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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.

Seyhmus Güngören
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1 Answers1

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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:

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

D. Thomine
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  • Isn't the title of Villani's book: "Optimal transport: Old and New" and basically it is not really about coupling? – Arash Apr 20 '14 at 21:35
  • Can this be useful: http://math.stackexchange.com/questions/503504/creating-a-new-random-variable-with-the-same-distribution? – Arash Apr 20 '14 at 21:35
  • 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 $X<Y$. – Einar Rødland Apr 20 '14 at 21:38
  • @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. – D. Thomine Apr 20 '14 at 21:39
  • @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: http://math.stackexchange.com/questions/503504/creating-a-new-random-variable-with-the-same-distribution – Arash Apr 20 '14 at 21:42
  • I have a few remarks. Just at the beginning you say that "Note that $X$ and $Y$ are continuous, so that ..." if they are discrete as long as $X\not\equiv 0$ or $Y\not\equiv 0$ those events are well defined or are there some other possibilities? Another point is that if $X\equiv 0$, then its distribution is dirac delta at zero. Hence, $G(y)>F(y)$ cannot hold. But of cource $G(y)\geq F(y)$ is more general and nothing need to be changed. I also wonder what could be other possiblities other than $X\equiv 0$ or $Y\equiv 0$ s.t. the events are not well defined. – Seyhmus Güngören Apr 24 '14 at 12:36
  • @Seymus Güngören: yes, as long as we don't have $X \equiv 0$ or $Y \equiv 0$, that's fine. I used the continuity to exclude those cases, but $G>F$ is enough (actually, the fact that $X$ and $Y$ are unbounded is also enough). – D. Thomine Apr 24 '14 at 15:41
  • yes exactly. 1 last question. This is not directly related to this question but very closely related and I would probably need to post a new question about it. Very simply it seems that we also have $E[\tau_A^{Y^{'}}]\geq E[\tau_A^{X^{'}}]$ and $E[\tau_B^{Y^{'}}]\leq E[\tau_B^{X^{'}}]$. I claim that for most of the choice of $A,B$ (if not all) and distributions of $X$ and $Y$ we also have $E[\tau_A^{X^{'}}]+E[\tau_B^{X^{'}}]\geq E[\tau_A^{Y^{'}}]+E[\tau_B^{Y^{'}}]$. Do you have any idea how to approach this problem? I am interested in the minimal conditions such that the relation is tue. – Seyhmus Güngören Apr 25 '14 at 22:33
  • one more remark: I tested mean shifted Gaussian densities for this relation. I changed $A$ and $B$ and I never saw that the relation which I gave above didnt hold. – Seyhmus Güngören Apr 25 '14 at 22:35
  • @Seymus Güngören: if $X$ and $Y$ have negative means, then $\mathbb{E} (\tau_A^{X'})$ and $\mathbb{E} (\tau_A^{Y'})$ are both infinite, as per the strong law of large numbers (because $\tau_A^{X'}$ and $\tau_A^{Y'}$ are both infinite on a set of positive measure). – D. Thomine Apr 26 '14 at 22:10
  • @D.Thomine I think I have a little bit misunderstanding. Lets take $A=2$ and $B=-2$ and a Gaussian density with $\mu=-1$ and $\sigma=1$ for the distribution of $X$. I make Monte-Carlo simulations say with $10^6$ runs. I calculate $S_n=\sum_i X_i$ and whenever $S_n$ exceeds $A$ I record $n$ and whenever it exceeds $B$, I record again the number $n$. Then I calculate the average of $n$ values corresponding to $A$ and $B$. They are both finite. What is the difference from $\tau_A^{X^{'}}$ and $\tau_A^{Y^{'}}$? how should I write what I am saying? and last did you understand what I want to have? – Seyhmus Güngören Apr 27 '14 at 17:51
  • @Seymus Güngören: $\mathbb{E} (\tau_A^{X'})+\mathbb{E} (\tau_B^{X'})$ is infinite, but that's not what you are looking at. What you are looking at is $\mathbb{E} (\min { \tau_A^{X'}, \tau_B^{X'} })$, which is finite. And I think that you don't have in general $\mathbb{E} (\min { \tau_A^{X'}, \tau_B^{X'} }) \geq \mathbb{E} (\min { \tau_A^{Y'}, \tau_B^{Y'} })$, although perhaps working with Gaussians gives you enough rigidity for this result to hold. I'm pretty sure I can find slightly more complicated counter-examples, though (if $S_n X$ has a fair chance of going above $A$). – D. Thomine Apr 27 '14 at 23:47
  • @D.Thomine okay you are right. Sorry for confusion. What I am looking for is exactly as you said $E[\min{\tau_A^{X^{'}},\tau_B^{X^{'}}}]\geq E[\min{\tau_A^{Y^{'}},\tau_B^{Y^{'}}}]$. Although I wonder very much for a counterexample on $\mathbb{R}$, I am also interested in conditions where it works rather than it fails. For example an answer for: what must I do with the choice of $F$ and $G$ s.t. this property holds? – Seyhmus Güngören Apr 28 '14 at 14:45
  • Seymus Güngören: I don't see any nice answer. Perhaps you should ask it as a separate question. – D. Thomine Apr 28 '14 at 18:24
  • @D.Thomine I asked it as a separate question http://math.stackexchange.com/questions/775498/comparing-the-expected-stopping-times-of-two-stochastically-ordered-random-proce and claim 2 was still open. I worked on it and then I asked it in mathoverflow http://mathoverflow.net/questions/165835/comparing-the-expected-stopping-times-of-two-stochastically-ordered-random-proce even after puting a bounty I couldnt get any response. – Seyhmus Güngören May 23 '14 at 20:10