0
$\begingroup$

The correlation coefficient is given by

$$\rho_{XY}=\frac{R_{XY}-\mu_X \, \mu_Y}{\sigma_X \, \sigma_Y}$$

If the product $\mu_X \, \mu_Y \neq 0$ and $\rho_{XY}\neq 0$, then we can have two cases:

  1. $R_{XY}= 0$ when $X$ and $Y$ are orthogonal;

  2. $R_{XY}\neq 0$ when $X$ and $Y$ are not orthogonal to each other;

So I see from Case $1$ that $R_{XY}= 0$ is possible to have when $\rho_{XY}\neq 0$. Is my reasoning correct? Is this not counterintuitive?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

Yes, it is possible.

Following this very good paper on the topic, consider two random variables $X$ and $Y$ with the following realizations:

$$X = (1, -5, 3, -1), Y = (5, 1, 1, 3)$$

You have:

$$R_{XY} = E(XY') = E(1\times5 - 5\times1 + 3\times1 - 1\times3) = 0$$

but:

$$\mu_{X} = -\frac{1}{2}, \mu_{Y} = \frac{5}{2} \Rightarrow \rho_{X,Y} \neq 0$$

In general, recall that while both orthogonality and uncorrelation imply linear independence, there's no implication between orthogonality and uncorrelation themselves.

EDITING TO ANSWER COMMENT:

When it comes to stochastic processes, following your notation, we say that $(X_{t})_{t \geq 1}$ $(Y_{t})_{t \geq 1}$ are uncorrelated if:

$$\forall t_{1}, t_{2}, COV_{X,Y}(t_{1}, t_{2}) = R_{XY}(t_{1}, t_{2}) - \mu_{X}(t_{1})\mu_{Y}(t_{2}) = 0 $$

while we say that they're orthogonal if:

$$\forall t_{1}, t_{2}, R_{XY}(t_{1}, t_{2}) = E[X(t_{1})Y(t_{2})'] = 0$$

so the same reasoning as before applies. Here for a broader analysis.

$\endgroup$
6
  • $\begingroup$ Great! Also $\sigma_x$ and $\sigma_y$ are non-zero here. When you are calculating $R_{XY}$, you are doing it at lag zero i.e. no shift between X and Y. Is there a special reason. Perhaps a similar reasoning as evaluating an autocovariance which gives the variance? $\endgroup$
    – macy
    Commented Mar 29, 2019 at 5:26
  • $\begingroup$ See edited answer. $\endgroup$
    – Nicg
    Commented Mar 29, 2019 at 10:32
  • $\begingroup$ In your example, why is $R_{XY}$ computed at lag $0$? $\endgroup$
    – macy
    Commented Mar 29, 2019 at 11:58
  • $\begingroup$ Maybe I am missing what you mean by lag, but between $t_{2}$ and $t_{1}$ there's actually a one period lag. Perhaps it helps to see it using a single time series: Let $(X_{t})_{t \geq 0}$ be a time series. Then $$COV(X_{t + h}, X_{t}) = E[X_{t + h}X_{t}] - E(X_{t + h})E(X_{t})$$ is the autocovariance function, with $h$ being the lag, $1$ in our case. Hope this clarifies. $\endgroup$
    – Nicg
    Commented Mar 29, 2019 at 13:44
  • $\begingroup$ By lag, I mean the shift in one sequence relative to the other one when doing the correlation. I say lag or shift zero because you take the first sample of $X$ and multiply it by the first sample of $Y$ in your example etc. i.e. you are doing $X_1Y_1+X_2Y_2+X_3Y_3+ X_4Y_4$. So your $h$ in your explanation is $0$. For instance the first product is $X_{1+0}X_1$ with your $h$ as $0$. Your $h$ is what I called lag or shift. In effect you are computing $R_{XX}(0)$ with the argument $h=0$. We can have $R_{XX}(1)$, $R_{XX}(2)$ and $R_{XX}(3)$ i.e. by shifting one sequence by $1$, $2$ or $3$ unit. $\endgroup$
    – macy
    Commented Mar 30, 2019 at 11:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .