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I need a bit of help proving the following:

$$Cor(X,Y) = \pm 1 \Leftrightarrow Y = a +BX$$ with probability 1, for some constants a and b. Here $Cor(X,Y)$ is the Correlation of $X$ and $Y$.

I can manage to prove one way, that is if $Y = a +bX$, then $Cor(X,Y) = \pm1$, however I'm having trouble proving the opposite way. Any help is appreciated.

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    Can you tell us about $X$ and $Y$? More precisely, are you hoping for a proof where $X$ and $Y$ are random variables on an arbitrary probability space? Or are they random variables on some nice finite probability space like "the set of possibilities when I flip three coins"? – 2017-01-08
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    @JohnHughes Why do you ask? Unless specified otherwise, one is "hoping for a proof where X and Y are random variables on an arbitrary probability space"... – 2017-01-08
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    Hint: Assume without loss of generality that $E(X)=E(Y)=0$ and that the correlation is $+1$, then this corresponds to the equality case in Cauchy-Schwarz inequality. Thus... – 2017-01-08
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    @Did: you might well be right. But given the nature of the question, it's possible that OP is just starting probability theory, and has not even heard of a probability space (think "is taking a first discrete-math course" rather than "already knows a good deal of measure theory"). In that case, mentioning Cauchy-Schwartz may not be helpful (indeed, OP may not know C-S). I was trying to gauge the OP's level of sophistication so that I could give an answer appropriate to that level. But please, go ahead and take over: you'll have an easier time providing an answer than I will. – 2017-01-08
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    @JohnHughes Re-reading your comment, what is mysterious to me in your queries is that I know no proof of the desired property for "random variables on (every) nice finite probability space" that does not work for "random variables on (every) probability space" as well. Anyway, as you say, it might be better to wait for the OP's input now... – 2017-01-08
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    Well...one proof will only involve finite sums. Of course, it's readily generalizable ... but the generalization is informative only if you know what all the terms mean, while the finite-sum version should be comprehensible to anyone who knows enough algebra. (It's possible, albeit unlikely, that OP doesn't even know any calculus or linear algebra.) – 2017-01-08
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    @JohnHughes I am just told that X and Y are random variables.I am familiar with Cauchy-Schwartz, but really I'm just looking for any sort of proof right now and then I can judge whether the methods used are beyond the scope of the class. – 2017-01-08
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    ?? If you know Cauchy-Schwarz (with no "t") then apply my comment and basta. Did you? – 2017-01-08

2 Answers 2

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Following @Did's hint:

Step 1:

Let $X' = X - E(X), Y' = Y - E[Y]$ (I suppose I'm assuming here that they actually have expected values...). Then $$ Cor(X', Y') = Cor(X, Y) $$ If we can show that $Y' = c X'$, then we have $$ Y = E[Y] - cE[X] + cX $$ so letting $$ a = E[Y] - c E[X] \\ b = c $$ we then have $Y = a + bX$ as needed. In short: we need only consider the case where $X$ and $Y$ have mean zero.

Step 2: Let's examine the case where the correlation is $+1$. (The $-1$ case is very similar, and I leave it to you).

By definition of the correlation, we have $$ \frac{E(X'Y')}{\sigma_X' \sigma_Y'} = 1 $$ so that $$ E(X'Y') = \sqrt{E(X'^2) E(Y'^2)} $$ Now noting that $\langle \rangle = E(X'Y')$ is an inner product on the space of mean-zero random variables, the Cauchy-Schwarz-Bunyakovsky inequality tells us that in general, $$ E(X'Y') \le \sqrt{E(X'^2) E(Y'^2)} $$ with equality only in the case where $Y'$ is a multiple $cX'$ of $X'$.

I'm being a little sloppy here: what's really required is that $Y'$ and $X'$ are linearly dependent, but since $X'$ is nonzero, that's the same as $Y'$ being a multiple of $X'$. I'm being sloppy in another way: if you change the value of $X'$ or $Y'$ on a set of probability zero, then equality still holds. So what I can really conclude is that $$ P(Y' - cX' \ne 0) = 0. $$

But that's exactly what we needed to prove that $$ Pr (Y = a + bX) = 1. $$

(I'm pretty sure that the sloppiness here will offend some folks, but it gets across the main ideas, I believe; in the case where the domain of the random variables $X$ and $Y$ is finite, and where there are no massless atoms, what I've said is correct.)

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    Fixed, thanks. With a small addition. :) – 2017-01-08
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WLOG, assume $\mathbb{E}X=\mathbb{E}Y=0,\mathbb{E}X^2=\mathbb{E}Y^2=1,\text{Corr}[X,Y]=1$.

  • Then $\text{Cov}[X,Y]=\mathbb{E}XY=1$.

  • Now note $\mathbb{E}(X-Y)^2=\mathbb{E}X^2-2\mathbb{E}XY+\mathbb{E}Y^2=1-2+1=0$.

  • As $(X-Y)^2\ge0$, we can infer that $(X-Y)^2\equiv0$ a.s., i.e. $Y=X$.

So, in generality, $\text{Corr}[X,Y]=1$ implies that the rescaled versions of $X,Y$ are equal, i.e. $Y=a+bX$ for some $b>0$.

For $\text{Corr}[X,Y]=-1$, one follows the same line of reasoning with $(X+Y)^2$ in place of $(X-Y)^2$.