Question: "Let X be a continuous random variable, and $Y := g(X)$ where $g: \mathbb{R} \rightarrow \mathbb{R}$ is a non-decreasing, Borel-regular function.
Prove that the correlation coefficient $\rho(X,Y) := \frac{Cov(X,Y)}{\sqrt{Var(X) Var(Y)}} = 1$ "
My approach: So if we show that Cov(X,Y) = $\sqrt{\text{Var(X) Var(Y)}}$, then we're done.
I wrote out:
Var(X) Var(Y) $= (\mathbb{E}(X^2)-\mathbb{E}(X)^2)(\mathbb{E}(Y^2)-\mathbb{E}(Y)^2)$ and expanded it.
Then I did the same for: Cov(X,Y) $= \mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)$.
And then, $\text{Cov(X,Y)}^2 = \left( \mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y) \right)^2$.
But I'm not sure how to proceed from here. Any help would be appreciated, thank you!
This is not true. Let $X \sim U[0,1]$ and $y=X^{2}=g(x)$ where $g(x) =x^{2}$ for $x \geq 0$ and $g(x)=0$ for $x <0$. I will let you compute the correlation coefficient and see that it is not $1$.