Which Of The Following Are Valid Probability Distributions

Introduction

When you encounter a list of functions or tables and wonder which of the following are valid probability distributions, the answer lies in a handful of fundamental rules that every distribution must satisfy. Whether you are dealing with discrete random variables, continuous densities, or mixed cases, the same core criteria apply: non‑negativity, proper normalization, and, when required, adherence to the support set. This article walks you through those criteria, examines common pitfalls, and evaluates several typical examples to illustrate how to determine the validity of a probability distribution. By the end, you will be able to assess any candidate function with confidence and avoid the subtle mistakes that often trip up students and professionals alike.

What Makes a Function a Probability Distribution?

1. Non‑negativity

For any random variable (X), the probability assigned to any outcome must be greater than or equal to zero:

• Discrete case: (P(X = x_i) \ge 0) for every possible value (x_i).
• Continuous case: The probability density function (pdf) (f(x) \ge 0) for all (x) in its domain.

If a candidate function yields a negative value for any point in its support, it is automatically disqualified It's one of those things that adds up..

2. Normalization (Total Probability Equals One)

The probabilities must sum (or integrate) to one:

• Discrete: (\displaystyle \sum_{i} P(X = x_i) = 1).
• Continuous: (\displaystyle \int_{-\infty}^{\infty} f(x),dx = 1).

Normalization guarantees that the distribution accounts for all possible outcomes of the experiment.

3. Proper Support

The support of a distribution is the set of values where the probability (or density) is non‑zero. A valid distribution must explicitly define this support and respect it:

• If the support is finite, the function must be zero outside that range.
• For infinite support (e.g., the normal distribution), the integral over the entire real line must still equal one.

4. Measurability and Integrability (Advanced)

In rigorous probability theory, the function must be measurable with respect to the underlying sigma‑algebra, and the integral used for normalization must be finite. For most introductory contexts, ensuring non‑negativity and proper normalization suffices.

Step‑by‑Step Checklist

When presented with a list of candidates, follow this systematic checklist:

1. Identify the type (discrete vs. continuous).
2. Write down the support explicitly.
3. Test non‑negativity on the entire support.
4. Compute the total probability (sum or integral).
5. Verify that the total equals one (within acceptable tolerance for numerical examples).
6. Check for hidden constraints (e.g., parameters that must be positive).

If any step fails, the candidate is not a valid probability distribution.

Common Examples and Their Evaluation

Below are several frequently encountered candidates. For each, we apply the checklist and explain why it passes or fails.

Example 1 – Discrete Distribution

[ P(X = k) = \frac{2k}{n(n+1)}, \qquad k = 1,2,\dots ,n ]

1. Support: ({1,2,\dots ,n}) – clearly defined.
2. Non‑negativity: (2k \ge 0) for all (k\ge 1); denominator is positive, so the fraction is non‑negative.
3. Normalization:

[ \sum_{k=1}^{n} \frac{2k}{n(n+1)} = \frac{2}{n(n+1)}\sum_{k=1}^{n} k = \frac{2}{n(n+1)}\cdot\frac{n(n+1)}{2}=1. ]

All criteria are satisfied; this is a valid discrete probability distribution (in fact, it is the distribution of the rank of a uniformly random permutation) Simple, but easy to overlook. Worth knowing..

Example 2 – Continuous Density with a Parameter

[ f(x)=\begin{cases} \lambda e^{-\lambda x}, & x\ge 0,\[4pt] 0, & x<0, \end{cases} \qquad \lambda>0. ]

1. Support: ([0,\infty)).
2. Non‑negativity: Exponential function is always positive; multiplied by (\lambda>0) remains positive.
3. Normalization:

[ \int_{0}^{\infty} \lambda e^{-\lambda x},dx = \lambda\Big[-\frac{1}{\lambda}e^{-\lambda x}\Big]_{0}^{\infty}=1. ]

Thus, the exponential pdf is a valid continuous probability distribution for any (\lambda>0) And that's really what it comes down to..

Example 3 – Improperly Normalized Function

[ g(x)=\frac{1}{1+x^2}, \qquad -\infty < x < \infty. ]

1. Support: Entire real line.
2. Non‑negativity: Denominator is always positive, so (g(x)\ge 0).
3. Normalization:

[ \int_{-\infty}^{\infty} \frac{1}{1+x^2},dx = \pi \neq 1. ]

Because the integral equals (\pi), not 1, (g(x)) is not a probability density. It can be turned into one by dividing by (\pi), yielding the standard Cauchy density.

