Let $(X,Y)$ be a bivariate random variable, where $X$ is an exponential r.v. with mean $1$. Also let $ \operatorname{Cov}(X,Y) = -2$, $E[Y]=-2$, and $ \operatorname{Var}(Y) = 4$. Find the cdf of $Y$.
All I can get is $E[XY]=-4$ and $E[Y^2]=8$. How can these condition get the cdf of $Y$?
The key observation is that $$ -\sqrt{\text{Var}(X)}\sqrt{\text{Var}(Y)}=-2=\text{Cov}(X,Y). $$This holds since $E[X^2]=\int_0^\infty x^2e^{-x}dx = \Gamma(3)=2$ and $\text{Var}(X)=E[X^2]-[EX]^2 = 1$. By Cauchy-Schwarz inequality (and its equality condition) it holds that $Y-EY=c(X-EX)$ almost surely for some constant $c\in\mathbb{R}$. We can see that $c=-2$ from $\text{Cov}(X,Y)=-2\text{Var}(X)$. This gives the distribution of $Y=-2X$ as follows. $$ P(Y\le y)=P(-2X\le y)=P(X\ge -\frac{y}{2})=\begin{cases}1,\quad y\ge 0\\e^{\frac{y}{2}},\quad y<0\\ \end{cases}. $$