Question

Question about correlation!

Show correlation = +/- 1 iff Y = aX+b

Answer 1

Show \(|\rho (X,Y)|=1\) iff Y=aX+b, where \(\rho(X,Y)\) is the correlation coefficient of random variables X and Y, and a,b are constants with a \(\neq 0\).

I am able to show \(\rho =1\) ,but unable that it can also be -1.

\(Cov(X,Y)\)
\(=Cov(X,aX+b)\)
\(=E[X(aX+b)]-E(X)E(aX+b)\)
\(=aE(X^2)+bE(X)-E(X)[aE(X)+b]\)
\(=aE(X^2)+bE(X)-a[E(X)]^2-bE(X)\)
\(=a[E(X^2)-[E(X)]^2]\)

\(=aVar(X)\)

Now,
\(\rho(X,Y)\)
\(= \frac{Cov(X,Y)}{\sqrt{Var(X)}\sqrt{Var(Y)}}\)

\(= \frac{aVar(X)}{\sqrt{Var(X)}\sqrt{Var(aX+b)}}\) the numerator using the above result

\(=\frac{aVar(X)}{\sqrt{Var(X)}\sqrt{a^2Var(X)}}\)

\(=\frac{aVar(X)}{\sqrt{Var(X)}\big (a\sqrt{Var(X)}\big )}\)

\(=\frac{aVar(X)}{aVar(X)} = 1\) ... I don't know to get it to be also either -1

Answer 2

Do I just need to stick an absolute bar on both sides of the equation for \(\rho (X,Y)\) and get \(\rho (X,y) = |1| = \pm 1\)? It seems trivial to do so for some reason o_O.

Answer 3

Actually I don't know what I wrote in my last reply... lol :(... |1| = 1 only lol

Answer 4

if a is negative, it will be negative on the top, but the bottom is the square root of a^2, which is always positive.

Answer 5

ohhh right!! \(\sqrt{a^2}=|a|\) I always forget that