Probability of $Y+b > X$ for iid normal variables conditioned on a rare event

Question

I am thinking of the (imprecise) claim that a small increase in the average ability in a population results in a large increase in the ability of the exceptionally able.

So I'm considering two independent random variables $X$ and $Y$, both normally distributed $N(0,\sigma^2)$ (same $\sigma$ for both), and my general curiosity is to compare the distributions of $X$ and $Y+b$ conditioned on $X\geq c$ and $Y\geq c$ (here $b$ and $c$ are positive constants).

For example, what happens if we keep $b$ constant and take $c$ to infinity?

For example of the example, what can we say about $p_{b,c}=P(Y+b > X \mid X\geq c,Y\geq c)$?

If the imprecise claim is true, I expect $p_{b,c}$ to increase for a fixed $b$ as $c$ increases. Does it increase? At what rate?

I've made it into a precise question about comparing $X$ and $Y+b$ conditioned on $X\geq c$ and $Y\geq c$. But if there's anything else that's interesting to know about this situation I'll be glad to know.

Answer 1

(Following is a major revision based on answers I received to a related question.)

Summary

• For i.i.d. Normal random variables, $p_{b,c}$ is monotone increasing in $c$, with $\lim\limits_{c\to\infty}p_{b,c}=1$ (proof below).
• For i.i.d. non-Normal random variables, $\lim\limits_{c\to\infty}p_{b,c}$ may be any value in the interval $\left[{1\over2},1\right]$ (proof below) and numerical integration indicates that $p_{b,c}$ may be either monotone increasing or decreasing in $c$.

Formula for $\lim_{c\to\infty}p_{b,c}$

In the general case of $X,Y$ i.i.d. with c.d.f $F$ and p.d.f. $f$ , not necessarily Normal, we have (letting $\bar F=1-F$):

$$\begin{align} p_{b,c}&:=P(Y+b>X\mid X>c,Y>c)\\[1ex] &=1-P(Y +b<X\mid X>c, Y>c)\\[1ex] &=1-{P(Y+b<X, X>c, Y>c)\over \bar F(c)^2}\\[1ex] &=1-{\int_{c+b}^\infty f(x) \int_c^{x-b} f(y)\,dy\, dx \over \bar F(c)^2}\\[1ex] &=1-{\int_{c+b}^\infty f(x)\left(F(x-b)-F(c)\right)\, dx \over \bar F(c)^2}\\[1ex] &=1-{\int_{c+b}^\infty f(x) F(x-b)\,dx-\bar F(c+b)F(c) \over \bar F(c)^2}\tag{1}\\[1ex] \end{align}$$

Therefore, applying the same approach as in @BJM's answer, we have

$$\begin{align} \lim_{c\to\infty}p_{b,c}&=1-\lim_{c\to\infty}{\int_{c+b}^\infty f(x) F(x-b)\,dx-\bar F(c+b)F(c) \over \bar F(c)^2}\\[1ex] &=1-\lim_{c\to\infty}{0\cdot 1-f(c+b)F(c) - [-f(c+b)F(c) + \bar F(c+b)f(c)] \over -2\bar F(c)f(c)}\tag{2}\\[1ex] &=1-{1\over2}\lim_{c\to\infty}{\bar F(c+b) \over \bar F(c)}\tag{3}\\ &=1-{1\over2}\lim_{c\to\infty}{-f(c+b)\over -f(c)}\tag{4}\\[1ex] \end{align}$$

where (2) is obtained using L'Hospital's rule, differentiating the integral using Leibniz' integral rule, and (4) is by another application of L'Hospital's rule. Thus,

$$\bbox[10px, border:3px solid lightgrey]{\lim\limits_{c\to\infty}p_{b,c}=1-{1\over2}\lim_{c\to\infty}{f(c+b)\over f(c)}}\tag{5}$$

