0
$\begingroup$

Let $X$ and $Y$ be two dependent normally distributed continuous random variables (their marginals are $\mathcal{N}(0, 1)$). I would like to find an upper bound on the probability that one is greater than the other by a given threshold, i.e. find a $\theta$ such that $P(Y > X + \delta) < \theta$, $\delta, \theta \in \mathbb{R}$

If $X$ and $Y$ were independent it would be easy as I could directly compute their joint density, but in the case were they are allowed to be dependent all I could do is find a numerical solution by discretizing the joint density (which looked far from trivial).

1 Answers 1

2

If $X$ and $Y$ are jointly normal, $Y - X$ is normal with mean $0$ and variance $v = 2 - 2 \text{Cov}(X,Y)$, which can be anywhere from $0$ to $4$, and then $P(Y > X + \delta) = \Phi(-\delta/\sqrt{v})$, where $\Phi$ is the standard normal cdf.

If $X$ and $Y$ are not jointly normal, it's still true that $Y-X$ has mean $0$ and variance $v \in [0, 4]$. Now for any random variable $Z$ with mean $0$ and variance $v\le 4$, if $\delta > 0$ we have $$P(Z > \delta) \le \frac{v}{v + \delta^2} \le \frac{4}{4+\delta^2}$$
This comes from the fact that $$\frac{ (Z + v/\delta)^2}{(\delta + v/\delta)^2} \ge \cases{1 & if $Z \ge \delta$\cr 0 & otherwise\cr}$$ and $$E\left[\frac{(Z+v/\delta)^2}{(\delta+v/\delta)^2}\right] = \frac{ v + v^2/\delta^2}{(\delta+v/\delta)^2} = \frac{v}{v+\delta^2}$$

  • 0
    If $X$ and $Y$ are not jointly normal, I do not think it is true that $X - Y$ is jointly normal. For example, let $D$ be a -1,1 coinflip independent of $X$ and let $X$ be a standard normal random variable. Then $Z=XD$ is normally distributed, but $X - Z$ is not normally distributed (it takes on the value of zero with probability $\frac{1}{2}$).2012-03-19
  • 0
    I didn't mean to say $X-Y$ is normal.2012-03-19
  • 0
    Ok it makes sense, thank you very much!2012-03-19
  • 0
    Another bound: $$P(Y - X > \delta) \le P(Y > \delta/2) + P(X < -\delta/2) = 2 \Phi(-\delta/2)$$ This seems to be less than $4/(4+\delta^2)$ for all $\delta > 0$.2012-03-19
  • 0
    Ah? Could you please explain how you derived this second bound?2012-03-19
  • 0
    If $Y \le \delta/2$ and $X \ge -\delta/2$, then $Y-X \le \delta$. So $P(Y - X > \delta) \le P((Y > \delta/2) \cup (X < -\delta/2)) \le P(Y > \delta/2) + P(X < -\delta/2)$. Now $P(Y > \delta/2) = P(X < -\delta/2) = \Phi(-\delta/2)$.2012-03-19
  • 0
    I get all the steps but the first one, I don't see why $P(Y−X>\delta) \leq P((Y>\delta/2) \cup (X<−\delta/2))$ since $Y-X>\delta$ is not a subset of $(Y>\delta/2) \cup (X<−\delta/2)$. For example $Y=0, X=-2\delta$ is in the first one but not the second one.2012-03-19
  • 0
    $X = -2\delta$ is certainly in $X < -\delta/2$ if $\delta > 0$. Maybe it would be easier to see if we wrote $Z = -X$. I'm saying that if $Y + Z$ is more than $\delta$, then at least one of $Y$ and $Z$ is more than $\delta/2$.2012-03-19