Example 4 – Negative Values in a Discrete Table

[ P(X = k) = \frac{(-1)^k}{2}, \qquad k = 1,2. ]

1. Support: ({1,2}).
2. Non‑negativity:

• For (k=1): ((-1)^1/2 = -0.5) (negative).

Since a probability cannot be negative, this fails the non‑negativity test and is invalid.

Example 5 – Piecewise Continuous Function with Gaps

[ h(x)=\begin{cases} \frac{3}{4}x, & 0\le x < 1,\[4pt] \frac{3}{4}(2-x), & 1\le x \le 2,\[4pt] 0, & \text{otherwise}. \end{cases} ]

1. Support: ([0,2]).
2. Non‑negativity: Both pieces are linear with positive slope or decreasing but stay non‑negative on the given intervals.
3. Normalization:

[ \int_{0}^{1}\frac{3}{4}x,dx + \int_{1}^{2}\frac{3}{4}(2-x),dx = \frac{3}{4}\Big[\frac{x^{2}}{2}\Big]{0}^{1} + \frac{3}{4}\Big[2x-\frac{x^{2}}{2}\Big]{1}^{2} = \frac{3}{8} + \frac{3}{8}= \frac{3}{4}. ]

The total integral is (0.Think about it: 75), not 1. Multiplying the whole function by (\frac{4}{3}) would fix it, but as written (h(x)) is not a valid pdf.

Example 6 – Mixed Distribution (Discrete + Continuous)

[ P(X=0)=0.3,\qquad f(x)=0.7,e^{-x},; x>0. ]

1. Support: ({0}\cup (0,\infty)).
2. Non‑negativity: Both components are non‑negative.
3. Normalization:

[ P(X=0) + \int_{0}^{\infty}0.7e^{-x},dx = 0.Consider this: 3 + 0. 7 = 1.

All conditions are met, so this mixed distribution is valid. Mixed cases often appear in reliability engineering (probability of immediate failure vs. later failure) And that's really what it comes down to..

Frequently Asked Questions

Q1: Can a function that integrates to a value greater than 1 be salvaged?

Yes. Consider this: if the integral (or sum) equals a constant (c>0), dividing the entire function by (c) yields a properly normalized distribution. This is the principle behind constructing probability density functions from unnormalized kernels (e.Also, g. , the Cauchy kernel divided by (\pi)) Worth keeping that in mind..

Q2: What if a discrete distribution has infinitely many points?

The same normalization rule applies:

[ \sum_{i=1}^{\infty} p_i = 1. ]

A classic example is the geometric distribution (p_i = (1-p)^{i-1}p) for (i=1,2,\dots), whose infinite sum equals 1 for any (0<p\le1).

Q3: Do probability distributions have to be continuous or differentiable?

No. Continuity is required only for continuous densities; discrete pmfs are inherently discontinuous. Practically speaking, differentiability is not a requirement; many valid pdfs (e. In practice, g. , the uniform distribution on ([0,1])) are not differentiable at the endpoints Simple, but easy to overlook..

Q4: How do I handle distributions defined only on a subset of the real line, like ([a,b])?

Specify the support explicitly and ensure the integral over ([a,b]) equals 1. To give you an idea, the beta distribution is defined on ([0,1]) with density

[ f(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}, ]

where (B(\alpha,\beta)) is the beta function that guarantees normalization.

Q5: Is a negative probability ever acceptable in quantum mechanics or other advanced fields?

In standard probability theory, no. Quantum mechanics uses probability amplitudes that can be negative or complex, but observable probabilities are always the squared magnitudes, which are non‑negative. So, when translating to a classical probability distribution, negativity is still prohibited.

Practical Tips for Quick Validation

• Plug in boundary values: For piecewise definitions, evaluate the function at the edges of each interval.
• Use symmetry: If a function is symmetric around a point, you can often halve the integral and double the result, saving time.
• Check parameters: Many distributions contain parameters (e.g., (\lambda) in the exponential). Ensure the parameter constraints (positivity, integer, etc.) are stated and respected.
• take advantage of known families: If the function resembles a known distribution up to a constant, compare with the standard form to spot missing normalizing factors.
• Employ software for messy integrals: When the integral is not elementary, use symbolic tools (e.g., Mathematica, SymPy) to verify that it equals 1.

Conclusion

Determining which of the following are valid probability distributions boils down to three core principles: non‑negativity, normalization, and proper support. By methodically applying these rules, you can evaluate discrete tables, continuous densities, and mixed cases with confidence. Still, remember that a function that fails any of the criteria can often be rescued by adjusting a constant or redefining its support, but the essential properties must never be compromised. Mastering this checklist not only prepares you for exams and research but also equips you with a reliable mental framework for any probability‑related problem you encounter.