Questions: Distribution and Density Functions (Rigorous)

Question 1

A student claims that every random variable has a probability density function because 'you can always compute the probability of falling in any interval.' What is wrong with this claim?

• A. Not every random variable has well-defined probabilities for intervals — some distributions are not σ-finite
• B. A PDF requires the distribution to be absolutely continuous with respect to Lebesgue measure; discrete and singular distributions have no PDF even though interval probabilities are perfectly well-defined
• C. The student is essentially correct — every distribution has an associated PDF, though for discrete distributions it takes the form of a sum of delta functions
• D. PDFs are only defined for distributions supported on bounded intervals; unbounded distributions require a different formalism

Answer: (B)

The existence of a PDF is not about whether probabilities for intervals exist — it's about whether the distribution is absolutely continuous with respect to Lebesgue measure. A discrete distribution places all mass on countable points (e.g., P(X = k) = pₖ); interval probabilities are well-defined but no PDF exists because the distribution is singular with respect to Lebesgue measure. The Radon-Nikodym theorem guarantees a PDF exists if and only if the distribution μ_X satisfies μ_X ≪ λ (absolute continuity). Delta functions are a generalized-function tool that is not part of the rigorous measure-theoretic framework.

Question 2

The Cantor distribution has a CDF that increases continuously from 0 to 1 but has derivative zero almost everywhere. What does this imply?

• A. The distribution has a well-defined PDF equal to zero almost everywhere, which is consistent with a total probability of 1
• B. The distribution is singular continuous — it has no atoms and no PDF, because it is not absolutely continuous with respect to Lebesgue measure
• C. The Cantor distribution is actually discrete, with probability mass concentrated on the rational points of [0, 1]
• D. Since the CDF is continuous and non-decreasing, the PDF can be recovered by differentiating the CDF as usual

Answer: (B)

The Cantor distribution's CDF (the Devil's staircase) increases entirely on the Cantor set, which has Lebesgue measure zero. This means the distribution concentrates all its probability on a measure-zero set — it is not absolutely continuous with respect to Lebesgue measure. If a PDF f existed, we would need ∫_B f dλ = μ_X(B) for all Borel B, but the derivative is zero a.e., so any Radon-Nikodym density would be zero a.e. and integrate to zero, not 1 — a contradiction. The Cantor distribution is the canonical example of a third type: singular continuous (no atoms, but also no PDF).

Question 3

Any random variable whose CDF is everywhere continuous should have a probability density function.

• T. True
• F. False

Answer: False

A continuous CDF means the distribution has no atoms — no individual point x with positive probability P(X = x) > 0. But continuity of the CDF is weaker than absolute continuity of the distribution with respect to Lebesgue measure. The Cantor distribution is the definitive counterexample: its CDF is continuous, yet no PDF exists because the distribution is singular continuous, concentrating probability on the Cantor set (a measure-zero set with no isolated points). For a PDF to exist, you need the stronger condition that the distribution is absolutely continuous with respect to Lebesgue measure.

Question 4

The cumulative distribution function F(x) = P(X ≤ x) uniquely determines the probability distribution of X.

• T. True
• F. False

Answer: True

There is a one-to-one correspondence between CDFs (non-decreasing, right-continuous functions with F(−∞) = 0 and F(+∞) = 1) and probability measures on (ℝ, ℬ(ℝ)). Any function satisfying these three properties corresponds to a unique probability measure, and conversely every probability measure on ℝ has a unique CDF. This is why CDFs are the universal language for describing distributions — regardless of whether a PDF or PMF exists, the CDF always exists and fully encodes the distribution.

Question 5

What does it mean to say that a probability density function is the Radon-Nikodym derivative of the distribution with respect to Lebesgue measure, and why does this framing explain when a PDF fails to exist?

Model answer: A PDF exists when the distribution μ_X is absolutely continuous with respect to Lebesgue measure λ — meaning that any set of zero length also has zero probability. Under this condition, the Radon-Nikodym theorem guarantees a measurable function f such that P(X ∈ B) = ∫_B f dλ for all Borel sets B. The function f is the 'density of probability per unit length.' If the distribution is not absolutely continuous — as with a discrete distribution (all mass on countable points, a set of zero Lebesgue measure) or a singular continuous distribution (mass on a Cantor-like set of zero Lebesgue measure) — then the Radon-Nikodym hypothesis fails and no such f can exist. Intuitively: you cannot express probability as 'area under a curve' if probability doesn't spread continuously over intervals.

The Radon-Nikodym framing unifies the otherwise apparently different formulas for continuous and discrete distributions. P(a < X ≤ b) = ∫_a^b f(x) dx and P(X = k) = pₖ are both special cases of P(X ∈ B) = ∫_B dμ_X — in the first case μ_X is absolutely continuous so you can substitute f dλ; in the second case μ_X is a sum of point masses. The measure-theoretic framework is not just aesthetic: it provides the tools (Radon-Nikodym, Lebesgue decomposition) to handle all three types of distributions uniformly.