Which Of The Following Is A Valid Probability Distribution

Understanding Valid Probability Distributions: Rules, Examples, and Common Mistakes

A probability distribution is one of the most fundamental concepts in statistics, data science, and decision-making. At its core, it is a mathematical function that describes the likelihood of obtaining the possible values that a random variable can take. However, not every table or equation that claims to be a probability distribution actually qualifies. The critical question—"which of the following is a valid probability distribution?"—requires a systematic check against two non-negotiable rules. This article will demystify these rules, walk through concrete examples of both valid and invalid distributions, and explain why this knowledge is essential for interpreting data correctly and avoiding costly analytical errors.

The Two Golden Rules of a Valid Probability Distribution

Before evaluating any specific case, you must internalize the two universal conditions that all probability distributions must satisfy. These rules apply whether the random variable is discrete (like the roll of a die) or continuous (like daily temperature).

1. Non-Negativity: The probability of any individual outcome must be a number between 0 and 1, inclusive. Formally, for every possible value x, P(X = x) ≥ 0. Probabilities cannot be negative. A negative number has no interpretation as a chance or likelihood.
2. Total Probability Equals One: The sum of the probabilities for all possible outcomes must exactly equal 1. For a discrete random variable, this means Σ P(X = x) = 1 over all x. For a continuous random variable, the integral of the probability density function (PDF) over its entire range must equal 1. This rule embodies the certainty that some outcome from the defined set must occur.

These two rules are necessary and sufficient. A function that meets both is a valid probability distribution. A failure in either condition immediately invalidates it.

Analyzing Discrete Probability Distributions: Step-by-Step

Discrete distributions deal with countable outcomes. We often represent them with a table or a formula like a probability mass function (PMF). Let’s apply the rules to several scenarios.

Example 1: A Valid Discrete Distribution (Fair Six-Sided Die) Consider the random variable X = outcome of a fair die roll.

• Check Rule 1 (Non-Negativity): All probabilities are 1/6 (~0.167), which is ≥ 0. PASS.
• Check Rule 2 (Sum to 1): 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1. PASS. This is a perfectly valid distribution.

Example 2: Invalid Due to Sum ≠ 1

• Rule 1: All probabilities are non-negative. PASS.
• Rule 2: Sum = 0.3 + 0.3 + 0.3 = 0.9. This is less than 1. FAIL. The total probability is missing 0.1, meaning there’s a 10% chance of an outcome not listed (e.g., "Other" color), which violates the defined sample space. This is invalid.

Example 3: Invalid Due to Negative Probability

• Rule 1: P(Lose $5) = -0.2 is negative. FAIL. A negative probability is meaningless. Even though the sum (0.5 - 0.2 + 0.7 = 1.0) equals 1, the violation of non-negativity makes this entire distribution invalid.

Example 4: A Valid Discrete Distribution with a Formula Let X be the number of heads in two fair coin flips. The PMF is P(X = x) = C(2, x) * (0.5)^x * (0.5)^(2-x) for x = 0, 1, 2.

• P(X=0) = 1 * 1 * 0.25 = 0.25
• P(X=1) = 2 * 0.5 * 0.5 = 0.5
• P(X=2) = 1 * 0.25 * 1 = 0.25
• Rule 1: All values (0.25, 0.5, 0.25) are ≥ 0. PASS.
• Rule 2: Sum = 0.25 + 0.5 + 0.25 = 1. PASS. This is the valid Binomial distribution for n=2, p=0.5.

Continuous Probability Distributions: The Integral Rule

For continuous random variables (e.g., height, time, measurement error), we use a Probability Density Function (PDF), f(x). The probability of any single exact value is zero; we calculate probabilities over intervals. The rules adapt slightly:

1. Non-Negativity: f(x) ≥ 0 for all x in the domain.
2. Total Area Equals One: The integral of f(x) over its entire support (the interval where it’s defined) must equal 1. That is, ∫ f(x) dx = 1 over the support.

Example 5: Valid Continuous Distribution (Uniform on [0,1]) f(x) = 1 for 0 ≤ x ≤ 1, and f(x) = 0 otherwise.

• Rule 1: f(x) is 1 or 0, both ≥ 0. **

Example 5 (continued)
The integral of f(x) over its support is computed as
[ \int_{0}^{1} 1 ,dx = 1-0 = 1, ] so the total area equals 1, satisfying the second requirement. Hence this uniform density is a legitimate probability model.

Example 6: A Valid Exponential Model
Consider a random variable Y that measures the waiting time (in minutes) for the next bus, with density
[ g(y)=\begin{cases} \lambda e^{-\lambda y}, & y\ge 0,\[4pt] 0, & y<0, \end{cases} ]
where λ = 0.2.
Non‑negativity holds because the exponential term is always positive and the multiplier λ is positive. Total area is verified by integration:
[ \int_{0}^{\infty}\lambda e^{-\lambda y},dy = \Big[-e^{-\lambda y}\Big]_{0}^{\infty}=1. ]
Both conditions are met, so this is a proper continuous distribution.

Example 7: Failure Because the Integral Falls Short
Suppose a candidate density is defined as
[ h(x)=\begin{cases} 0.8, & 0\le x\le 1,\ 0, & \text{otherwise}. \end{cases} ]
The integral over its domain equals 0.8 × 1 = 0.8, which is less than 1. Consequently, the missing 0.2 of probability mass cannot be assigned to any interval, leaving the model incomplete. The distribution is therefore invalid.

Example 8: Violation of Non‑Negativity in a Piecewise Form
Let
[ k(x)=\begin{cases} 2-x, & 0\le x\le 2,\ 0, & \text{elsewhere}. \end{cases} ] While the integral of k(x) from 0 to 2 does equal 1, the function becomes negative when x>2 or when extending the definition beyond the intended interval (e.g., if one mistakenly allows x up to 3, then k(3) = -1). Any negative value breaches the non‑negativity rule, rendering the entire specification unacceptable as a probability density.

Summary of Key Takeaways

• For discrete models, every point‑mass must be a non‑negative number and the collection of all masses must add up to exactly one.
• For continuous models, the height of the density curve must never dip below zero, and the total area under the curve across its entire support must be one. - A common source of error is overlooking the need to normalize a formula; scaling a function so that its integral equals one is often the final step in constructing a valid model.
• Even when the sum or integral happens to be one, any negative component instantly disqualifies the candidate, because probabilities cannot be sub‑zero.
• Recognizing these two constraints allows practitioners to quickly validate candidate distributions and to correct them—by renormalizing, truncating, or discarding—before using them for inference or simulation.

By consistently applying these checks, one can ensure that any proposed probability model—whether discrete or continuous—adheres to the foundational axioms of probability theory.