Does $\rho>0$ imply $P(X>a,Y>a)$$>$$P(X>a) P(Y>a)$? Consider two stochastic variables X and Y that are distributed $N(0, \sigma_1)$ and $N(0, \sigma_2)$ and are correlated $\rho>0$.  
Is it true that $P(X>a,Y>a)>P(X>a)$$P(Y>a)$?
Does the answer depend on whether $a>0$ or $a<0$?
I have tried to find the answer to the solution by starting with the definition of the correlation coefficient, $\rho>0$. This provides me with $E(XY)$$>$$E(X)$$E(Y)$. However, after this I am stuck.
Any help or hint is much appreciated.
 A: From the formula of conditional probability we have that $$P(X>a|Y>a)=\frac{P(X>a,Y>a)}{P(Y>a)}$$ (There is no problem to divide by $P(Y>a)$, since normal probability distribution gives positive probabality to every interval. That is also why the precise value of a, i.e. whether positive or negative, does not affect your answer). Since $X,Y$ are positively correlated we have that $$P(X>a|Y>a)>P(X>a)$$ As David L intuitively stated, the positive correlation implies that when $Y>a$ it is more likely that also $X>a$ compared to when we have no information about $Y$ (in other words $X,Y$ are smaller together and bigger together). Combining the two equations $$\frac{P(X>a,Y>a)}{P(Y>a)}>P(X>a) \implies P(X>a,Y>a)>P(Y>a)P(X>a)$$ we receive the result.
A: Assume without loss of generality that $\sigma^2_2\geqslant\sigma_1^2$ and note that 
$$Y=(\sigma_2\rho/\sigma_1)\cdot X+\sqrt{1-\rho^2}\cdot Z,
$$ where $Z$ is normal $(0,\sigma_2^2)$ and independent of $X$. Furthermore, $\rho\geqslant0$ hence, if
$X\gt a$ and $Z\gt a$, then $Y\gt\theta a$,
where
$$
\theta=(\sigma_2\rho/\sigma_1)+\sqrt{1-\rho^2}\geqslant\rho+\sqrt{1-\rho^2}\geqslant1.
$$
Thus, if $a\gt0$ then $\theta a\geqslant a$ hence
$$
P(X\gt a)P(Y\gt a)=P(X\gt a,Z\gt a)\leqslant P(X\gt a,Y\gt\theta a)\leqslant P(X\gt a,Y\gt a).
$$
Finally the inequality is strict because $\theta\gt1$ except when $\rho=0$ (but this case is excluded) and $\rho=1$ (but then $Y=\sigma_2 X/\sigma_1$ and a direct argument works).
A: If you do not assume, that $X$ and $Y$ have jointly Gaussian distribution, then the inequality $P(X>a,Y>a)>P(X>a)P(Y>a)$ need not hold, even if $X$ and $Y$ are positively correlated. 
For example, let $\sigma_1 =1$ , and for $X$ with distribution $N(0,1)$ take 
$$Y=I_{\{|X| \leq n\}}X  -  I_{\{|X| > n\}}X.$$ Then $Y$ has Gaussian law $N(0,1)$, and $\text{E} XY  >0$ for big enough $n$. However, 
$$P(X>n,Y>n)  =  0 < P(X>n)P(Y>n)$$
If $X$ and $Y$ have jointly Gaussian distribution, then the way of reasoning provided by Did seems to be correct.
