Although independence implies zero correlation, zero correlation does not necessarily imply independence.
While I understand the concept, I can't imagine a real world situation with zero correlation that did not also have independence.
Can someone please give me an example so I can better understand this phenomenon?
Consider the following betting game.
Flip a fair coin to determine the amount of your bet: if heads, you bet \$1, if tails you bet \$2. Then flip again: if heads, you win the amount of your bet, if tails, you lose it. (For example, if you flip heads and then tails, you lose \$1; if you flip tails and then heads you win \$2.) Let $X$ be the amount you bet, and let $Y$ be your net winnings (negative if you lost).
$X$ and $Y$ have zero correlation. You can compute this explicitly, but it's basically the fact that you are playing a fair game no matter how much you bet. But they are not independent; indeed, if you know $Y$, then you know $X$ (if $Y = -2$, for instance, then $X$ has to be 2.) Explicitly, the probability that $Y=-2$ is $1/4$, and the probability that $X=2$ is $1/2$, but the probability that both occur is $1/4$, not $1/8$. (Indeed, in this game, there is no event with probability $1/8$.)
Zero correlation will indicate no linear dependency, however won't capture non-linearity. Typical example is uniform random variable $x,$ and $x^2$ over [-1,1] with zero mean. Correlation is zero but clearly not independent.
I will give a geometric example involving random points in the plane. These come up in real life all the time if there is a mechanism by which points are distributed. (For example, it could be the location of a house or something)
Choose a random point $(X,Y)$ in the plane chosen uniformly from the unit circle $x^2 + y^2 = 1$ (by this I mean, the probability of $(X,Y)$ being contained in an arc of the circle is proportional to the length of the arc...you could also choose $\theta$ uniformly distributed in $[0,2\pi)$ and put $X=\cos(\theta), Y=\sin (\theta)$)
Now, the random variables $X$ and $Y$ are uncorrelated. Indeed, for any given value of $X=x$ there are always exactly two possible values of $Y$ that fit, namely $+\sqrt{1-x^2}$ and $-\sqrt{1-x^2}$. These are equally likely so both have probability $\frac{1}{2}$. Hence $E(XY|X=x) = \frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}x (-\sqrt{1-x^2})=0$. From here, you should be able to see that they are uncorrelated.
However these are not independenet! There are many ways to see why. Here is one "certificate" that shows they are not independent. (Although this doesn't really clear up the intuition of why they arent independent, you will have to think about that one). Notice $P(X>\frac{\sqrt{2}}{2}, Y>\frac{\sqrt{2}}{2})=0$ since $X^2+Y^2=1$ always. However, each probability $P(X>\frac{\sqrt{2}}{2})$ and $P(Y>\frac{\sqrt{2}}{2})$ are non-zero, so it is impossible that $P(X>\frac{\sqrt{2}}{2}, Y>\frac{\sqrt{2}}{2})=P(X>\frac{\sqrt{2}}{2})P(Y>\frac{\sqrt{2}}{2})$
Let $X$ be any random variable. Let $P\{I = 1\} = P\{I = -1\} = 1/2$, with $I$ independent of $X$. Let $Y = IX$. (Thus, $Y = \pm X$, each with probability $1/2$, independent of the value of $X$.) Then $X$ and $Y$ are uncorrelated but not independent. We could replace $I$ by any zero-mean random variable independent of $X$. [Could someone please tell me how to insert that first equation correctly?]
\{ and \}. I edited them in for you. But I think most people write parentheses instead: $P(I = 1)$ etc.
– Nate Eldredge
Jul 15 '13 at 19:38
Think about this problem visually. If the joint distribution is symmetric to the origin, then the correlation could be $0$. But the joint distribution may not be uniform in the domain. e.g. $$p(x,y) = \frac{|x| + |y|}{4}, \text{ for }x,y \in [-1,1].$$ Then the marginal densities are $p(x) = |x|/2 + 1/4$, and $p(y) = |y|/2 + 1/4$. Obviously, these two variables are not independent since they tend to appear at the corners. Strictly speaking, $p(x,y) \neq p(x)p(y)$. Brighter color means higher probability.
Next, it is obvious that the correlation is $0$ as $\text{Cov}(X,Y)= E(XY) -EX\cdot EY = 0$.
Another example is $p(x,y) = 1/2$, for $|x|+|y| < 1$, which is a uniform distribution over the diamond domain. Then $p(x) = 1 - |x|$, and $p(y) = 1 - |y|$. The reasoning is similar to the above example.
Consider these two physical variables:
Let's say velocity takes values between $-50$ and $+50$ miles an hour with equal probability, average velocity $0$. When velocity is $-50$ kinetic energy is $1250m$ and when velocity is $+50$ kinetic energy is also $1250m$. Because the mean velocity is zero and all velocities are equally likely, the correlation is simply proportional to the sum of the products of velocity and kinetic energy (an integral rather than a sum, in fact, because velocity is continuous). The product $KV$ when $V=-50$ is $-62500m$, and the product $KV$ when $V=50$ is $62500m$, and these terms cancel each other out in the sum. And because negative velocities occur just as much as positive velocities, the sum is composed of equal and opposite pairs, which cancel out, and the correlation is zero.
