If we assume Var(x) is not = 0, show that p(Y, 7Y+3) = 1.
To solve this question it is my understanding that we use the correlation formula,
ρ(X, Y ) = Cov(X, Y) / sqrt[Var(X)] sqrt[Var(Y)].
The theory behind it makes sense but I can't figure out how to show the work here. Thanks for your help!
Leonard
The result is intuitive, since this is basically correlating a variable with itself (which is not affected by the linear combination of 7Y+3).
In your equation, let the "X" be "Y", and the "Y" be "7Y+3".
so Var(7Y+3)=7^2 Var(Y)=49Var(Y). And then the denominator of your equation is
sqrt[(Var(Y)] sqrt[49Var(Y)] = 7 Var(Y).
Cov(Y,7Y+3)=E(Y(7Y+3)) - E(Y)E(7Y+3) = 7E(Y^2) +3E(Y) - 7E(Y)^2 - 3E(Y) =
7Var(Y) (since Var(Y)=E(Y^2) - E(Y)^2).
And the result follows, since 7Var(Y)/7Var(Y)=1.