Olympiad Maths Prep

Track / Stage 6 / 238 of 400 #1238 of 2000

Problem 1238

National olympiad, first round
Algebra Difficulty 6.4 Prove it

7. Let ξ\xi and η\eta be two random variables, Dξ>0,Dη>0\mathrm{D} \xi>0, \mathrm{D} \eta>0 and ρ=\rho= ρ(ξ,η)\rho(\xi, \eta) be their correlation coefficient. Show that ∣ρ∣⩽1|\rho| \leqslant 1. Moreover, if ∣ρ∣=1|\rho|=1, then there exist constants aa and bb such that η=aξ+b\eta=a \xi+b. Furthermore, if ρ=1\rho=1, then

η−EηDη=ξ−EξDξ \frac{\eta-\mathrm{E} \eta}{\sqrt{D \eta}}=\frac{\xi-\mathrm{E} \xi}{\sqrt{D \xi}}

(and, therefore, in the case ρ=1\rho=1 the constant aa is positive), and if ρ=−1\rho=-1, then

η−EηDη=−ξ−EξDξ \frac{\eta-\mathrm{E} \eta}{\sqrt{\mathrm{D} \eta}}=-\frac{\xi-\mathrm{E} \xi}{\sqrt{D \xi}}

(and, therefore, in this case a<0a<0).

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution. Let

ξ~=ξ−EξDξ,η~=η−EηDη \widetilde{\xi}=\frac{\xi-\mathrm{E} \xi}{\sqrt{D \xi}}, \quad \widetilde{\eta}=\frac{\eta-\mathrm{E} \eta}{\sqrt{\mathrm{D} \eta}}

Then Eξ~=Eη~=0,Dξ~=Eξ~2=Dη~=Eη~2=1\mathbf{E} \widetilde{\xi}=\mathbf{E} \widetilde{\eta}=0, \mathbf{D} \widetilde{\xi}=\mathbf{E} \widetilde{\xi}^{2}=\mathbf{D} \widetilde{\eta}=\mathbf{E} \widetilde{\eta}^{2}=1,

Eξ~η~=E[(ξ−Eξ)(η−Eη)]DξDη=ρ(ξ,η)=ρ \mathrm{E} \widetilde{\xi} \widetilde{\eta}=\frac{\mathrm{E}[(\xi-\mathrm{E} \xi)(\eta-\mathrm{E} \eta)]}{\sqrt{\mathrm{D} \xi} \sqrt{\mathrm{D} \eta}}=\rho(\xi, \eta)=\rho

We use the Cauchy-Bunyakovsky inequality:

∣ρ∣=∣Eξ~η~∣⩽E∣ξ~η~∣⩽Eξ~2Eη~2=1 |\rho|=|\mathrm{E} \widetilde{\xi} \widetilde{\eta}| \leqslant \mathrm{E}|\widetilde{\xi} \widetilde{\eta}| \leqslant \sqrt{\mathrm{E} \widetilde{\xi}^{2} \mathrm{E} \widetilde{\eta}^{2}}=1

In the introduced notation, it remains to show that if ρ=1\rho=1, then ξ~=η~\widetilde{\xi}=\widetilde{\eta}, and if ρ=−1\rho=-1, then ξ~=−η~\widetilde{\xi}=-\widetilde{\eta}:

ρ=1⇔Eξ~η~=1⇔2Eξ~η~=Eξ~2+Eη~2⇔⇔E(ξ~−η~)2=0⇔ξ~=η~ρ=−1⇔Eξ~η~=−1⇔−2Eξ~η~=Eξ~2+Eη~2⇔⇔E(ξ~+η~)2=0⇔ξ~=−η~ \begin{aligned} \rho=1 \Leftrightarrow \mathrm{E} \widetilde{\xi} \widetilde{\eta}=1 & \Leftrightarrow 2 \mathrm{E} \widetilde{\xi} \widetilde{\eta}=\mathrm{E} \widetilde{\xi}^{2}+\mathrm{E} \widetilde{\eta}^{2} \Leftrightarrow \\ & \Leftrightarrow \mathrm{E}(\widetilde{\xi}-\widetilde{\eta})^{2}=0 \Leftrightarrow \widetilde{\xi}=\widetilde{\eta} \\ \rho=-1 \Leftrightarrow \mathrm{E} \widetilde{\xi} \widetilde{\eta}=-1 & \Leftrightarrow-2 \mathrm{E} \widetilde{\xi} \widetilde{\eta}=\mathrm{E} \widetilde{\xi}^{2}+\mathrm{E} \widetilde{\eta}^{2} \Leftrightarrow \\ & \Leftrightarrow \mathrm{E}(\widetilde{\xi}+\widetilde{\eta})^{2}=0 \Leftrightarrow \widetilde{\xi}=-\widetilde{\eta} \end{aligned}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.