Examples

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If $X,Y$ i.i.d. Normal$(\sigma)$, then $$\lim_{c\to\infty}p_{b,c}=1$$ because $${f(c+b)\over f(c)}={\exp\left(-{1\over2}{\left({c+b\over\sigma}\right)^2}+{1\over2}{\left({c\over\sigma}\right)^2}\right)}=\exp\left({-2bc-b^2\over2\sigma^2}\right)\to0\ \ \text{as $c\to\infty$}.$$

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If $X,Y$ i.i.d. Laplace$(\sigma=\sqrt{2}\,\alpha)$, then $$\lim_{c\to\infty}p_{b,c}=1-{1\over2}\exp\left(-{b\over\alpha}\right)$$ because $${f(c+b)\over f(c)}={\exp\left(-{|c+b|\over\alpha}+{|c|\over\alpha}\right)}=\exp\left(-{b\over\alpha}\right)$$

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If $X,Y$ i.i.d. Cauchy$(\sigma=\infty)$, then $$\lim_{c\to\infty}p_{b,c}={1\over2}$$ because $${f(c+b)\over f(c)}={(1+(c+b)^2)^{-1}\over (1+c^2)^{-1}}\to1$$

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Here are typical plots for these examples, obtained by numerical integration using Eq.(1) (confirmed also by Monte Carlo simulation):

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NB: In general, $p_{b,c}$ can't be less than ${1\over2}$, because the bivariate p.d.f.s are symmetrical in their arguments (i.e. $f(x,y)=f(y,x)$); furthermore, the above three examples suffice show that $\lim_{c\to\infty}p_{b,c}$ may be any value in the interval $\left[{1\over2},1\right]$.

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Proof (sketch) that, in the Normal case, $p_{b,c}$ is monotone increasing in $c$

Defining $(\beta,\gamma):=(b/\sigma, c/\sigma),$ and $p(\beta,\gamma):=p_{b/\sigma,c/\sigma},$ we can prove that ${\partial p(\beta,\gamma) \over \partial \gamma}>0$ for all $c$ by taking the derivative of Eq.(1) using Leibniz' rule, which reduces to

$${\partial p(\beta,\gamma)\over\partial\gamma}={\phi(\gamma)\over\bar\Phi(\gamma)^2}\left(\bar\Phi(\gamma+\beta)-{2\,I(\beta,\gamma)\over \bar\Phi(\gamma)}\right)\tag{6} $$

where $\phi$ and $\Phi$ are the standard normal p.d.f and c.d.f., respectively, and

$$I(\beta,\gamma) :=\int_{\gamma+\beta}^\infty\phi(z_1) \int_\gamma^{z_1-\beta}\phi(z_2)\,dz_2\, dz_1 .$$

Therefore, ${\partial p(\beta,\gamma)\over\partial\gamma}>0$ if

$$I(\beta,\gamma) < {1\over2}\,{\bar\Phi(\gamma+\beta)\,\bar\Phi(\gamma)}.\tag{7} $$ Now (7) is a consequence of the following fact:

If a continuous bivariate distribution has circular symmetry about the origin, then
$$x' \lesseqqgtr y'\implies P(A) \lesseqqgtr P(B)$$ or equivalently, $$x' \lesseqqgtr y'\implies P(A)\lesseqqgtr{1\over2}P(Q)\implies P(B)\gtreqqless{1\over2}P(Q)$$ where $A$ and $B$ are the upper and lower halves, respectively, of the "shifted quadrant" $Q(x',y'):=\{(x,y)\in\Bbb R^2: x>x', y>y'\}$ bisected by its diagonal through point $(x',y')$.

To apply this, note that $I(\beta,\gamma)=P(B)$ is the probability of the lower half ($B$) of the shifted quadrant $Q(x'=\gamma+\beta,y'=\gamma)$ bisected by its diagonal. Since $x'=\gamma+\beta>y'=\gamma$, we have the desired result:

$$I(\beta,\gamma)=P(B)< {1\over2}P(Q)={1\over2}{\bar\Phi(\gamma+\beta)\,\bar\Phi(\gamma)} .$$