Correlation between two random variables

Question

Show that if $\rho_{XY} = +1$ then $X=a+bY$ for some constants $a,b$ and $b>0$.

How would I go about showing this?

Note: $$ \rho_{XY} = \frac{\mbox{Cov}(X,Y)}{\sqrt{\mbox{Var}(X)\mbox{Var}(Y)}} $$

Answer 1

Let $\hat{X}= X-E(X)$ and $\hat{Y}= Y-E(Y)$ denote the de-meaned variables. Then if $\alpha$ and $\beta$ are any constants

$$\int_{\Omega} (\alpha\hat{X}-\beta\hat{Y})^2dP \geq 0.$$

Let $\alpha = \sqrt{var(Y)}$ and $\beta = \sqrt{var(X)}$. Then

$$\alpha^2\int_{\Omega} \hat{X}^2dP -2\alpha\beta\int_{\Omega} \hat{X}\hat{Y}dP +\beta^2\int_{\Omega}\hat{Y}^2dP \geq 0.$$

Whence, it follows that $$2var({X})var({Y})-2\sqrt{var(X)}\sqrt{var(Y)}\int_{\Omega} \hat{X}\hat{Y}dP \geq 0.$$

Dividing both sides by $2$ and the product of non-zero variances we get

$$1-\frac{\int_{\Omega} \hat{X}\hat{Y}dP}{\sqrt{var(X)}\sqrt{var(Y)}} = 1-\rho_{XY} \geq 0.$$ which represents one side of the known correlation constraint $|\rho_{XY}| \leq 1$. In order for equality to hold (i.e., $\rho_{XY}=1$), then the first integral must vanish and the non-negative integrand $(\alpha\hat{X}-\beta\hat{Y})^2$ must almost surely equal $0$:

$$(\alpha\hat{X}-\beta\hat{Y})^2= 0$$

This implies that $X$ and $Y$satisfy the following linear relationship with probability $1$,

$$X-E(X)= \frac{\sqrt{var(X)}}{\sqrt{var(Y)}}[Y-E(Y)]$$