Introduction
When you encounter a list of candidate functions or tables and are asked “which of the following are valid probability distributions?Practically speaking, whether the candidates are discrete probability mass functions (PMFs), continuous probability density functions (PDFs), or mixed distributions, they all share the same core requirements: non‑negativity, normalization, and, when applicable, proper support. But ”, the answer hinges on a handful of fundamental rules that any probability distribution must obey. In this article we will unpack these criteria, illustrate them with common examples, and walk through a systematic checklist that lets you decide quickly and confidently which candidates are legitimate probability distributions.
1. Core Properties of All Probability Distributions
1.1 Non‑negativity
For any outcome (x) in the sample space (\mathcal{X}),
[ p(x) \ge 0 \qquad\text{(discrete)} \qquad\text{or}\qquad f(x) \ge 0 \qquad\text{(continuous)}. ]
A negative value instantly disqualifies a candidate because probabilities cannot be less than zero Most people skip this — try not to. Still holds up..
1.2 Normalization (Total Probability Equals One)
- Discrete case:
[ \sum_{x\in\mathcal{X}} p(x) = 1. ]
- Continuous case:
[ \int_{-\infty}^{\infty} f(x),dx = 1. ]
If the sum or integral does not equal one, the function can be rescued only by multiplying it by a constant (the normalising constant). g.That's why if such a constant does not exist (e. , the integral diverges), the candidate is invalid That alone is useful..
1.3 Proper Support
The support is the set of points where the distribution assigns positive probability. For a discrete distribution, the support must be a countable set; for a continuous distribution, it is usually an interval or a union of intervals. A candidate that claims non‑zero probability outside its declared support is inconsistent Worth keeping that in mind..
1.4 Measurability (Technical Note)
In rigorous probability theory the function must be measurable with respect to the underlying sigma‑algebra. In most elementary contexts this condition is automatically satisfied, but it matters when dealing with exotic or pathological functions That's the part that actually makes a difference..
2. Checklist for Validating a Candidate
| Step | Question | Action |
|---|---|---|
| 1 | Is the function defined for all (x) in its claimed domain? | |
| 2 | Are all values (\ge 0)? | Any negative value → invalid. |
| 3 | Does the sum (discrete) or integral (continuous) equal 1? | |
| 4 | Does the support match the description? | Compute; if it equals a finite constant (C\neq1), multiply by (1/C). Consider this: if the sum/integral diverges, invalid. And |
| 5 | For mixed distributions, do the discrete and continuous parts together sum to 1? | Verify that probabilities outside the support are zero. Consider this: |
Applying this checklist to each option in a multiple‑choice list will quickly reveal the valid distributions That's the part that actually makes a difference. Took long enough..
3. Common Pitfalls and How to Spot Them
3.1 Forgetting the Normalisation Constant
A classic mistake is presenting a function like
[ f(x)=\frac{1}{x^2},\qquad x\ge 1, ]
and calling it a PDF. The integral (\int_{1}^{\infty} \frac{1}{x^2}dx = 1), so this one is valid. Even so, the same functional form on ([0,\infty)) would give an infinite integral, making it invalid. Always check the limits Not complicated — just consistent..
3.2 Implicit Negative Values
Consider
[ p(k)=\frac{(-1)^k}{2^k},\qquad k=0,1,2,\dots ]
Although the absolute values sum to a finite number, the alternating sign produces negative probabilities for odd (k). This violates non‑negativity.
3.3 Overlooking Discrete Mass in Mixed Distributions
A mixed distribution might be defined as
[ P(X=0)=0.3,\qquad f(x)=0.7,e^{-x},;x>0. ]
Here the discrete mass (0.3) plus the integral of the exponential density ((0.7)) sum to 1, so the distribution is valid. Forgetting to add the discrete part would lead to the mistaken belief that the density alone must integrate to 1.
3.4 Using Improper Support
A candidate that says
[ p(k)=\frac{1}{6},\qquad k\in{1,2,3,4,5} ]
leaves out the outcome “6”. The probabilities sum to (5/6), not 1, and the support is incomplete. The correct PMF for a fair die must include (k=6).
4. Illustrative Examples
Below we examine a set of typical candidates and determine their validity using the checklist That's the part that actually makes a difference..
Example 1 – Discrete PMF
[ p(k)=\frac{k}{10},\qquad k=1,2,3. ]
Non‑negativity: All values are positive.
Normalization: (\sum_{k=1}^{3} \frac{k}{10}= \frac{1+2+3}{10}=0.6) And it works..
Because the total is not 1, we can renormalize: multiply by (1/0.6 = \frac{5}{3}). The adjusted PMF becomes
[ p'(k)=\frac{5k}{30}=\frac{k}{6},\qquad k=1,2,3, ]
which now sums to 1. Hence the original shape is acceptable, but the given numbers are not a valid PMF until normalized.
