A probability problem by ابراهيم فقرا

$ρ_{X,Y} = \frac{EXY -EXEY}{\sqrt{{σ_X}^2*{σ_Y}^2}}=\frac{-1-0*0}{\sqrt{{1}^2*{1}^2}}=-1$

So we now can assume that $Y=aX+b$ because the dependence between them is linear ,

let's find the value of $a$

$Cov(X,Y)=Cov(X,aX+b)=Cov(X,aX)=aCov(X,X)=aVar(X)=a*1=a$

On the other hand :-

$Cov(X,Y)=EXY -EXEY=-1-0*0=-1$

so we got that $a=-1$ .

let's find the value of $b$ :

$EY=E(aX+b)=E(-X+b)=b+E(-X)=b-EX=b-0=b$

On the other hand :-

$EY=0$

so we got that $b=0$ .

So $Y=aX+b=-1*X+0=-X$

Since $E[(X+Y)^2] = E[X^2] + 2E[XY] + E[Y^2] = \mathrm{Var}[X] - 2 + \mathrm{Var}[Y] = 0$ we see that $X + Y = 0$ , and so $Y = -X$ , almost surely (in that $P[X+Y \neq 0] = 0$ ).