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I need to prove that $$\rho(X,Y) = \pm 1 \implies Y = aX+b,$$ for some constants $a, b$.

With the help of this thread and this document here's what I've got so far:

Let $X, Y$ be random variables and $a$ some constant. $aX + Y$ is also a random variable whose variance by definition is non-negative $$V(aX+Y) \ge 0.$$ From variance properties: $$a^2 V(X)+2a\text{Cov}{X,Y}+V(Y)\ge 0$$ This is a quadratic as a function of $a$, which has a maximum of 1 root. It has a root only for $a$ that satisfies $$a^2 V(X)+2a\text{Cov}{X,Y}+V(Y)=0$$ let it be $a_0$. That means that $$V(a_0X+Y)=0$$

How can I continue from here?

Thank you.

sam wolfe
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Netanel
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2 Answers2

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Hint: In which cases do we have equality in the Cauchy-Schwarz inequality?

Further hint: Here, the Cauchy-Schwarz inequality is used in the form $$\operatorname{Cov}(X,Y)^2\le\operatorname{Var}(X) \operatorname{Var}(Y)$$

Maximilian Janisch
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  • It happens for $Y = aX + b$ because: $$Cov^2 (X,aX+b) = (E[(X-μ_X )(aX+b-E(aX+b)])^2 = (E[(X-μ_X )(aX-aμ_X ])^2=a^2 (E[(X-μ_X )^2 ])^2=a^2 (V(X))^2=V(X)V(aX+b)=V(X)V(Y)$$ But what makes it the only option? – Netanel Dec 16 '19 at 20:11
  • @Netanel From the Wikipedia article on the Cauchy-Schwarz inequality: „ Moreover, the two sides are equal if and only if [the two vectors] are linearly dependent“ In our case this means that we have equality if and only if $$X-\operatorname E (X)=c\cdot (Y-\operatorname{E}(Y))$$ for some constant $c$. Can you take it from here? – Maximilian Janisch Dec 16 '19 at 23:29
  • Need more clarification on the connection between the vector version of the inequality and the probability version if that's not a trouble – Netanel Dec 17 '19 at 06:58
  • @Netanel Sure, in the general case, we have an inner product space $(V,\langle\cdot,\cdot\rangle)$. In our case, we use $$V={\text{All square integrable random variables}}=L^2(\Omega,\mathsf P)$$ and the inner product $$\langle f,g\rangle =\int_\Omega f(\omega)\cdot g(\omega),\mathrm d\mathsf P(\omega)=\mathsf E(f\cdot g)$$ For further reading see http://mathworld.wolfram.com/L2-Space.html Notice that we have $$\operatorname{Cov}(X,Y)=\langle X-\mathsf E(X),Y-\mathsf E(Y)$$ – Maximilian Janisch Dec 17 '19 at 09:03
  • @Netanel About my notation: $\mathsf P$ is the probability measure, $\Omega$ is the set of events, and I forgot the closing angle in the last line – Maximilian Janisch Dec 17 '19 at 09:09
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Here is a slightly different proof. Let $X':=X/\sigma_X$ and $Y':=Y/\sigma_Y$. Then \begin{align} &\operatorname{Var}(X'+Y')=2(1+\rho_{X,Y}) \quad\text{and} \\ &\operatorname{Var}(X'-Y')=2(1-\rho_{X,Y}). \end{align} If $\rho_{X,Y}=1$, then $\operatorname{Var}(X'-Y')=0 \Leftrightarrow X'-Y'=\mathsf{E}X'-\mathsf{E}Y'$ a.s., which implies that $$ Y=aX+\left(\mathsf{E}Y-a\mathsf{E}X\right) \quad\text{a.s.}, $$ where $a\equiv\sigma_Y/\sigma_X$. Similarly, when $\rho_{X,Y}=-1$, $$ Y=-aX+\left(\mathsf{E}Y+a\mathsf{E}X\right) \quad\text{a.s.} $$