"For the uniform , the density is ."
Wrong: it is , the reciprocal of the interval width, so that the rectangle of area (width height) equals 1.
"The Gaussian PDF value is the probability of landing exactly at the mean."
Wrong: it is the peak density, not a probability. Any exact point has probability 0; only intervals carry probability.
"For a Binomial, ."
Missing the factor. That single number is the probability of one specific ordering; you must multiply by the count of orderings that give heads.
" can be negative as long as the total area is 1."
Wrong: non-negativity is a hard requirement. A negative density would imply a negative interval probability, which is meaningless.
Why questions
Why do we weight each value by its probability when computing ?
Because in a long run of trials, a value appears about times, so it contributes proportionally to its probability in the sample average.
Why does the "" in matter dimensionally?
carries units of probability-per-unit-length; multiplying by (a length) yields an actual probability, which is what we sum.
Why is for the uniform distribution, without doing the integral?
By symmetry — the density is flat, so mass is balanced equally on both sides of the midpoint, placing the mean at the center.
Why is for a Binomial provable without touching the PMF?
A Binomial is a sum of independent Bernoulli() variables; expectation is linear, so . See Expectation and Variance.
Why can't we just use a PMF for a continuous quantity like temperature?
A PMF assigns positive mass to specific points, but there are uncountably many temperature values; positive mass at each would blow the total past 1. We need a density instead.
Why does the Gaussian use rather than, say, ?
The squared term is the maximum-entropy answer when only mean and variance are fixed, and it gives the smooth bell that arises as the Central Limit Theorem limit of sums. See Gaussian Distribution.
Why is the CDF useful when we already have the PMF or PDF?
It answers "" directly, handles both discrete and continuous RVs uniformly, and lets us compute any interval as without re-integrating.
Edge cases
What is for a Bernoulli with ?
. The coin never shows heads, so is always 0 and its average is 0 — a degenerate (constant) random variable.
Is a constant (always the same value) a random variable?
Yes — it's the degenerate RV with , , and zero variance. Every outcome maps to the same number.
For with , does the PDF make sense?
No — the width makes undefined. This collapses to a degenerate point mass at , which needs a discrete description, not a density.
Can a single random variable be neither purely discrete nor purely continuous?
Yes — mixed RVs exist, e.g. rainfall that is 0 with positive probability (a point mass) but continuous when positive. Its distribution combines a PMF-like spike with a PDF.
What happens to the Gaussian as ?
The bell narrows and grows taller, its peak density , and all mass concentrates at — approaching a degenerate spike at the mean while area stays 1.
As with fixed, what shape does the Binomial approach?
A Gaussian, by the Central Limit Theorem, since it is a sum of many independent Bernoulli trials — the discrete bars trace out a bell curve.
Recall Quick self-check
Which quantity — or — is allowed to exceed 1, and why? ::: , because it is a density (per-unit-length), constrained only in total area; is a genuine probability capped at 1.