I found an example that shows how two uncorrelated random variables can be dependent: a normally distributed variable $X$ is not correlated with its square $Y=X^2$. What can be $X$ and what can be $Y$ (in finance terms) so that they represent a shape close to a parabola when plotted in $(x,y)$ plane (both branches present)? This would give 0 correlation, but not independence. Is there such an example?

  • $\begingroup$ Not sure I understand. Why would you give a reference to a textbook on portfolio theory? $\endgroup$
    – Qwerty
    Commented Nov 1, 2020 at 3:07
  • $\begingroup$ You're right, that book was too hard. Here is a suppy and demand curve example from a high school economics course. oocities.org/vuumanj/BusinessAlgebra/Quadratic.html $\endgroup$ Commented Nov 1, 2020 at 3:24
  • $\begingroup$ What about an asset‘s return and it’s squared return? $\endgroup$ Commented Nov 1, 2020 at 6:04

3 Answers 3


The simplest example might be Y= realized variance of a stock and X= return on the stock. Clearly these are dependent since they are both calculated from daily stock prices. X can be positive or negative , but Y is always positive. If large moves in the stock occur (up or down) , we would expect to measure high realized volatility. This might give a close to zero correlation for X and Y.


A correlation and a dependence cannot be interchanged. The dependence is more general term that two radnom variables are somehow linked. The correlation concerns linear dependence only. So, in your example variables $X$ and $Y$ are dependent because $Y=X^2$. As you pointed out, this is a quadratic dependency, not linear, hence there is no correlation.

A general measure for normally distributed random variables measuring how much are two variables linked is called covariance and it is defined as $$ \text{cov}(X,Y)=\text{E}\{[X-\text{E}(X)][Y-\text{E}(Y)]\}, $$ where $\text{E}(.)$ means expected value.

Here are some other measures of dependence.

  • 3
    $\begingroup$ Covariance is not a general measure for how much random variables are linked. Basically, covariance is just the scaled (Pearson) correlation coefficient. A standard example (you mention it yourself): $\mathrm{cov}(X,X^2)=0$ if $X\sim\mathrm{Uni}(-1,1)$, see here. Zero covariance implies $X$ and $Y$ are uncorrelated but not necessarily independent (unless they are e.g. normal). $\endgroup$
    – Kevin
    Commented Nov 1, 2020 at 7:44
  • $\begingroup$ @Kevin: Thank for pointing that out. I edited my answer. $\endgroup$ Commented Nov 2, 2020 at 10:54

I guess there are examples in options trading, whenever things depend on Gamma (which is essentially a squared term). For instance, delta hedging: the strategy is, in the textbook version, long the option and short delta times the underlier. If you follow the changes in profit/loss over time and plot them against changes in the underlier, you can often see a u-shaped curve.

An example (R-code):

steps <- 100

## simulate a path of the underlier
S <- gbm(npaths = 1, timesteps = steps,
         S0 = 100, v = 0.3^2, tau = 1, r = 0)

## compute option value + delta
option <- vanillaOptionEuropean(S = S,
                                X = 100,
                                tau = seq(1, 0.1, length.out = steps + 1),
                                r = 0,
                             v = 0.3^2)
plot(diff(S), -diff(S) * option$delta[-length(option$delta)] +
     xlab = "Change in S", ylab = "PL of delta-hedged position")

PL of delta-hedged position

  • $\begingroup$ Your plot does not seem correct to me. I would expect that if you're making money on the large moves, you should be losing money on small moves (in your case, <~3 change in S) from the theta of the option. $\endgroup$
    – will
    Commented Nov 1, 2020 at 13:26
  • $\begingroup$ But it is losing money on the small moves: the points "in the middle" are all below zero. Note that the PL is computed along a path, so delta/gamma changes as a result of the simulated spot price. $\endgroup$ Commented Nov 1, 2020 at 14:48
  • $\begingroup$ haha. My bad. I didn't even read the scale and just assumed that zero was the bottom of the plot / that there'd be an axis on the plot. Where it crosses through zero lines up with expectation too, appologies. $\endgroup$
    – will
    Commented Nov 1, 2020 at 16:49

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