Let $X_i$ and $Y_i$ be two continuous random variables on $\mathbb{R}$ having distribution functions $F$ and $G$, respectively satisfying $G(y)>F(y)$ for all $y$. Let futhermore $S^X_n=\sum_{i=1}^n X_i$, $S^Y_n=\sum_{i=1}^n Y_i$, $A>0$, and $B<0$, Then, I wonder if there is a simple/standard proof for the following:

$$ \mathbb P(S^X_n\text{ hits }A\text{ before }B) > \mathbb P(S^Y_n\text{ hits }A\text{ before }B) $$

Note: Both $X$ and $Y$ have negative means.

Thanks alot.


First remark : the events "$S_n^X$ hits $A$ before $B$" are negligible. Let's be rigorous. Let $\tau_A := \inf \{n \geq 0: S_n \geq A\}$, and let $\tau_B := \inf \{n \geq 0: S_n \leq B\}$, both with values in $\mathbb{N} \cup \{+\infty\}$. Then you wan to prove that:

$$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) > \mathbb{P}_{(S_n^Y)} (\tau_A < \tau_B).$$

Essentially, I just want to point out that "hitting $A$" means "reaching $A$ or a higher value", and likewise for $B$. Note that $X$ and $Y$ are continuous, so that for each of them $\tau_A < +\infty$ a.s. or $\tau_B < +\infty$ a.s., so the event $\tau_A < \tau_B$ is well-defined up to a negligible subset (that wouldn't be the case if $X \equiv 0$, for instance).

Now, the idea is, as Did suggested, to use a coupling. The condition $G > F$ is equivalent to the fact that there exists a probability space $\Omega$ and a random variable $(X',Y')$ on $\Omega$ such that:

  • $X' = X$ in distribution;

  • $Y' = Y$ in distribution;

  • $X' > Y'$ a.s.

(Edit, thanks to Einar Rødland) If there is such a coupling, it is easy to see that $G>F$ everywhere. The converse - which is what we are interested in - is more delicate. The idea is to take a random variable $U$ with a uniform distribution on $[0,1]$, and put $(X',Y') := (F^{-1} (U), G^{-1} (U))$, where $F^{-1}$ and $G^{-1}$ are the generalized inverses of $F$ and $G$ respectively. Now, since $F^{-1} > G^{-1}$ on $(0,1)$, we've won.

I can't find a good reference now, see e.g. the first slide here. I guess it's done in Villani's Optimal tranport: Old and new, but there must be a probability-oriented book somewhere with the minimum on stochastic ordering...

Now, let us take a sequence of i.i.d. random variables $(X_n',Y_n')_{n \geq 1}$ such that $(X',Y') = (X_1',Y_1')$ in distribution. Then $(X_n')_{n \geq 1} = (X_n)_{n \geq 1}$ in distribution, and $(Y_n')_{n \geq 1} = (Y_n)_{n \geq 1}$. Let $S_n^{X'} := \sum_{i=1}^n X_i'$, and likewise for $Y'$. Then:

  • $(S_n^{X'})_{n \geq 1} = (S_n^X)_{n \geq 1}$ in distribution;

  • $(S_n^{Y'})_{n \geq 1} = (S_n^Y)_{n \geq 1}$ in distribution.

In addition, $X' > Y'$ a.s., so $X_i' > Y_i'$ for all $i$ a.s.. Hence:

  • $S_n^{X'} > S_n^{Y'}$ for all $n$ a.s..

Let $\tau_A^{X'} := \inf \{n \geq 0: S_n^{X'} \geq A\}$, and define $\tau_A^{Y'}$, $\tau_B^{X'}$, $\tau_B^{Y'}$ in the same way. Then $S_n^{Y'} \geq A$ implies that $S_n^{X'} \geq A$ for all $n$, so $\tau_A^{X'} \leq \tau_A^{Y'}$. In the same way, $\tau_B^{X'} \geq \tau_B^{Y'}$. Hence,

$$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) = \mathbb{P} (\tau_A^{X'} < \tau_B^{X'}) \geq \mathbb{P} (\tau_A^{Y'} < \tau_B^{Y'}) = \mathbb{P}_{(S_n^Y)} (\tau_A < \tau_B).$$

This is the standard proof. Proving a strict inequality is trickier. We have to find an event of positive measure on which $\tau_A^{X'} < \tau_B^{X'}$ but $\tau_A^{Y'} > \tau_B^{Y'}$. This relies more on the specific data of the problem, and thus is less standard.

Note that $G > F$ implies that $\mathbb{P} (Y < t) > 0$ for all $t$, and $\mathbb{P} (X > t) > 0$ for all $t$.

Let $M \in (A, +\infty]$ be such that $\mathbb{P} (X_1' \geq M) = \mathbb{P} (Y_1' \geq A)$. Such an $M$ exists, since both $X$ and $Y$ are continuous random variables. If $X_1' \in (A,M)$ then $Y_1' < A$ (this comes from the specific construction of the optimal coupling). In addition,

$$\mathbb{P} (X_1' \in (A,M)) = F(M)-F(A) = G(A)-F(A) > 0.$$

Now, let us consider the event "$X_1' \in (A,M)$ and $Y_2' < B-A$". On this event, we have $\tau_A^{X'} = 1 < \tau_B^{X'}$ and $\tau_A^{Y'} > \tau_B^{Y'} = 2$. Finally, this event has probability $(G(A)-F(A))G(B-A)$, whence:

$$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) - \mathbb{P}_{(S_n^Y)} (\tau_A < \tau_B) \geq (G(A)-F(A))G(B-A) > 0.$$

  • $\begingroup$ Isn't the title of Villani's book: "Optimal transport: Old and New" and basically it is not really about coupling? $\endgroup$ – Arash Apr 20 '14 at 21:35
  • $\begingroup$ Can this be useful: math.stackexchange.com/questions/503504/…? $\endgroup$ – Arash Apr 20 '14 at 21:35
  • $\begingroup$ If you let $U\sim\text{Uniform}[0,1]$, let $X=F^{-1}(U)$ and $Y=G^{-1}(U)$, and you have $X$ and $Y$ with the desired distributions and $X<Y$. $\endgroup$ – Einar Rødland Apr 20 '14 at 21:38
  • $\begingroup$ @Arash: thanks, that was a lapsus. If I remember well, there is a chapter (probably the first) on the basics of coupling, which (I guess) presents the optimal coupling on the real line. $\endgroup$ – D. Thomine Apr 20 '14 at 21:39
  • $\begingroup$ @EinarRødland, We do not know if $F^{-1}$ exist or not so there are some subtleties here. Look at the end of my answer here: math.stackexchange.com/questions/503504/… $\endgroup$ – Arash Apr 20 '14 at 21:42

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