Probability feels intuitive until you try to make a single sentence precise. The fastest way to learn what the definitions really mean is to watch one plausible inference fail in a clean, controlled setting. Counterexamples do not just refute; they expose the boundary of a concept, and they teach you how to reason without importing hidden assumptions.
A perfect example is the widespread belief that “uncorrelated” is basically the same as “independent.” In applications, people often check a covariance, see a zero, and conclude that there is no relationship. The counterexample below shows exactly what is true, what is not, and how to replace the false inference with a correct toolbox.
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The setup: a very simple probability space
Let $X$ be uniformly distributed on $[-1,1]$. Concretely, take the probability space $([-1,1],\mathcal{B},\mathbb{P})$ where $\mathcal{B}$ is the Borel $\sigma$-algebra and $\mathbb{P}$ is Lebesgue measure normalized so $\mathbb{P}([-1,1])=1$. Define the random variable
Now define a second random variable by
This is almost too simple, which is why it is such a good teacher.
A few basic facts are immediate:
- $Y$ is measurable because it is a continuous function of $X$.
- $Y$ takes values in $[0,1]$.
- Knowing $X$ determines $Y$, since $Y$ is a function of $X$.
Already you should feel tension: if one variable is a function of the other, how could they be “unrelated”? The answer is that the usual “unrelatedness test” based on covariance measures only linear dependence. The counterexample makes that statement exact.
Compute the covariance and see it vanish
First compute the means. By symmetry,
Next,
Now compute $\mathbb{E}[XY]$:
But the integrand $x^3$ is odd and the density is symmetric, so the integral is $0$. Therefore
So $X$ and $Y$ are uncorrelated.
At this point many people would say: “Great, the variables do not influence each other.” That inference is false.
Show they are not independent
Independence means that for all Borel sets $A,B$,
Because $Y=X^2$, the event $\{Y\le 1/4\}$ is exactly the same as $\{|X|\le 1/2\}$. Choose
Then
So
On the other hand,
If $X$ and $Y$ were independent, we would have
but the true value is $1/2$. Therefore $X$ and $Y$ are not independent.
This is the whole counterexample: uncorrelated does not imply independent.
What the counterexample is really saying
The key lesson is not a slogan; it is a structural fact about what covariance can see.
The covariance
measures whether $Y$ has a linear trend in $X$ after centering. In this example, the dependence of $Y$ on $X$ is quadratic, and symmetry cancels the linear term. Covariance is blind to that.
A more informative way to state the boundary is:
- Independence is a statement about products of events and factors of $\sigma$-algebras.
- Uncorrelatedness is a statement about one specific bilinear functional, the covariance.
Uncorrelatedness is weaker because it tests one moment identity, while independence tests an entire algebra of identities.
The $\sigma$-algebra view: dependence is about information
A good replacement for the false inference is to interpret dependence as shared information. Define
the $\sigma$-algebras generated by $X$ and $Y$. Independence of $X$ and $Y$ is equivalent to independence of $\mathcal{F}_X$ and $\mathcal{F}_Y$: every event determined by $X$ is independent of every event determined by $Y$.
In the present example, $\mathcal{F}_Y\subseteq \mathcal{F}_X$ because $Y$ is a function of $X$. That inclusion means: whatever you can learn from $Y$, you can learn from $X$. In particular, the events $\{Y\le t\}$ are events about $|X|$, and they are certainly not independent of events that also involve $|X|$.
That is the sharp contrast:
- Covariance $=0$ tells you a specific centered linear correlation vanishes.
- $\mathcal{F}_Y\subseteq \mathcal{F}_X$ tells you the variables are tied together by a deterministic constraint.
The counterexample is successful because it makes both facts simultaneously visible.
Conditional expectation makes the dependence quantitative
If you want a numeric test of “how much $Y$ depends on $X$,” conditional expectation is the right object.
Because $Y=X^2$ is a function of $X$,
So the conditional mean of $Y$ given $X$ is not constant; it varies with $X$. That is dependence in a very strong sense.
By contrast, if $X$ and $Y$ were independent, we would have
So a precise, checkable substitute for the false inference is:
- If $\mathbb{E}[Y\mid X]$ is almost surely constant, then $Y$ has no mean dependence on $X$.
- If $\mathbb{E}[Y\mid X]$ varies, dependence is present, even if covariance is zero.
In the counterexample, $\mathbb{E}[Y\mid X]$ varies maximally because it equals $Y$.
A “moment upgrade” that really does imply independence
You might ask: if covariance is too weak, what can you check instead?
For Gaussian random vectors, uncorrelated does imply independent. The reason is not magic; it is structure. The joint law of a multivariate Gaussian is determined completely by its mean vector and covariance matrix. So if the covariance matrix is block-diagonal, the joint density factors, and independence follows.
The correct way to carry that lesson is:
- In special families of distributions, low-order moments can determine the entire law.
- Outside those families, moments can miss nonlinear dependence.
This distinction explains why “uncorrelated implies independent” appears true in some data workflows: those workflows assume a near-Gaussian model, often implicitly.
How to build your own counterexamples
This example is the simplest member of a larger pattern.
If you want uncorrelated but dependent variables, a reliable method is:
- Start with a symmetric $X$ with $\mathbb{E}[X]=0$.
- Let $Y=g(X)$ where $g$ is an even function, so $Xg(X)$ is odd and $\mathbb{E}[Xg(X)]=0$.
- Choose $g$ nonconstant so that $Y$ genuinely depends on $X$.
For instance, $Y=X^2$, $Y=|X|$, or $Y=\mathbf{1}_{\{|X|>1/2\}}$ all work with symmetric $X$. The symmetry forces zero covariance, while the functional relation forces dependence.
This method teaches a deeper habit: when you see a cancellation, ask whether it is structural (true independence) or geometric (symmetry).
What this changes in practice
The counterexample has practical consequences in modeling, statistics, and even pure probability.
Correlation tests are not dependence tests
If you test dependence by correlation alone, you can miss:
- Quadratic or higher-order relationships
- Threshold effects (indicator functions)
- Mixtures where two regimes cancel linearly
The right fix depends on your goal. If you need full independence, you must test or justify it structurally. If you only need linear decorrelation, covariance is appropriate, but you should say that explicitly.
The right tool depends on what you need to control
Different tasks demand different notions.
| Goal | Notion that matches | Typical tool |
|—|—|—|
| Control linear prediction | Uncorrelatedness | covariance, least squares |
| Control mean dependence | $\mathbb{E}[Y\mid X]$ constant | conditional expectation |
| Control event factoring | Independence | product $\sigma$-algebras |
| Control tail interaction | Weak dependence | mixing, coupling, concentration |
The counterexample forces you to pick your notion rather than rely on a vague word like “unrelated.”
A short takeaway that is actually correct
The result is not “correlation is useless.” The correct takeaway is:
- Correlation is a measurement of linear alignment in $L^2$.
- Independence is a measurement of factorization of information.
When you move from one to the other without a theorem, you are smuggling in a model class.
If you remember only one thing, let it be this: a zero covariance can be caused by symmetry, not by separation. The counterexample $Y=X^2$ with symmetric $X$ is small enough to hold in your head, yet rich enough to calibrate your reasoning whenever probability starts to feel too intuitive.
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