Example 2 – Continuous PDF
[ f(x)=\frac{3}{4}(1-x^2),\qquad -1\le x\le 1. ]
Non‑negativity: The quadratic (1-x^2) is non‑negative on ([-1,1]).
Normalization:
[ \int_{-1}^{1}\frac{3}{4}(1-x^2)dx = \frac{3}{4}\Bigl[ x - \frac{x^3}{3}\Bigr]_{-1}^{1}= \frac{3}{4}\bigl[(1-\tfrac13)-(-1+\tfrac13)\bigr]=1. ]
All conditions are satisfied, so this is a valid continuous distribution (the semicircle distribution) Worth keeping that in mind..
Example 3 – Invalid Continuous Candidate
[ f(x)=\frac{2}{x},\qquad 0<x\le 1. ]
Non‑negativity: Positive on the interval.
Normalization:
[ \int_{0}^{1}\frac{2}{x}dx = 2\int_{0}^{1}\frac{1}{x}dx = 2[\ln x]_{0}^{1}= \infty. ]
The integral diverges, so no finite normalising constant exists. Invalid.
Example 4 – Mixed Distribution
[ P(X=0)=0.4,\qquad f(x)=0.6,\lambda e^{-\lambda x},;x>0,;\lambda=2. ]
Discrete part: 0.4.
Continuous part:
[ \int_{0}^{\infty}0.6\cdot2e^{-2x}dx =0.6\cdot2\Bigl[\frac{-e^{-2x}}{2}\Bigr]_{0}^{\infty}=0.6. ]
Total = 0.4 + 0.6 = 1 → valid Not complicated — just consistent..
Example 5 – Discrete with Improper Support
[ p(k)=\frac{1}{2^k},\qquad k\in\mathbb{Z}. ]
The series (\sum_{k=-\infty}^{\infty} 2^{-k}) diverges because terms for negative (k) grow without bound. Hence the candidate is invalid unless the support is restricted to non‑negative integers Most people skip this — try not to. Practical, not theoretical..
5. Frequently Asked Questions
Q1: Can a probability distribution have values greater than 1?
A: Yes, a density (f(x)) can exceed 1 as long as the integral over its support equals 1. For a mass function (p(x)), any single probability must be ≤ 1, but the sum of all probabilities must be exactly 1.
Q2: What if the integral of a continuous candidate equals 0.5?
A: Multiply the function by the reciprocal of 0.5 (i.e., by 2). The resulting function will integrate to 1 and become a valid PDF No workaround needed..
Q3: Do cumulative distribution functions (CDFs) need to be checked?
A: A CDF must be non‑decreasing, right‑continuous, start at 0, and approach 1 as (x\to\infty). If you are given a CDF, verify these properties; the associated PDF/PMF can then be derived by differentiation (continuous part) or by jumps (discrete part).
Q4: How to handle piecewise definitions?
A: Treat each piece separately, ensuring non‑negativity on its interval and that the sum of the integrals (or sums) across all pieces equals 1. Pay special attention to overlap or gaps in the pieces.
Q5: Is a Dirac delta “function” a probability distribution?
A: In the framework of generalized functions, the Dirac delta (\delta(x-a)) represents a degenerate distribution that places all probability mass at a single point (a). It satisfies (\int_{-\infty}^{\infty}\delta(x-a)dx = 1) and is considered a valid (though not ordinary) probability distribution.
6. Practical Tips for Exams and Assignments
- Write down the normalization condition first. Even before checking signs, compute the sum or integral; if it diverges, you can stop early.
- Sketch the function. A quick graph reveals negative regions or unbounded growth that may cause divergence.
- Keep a table of common normalising constants. To give you an idea, the Beta distribution (f(x)=x^{\alpha-1}(1-x)^{\beta-1}) on ([0,1]) normalises with (B(\alpha,\beta)). Recognizing the pattern saves time.
- Don’t forget the endpoints. For continuous PDFs, the probability of a single point is zero, but the density at the endpoint can be non‑zero; ensure the limits are inclusive/exclusive as specified.
- Check units in applied problems. If the variable represents a physical quantity (e.g., time in seconds), the density’s units are “per second.” A mismatch often signals a missing normalising factor.
7. Conclusion
Determining whether a given function or table constitutes a valid probability distribution is a matter of applying three fundamental rules: non‑negativity, normalization to one, and correct support. Think about it: by systematically checking these criteria—preferably with the concise checklist provided—you can swiftly filter out invalid candidates and, when possible, rescue near‑valid ones through proper normalisation. Mastery of this process not only prepares you for textbook exercises and examinations but also equips you with a solid foundation for more advanced work in statistics, machine learning, and stochastic modeling, where the integrity of probability models is very important And it works..
