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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 $Y$?

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    $\begingroup$ I don't think it is possible to determine the cdf of $Y$ from the given information. $\endgroup$ – Kabo Murphy Jan 14 at 5:31
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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. $$ P(Y\le y)=P(-2X\le y)=P(X\ge -\frac{y}{2})=\begin{cases}1,\quad y\ge 0\\e^{\frac{y}{2}},\quad y<0\\ \end{cases}. $$

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  • $\begingroup$ could you explain how you use Cauchy Schwarz inequality to get the equation Y−EY=c(X−EX)? $\endgroup$ – clement Jan 14 at 23:29
  • $\begingroup$ @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. $\endgroup$ – Song Jan 14 at 23:33

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