# Let $X \sim \exp(1)$, $\operatorname{Cov}(X,Y)=-2$, $E[Y]=-2$, and $\operatorname{Var}(Y)=4$. Find the cdf of $Y$

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 $$Y$$?

• I don't think it is possible to determine the cdf of $Y$ from the given information. – Kabo Murphy Jan 14 at 5:31

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. $$P(Y\le y)=P(-2X\le y)=P(X\ge -\frac{y}{2})=\begin{cases}1,\quad y\ge 0\\e^{\frac{y}{2}},\quad y<0\\ \end{cases}.$$
• @YibeiHe Sure. If you are not familiar with the argument, I'll directly show that. Let $X'=X-EX$ and $Y'=Y-EY$. Observe that $$E[(Y'+2X')^2]=\text{Var}(Y)+4\text{Cov}(X,Y)+4\text{Var}(X) = 4-8+4=0.$$ This implies $Y'+2X'=0$ almost surely and the claim is proved. – Song Jan 14 at 23:33