0
$\begingroup$

This question is slightly unusual because I've obtained the answer but I am looking for an interpretation of it as it's not intuitive. If we have $X,Y : Uniform(0,1)$ and independent, let $Z=max(X,Y) , W=min(X,Y)$. Find $Cov(Z,W)$. The answer can be found here (or here), and it is $Cov(Z,W)=\frac{1}{36}$. I went ahead and computed the correlation coefficient and it's $\rho(Z,W)=0.5$.

If we take a step back, this result is not intuitive at all: if we think of sampling intervals from $[0,1]$ then $(W,Z)$ represent the lower bound and upper bound, respectively. As we get more and more data, the overall lower bound will keep moving to 0, while the overall upper bound will keep moving towards 1, and there's no reason to assume that high values of $W$ will be associated with high values of $Z$. Therefore, my sampling argument is wrong (since the correlation is positive), but I don't see how.

Any help or clarification would be greatly appreciated, especially a geometric one.

$\endgroup$
2
$\begingroup$

It is true that if you observe $U_1,\cdots,U_n\in[0,1]$, then $\min(U_1,\cdots,U_n)$ converges to $0$ when you get more and more date, that is $n\to+\infty$, and $\max(U_1,\cdots,U_n)$ converges to $1$.

However, here you look at $Z=\max(X,Y)$ and $W=\min(X,Y)$, which corresponds in a situation in which only two observations $X$ and $Y$ on the intervals $[0,1]$ are made. So there is no connection to make with the idea of getting more and more data because you are in the context of two observations only (no more, no less).

Let us come back to the interpretation of $\rho$: it can be seen as a measure of the linear dependence between two random variables. When $\rho(Z,W)$ is positive (resp. negative), we expect that $Z$ takes higher values (resp. lower values) when $W$ takes higher values.

So the question here is: when $W$ is high, do we expect high values or low values of $Z$. Well, given that $W\le Z$, I would find it hard to believe that $Z$ can be low when $W$ is high. That is not in favour of a negative correlation.

Imagine two experiments in which you observe two values in the intervals $[0,1]$. In the first experiment, I tell you that the lowest value is $0.2$. In the second experiment, I tell you that the lowest value is $0.8$. In both cases, I ask you to guess what is the highest value. In the first case, the lowest value is $0.2$, so the highest value is between $0.2$ and $1$. If someone had to guess what the highest value is, I am sure he would say something like $0.6$ (the middle of $0.2$ and $1$). In the second case however, we are told that $0.8$ is the lowest value, so the highest value has no chance at all to be something like $0.6$: it has to be somewhere between $0.8$ and $1$, so if you had to guess, I think you would say something like $0.9$. What we just illustrated is that the higher $W$ is, the higher $Z$ is likely to be. Therefore, we expect a positive correlation.

$\endgroup$
1
+50
$\begingroup$

Here is a geometric explanation. See if it makes sense to you.

$(X,Y)$ is a uniform random point in the square $[0,1] \times [0,1]$.

But what is $(W,Z) \equiv (\min(X,Y), \max(X,Y))$?

Imagine folding the square along the $y=x$ line, aka $+45°$ line, folding the lower triangle onto the upper triangle. Then $(W,Z)$ is where $(X,Y)$ would end up:

  • If $Y>X$, the original point is in the upper triangle to begin with, and remains there (folding has no effect). In this case $(W,Z) = (X,Y)$.

  • If $X > Y$, the original point is in the lower triangle to begin with, and ends up in the upper triangle at $(Y,X)$ (that's the effect of folding). In this case $(W,Z) = (Y,X)$.

By symmetry, therefore, sampling $(X,Y)$ uniformly in the square and then computing $W=\min(X,Y)$ and $Z=\max(X,Y)$ is equivalent to sampling $(W,Z)$ uniformly from just the upper triangle.

I hope it is now obvious, because a triangle is not a square, that $W, Z$ will be dependent.

As for the sign of the correlation: Very roughly speaking positive correlation means the area is "biased / slanted with a positive slope" and negative correlation means the area is "biased / slanted with a negative slope". I hope it is also obvious that the upper triangle implies a positive correlation.

(Incidentally, as you probably know, dependence can still mean zero correlation, and some examples would be uniformly sampling from e.g. a circle, a $45°$ rotated square, etc.)

Sorry the geometric argument is kinda vague, but I thought this kind of "intuitive" picture is what your question is asking (since you already know the answer is $1/36$ and just need an explanation which is not the actual calculation of covariance). If you want a slightly less "visual" reason, simply note that, conditioned on $W=w, Z \sim Uniform(w,1)$, so as $w$ increases $Z$ certainly tends to also increase.

Re: your multiple samples argument: that doesn't quite apply. If you sample many $(W,Z)$ in the upper triangle, then sure $\min_j W_j \rightarrow 0$ and $\max_j Z_j \rightarrow 1$, but so what? That doesn't say anything about how in one particular sample, knowing $W$ gives you info about $Z$ and vice versa. (In fact, just for fun, it is also true that $\max_j W_j \rightarrow 1$ and $\min_j Z_j \rightarrow 0$, so what do you make of that? ;) )

$\endgroup$
  • $\begingroup$ What do you mean by this : "positive correlation means the area is "biased / slanted with a positive slope" and negative correlation means the area is "biased / slanted with a negative slope"" Is there a way to quantify this property of an area with some sort of formula? $\endgroup$ – baibo Mar 20 at 20:29
  • 1
    $\begingroup$ Quantifying this would probably be equivalent to calculating the covariance and correlation, and I presume that's not the kind of explanation you want. See if you find this helpful: check out the second picture ("Several sets of $(x,y)$ points...") at en.wikipedia.org/wiki/Pearson_correlation_coefficient -- especially the first row with the various ellipse-like blobs. What I'm saying is the upper triangle is kinda "similar" to an ellipse that is tilted with a positive slope. $\endgroup$ – antkam Mar 20 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.