# Comparing the expected stopping times of two stochastically ordered random processes (Added:(14.05.2014))

Information:

a-) $X$ and $Y$ are two continuous random variables on $\mathbb{R}$ having continuous distribution functions $F$ and $G$ with $G(y)\geq F(y)$ for all $y$.

b-) $S^X_n=\sum_{i=1}^n X_i$, $S^Y_n=\sum_{i=1}^n Y_i$, $A>0$, and $B<0$; where $X_i$ and $Y_i$ are i.i.d. replicas of $X$ and $Y$ respectively.

c-) $E[X]<0$ and $E[Y]<0$.

d-) $\partial F/\partial G$, (the ratio of densities) is an increasing function on its domain.

What I know:

By coupling (since $G\geq F$), I know that there exist a pair of random variables $(X^{'},Y^{'})$ such that $X=X^{'}$ in distribution, $Y=Y^{'}$ in distribution, and $X^{'}\geq Y^{'}$ almost surely. Using this result, I also have $S_n^{X^{'}}=\sum_{i=1}^n X^{'}_i\geq \sum_{i=1}^n Y^{'}_i=S_n^{Y^{'}}$. Since this holds for all $n$, I am able to compare the following:

$$\tau_A^{X^{'}}=\inf\{n\geq 0:S_n^{X^{'}}\geq A\}$$ $$\tau_A^{Y^{'}}=\inf\{n\geq 0:S_n^{Y^{'}}\geq A\}$$ $$\tau_B^{X^{'}}=\inf\{n\geq 0:S_n^{X^{'}}\leq B\}$$ $$\tau_B^{Y^{'}}=\inf\{n\geq 0:S_n^{Y^{'}}\leq B\}$$

with $\tau_A^{X^{'}}\leq \tau_A^{Y^{'}}$ since $S_n^{Y^{'}}\geq A$ implies $S_n^{X^{'}}\geq A$ and similarly $\tau_B^{X^{'}}\geq \tau_B^{Y^{'}}$. Please see (for details).

Question: I wonder if

$$E[\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}]\geq E[\min\{\tau_A^{Y^{'}},\tau_B^{Y^{'}}\}]$$

holds for any $(A,B)$ and $(X,Y)$.

I thought about using $$\min(a,b)=\frac{1}{2}(|a+b|-|a-b|)$$ the inequality became $$(E[\tau_A^{X^{'}}]-E[\tau_B^{Y^{'}}])-(E[\tau_A^{Y^{'}}]-E[\tau_B^{X^{'}}])\geq E[|\tau_A^{X^{'}}-\tau_B^{X^{'}}|]-E[|\tau_A^{Y^{'}}-\tau_B^{Y^{'}}|]$$ But I know $E[\tau_A^{X^{'}}]=\infty$ and therefore I dont know how one could go forward and what would be the role of increasing $\partial F/\partial G$.

Please correct me if I am mistaken. It seems that we also have $$E[\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}]=E[|\tau_A^{X^{'}}-\tau_B^{X^{'}}|]$$ $$E[\min\{\tau_A^{Y^{'}},\tau_B^{Y^{'}}\}]=E[|\tau_A^{Y^{'}}-\tau_B^{Y^{'}}|]$$

writing this result above we see that

$$E[\tau_B^{X^{'}}]-E[\tau_B^{Y^{'}}] \geq E[\tau_A^{Y^{'}}]-E[\tau_A^{X^{'}}]\quad\quad\quad\quad (*)$$

implies

$$E[\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}]\geq E[\min\{\tau_A^{Y^{'}},\tau_B^{Y^{'}}\}]$$

therefore to show that $(*)$ indeed holds will be enough. Once again $E[\tau_A^{X^{'}}]=\infty$ and $E[\tau_A^{Y^{'}}]=\infty$. If such a comparison is possible then I would guess that $E[\tau_A^{Y^{'}}]-E[\tau_A^{X^{'}}]\geq 0$ because of stochastical ordering. We also know $E[\tau_B^{X^{'}}]-E[\tau_B^{Y^{'}}]\geq 0$.

My idea was using $\tau_A^{X^{'}}=\tau_{A_{\lim_{B\rightarrow\infty}}}^{X^{'}}=\inf\{n\geq 0:S_n^{X^{'}}\geq A,S_n^{X^{'}}\leq B\}$ and the same for $\tau_A^{Y^{'}}$. I hope that this formulation could give me $$E[\tau_{A_{\lim_{B\rightarrow\infty}}}^{Y^{'}}]-E[\tau_{A_{\lim_{B\rightarrow\infty}}}^{X^{'}}]=E[\tau_A^{Y^{'}}]-E[\tau_A^{X^{'}}]<\infty$$

Can?

I asked the same question in mathstackexchange in two parts. I have a counterexample if $\partial F/\partial G$ is not necessarily increasing. However, I wonder very much if the claim holds with this monotonicity condition or not. Unfortunately, I seem to be hopeless to deal with this problem.

I will really appreciate any help. Thank you very much.

• I do not understand the notation in d), but I'll take it to mean that the ratio of densities is increasing. Here is what a counterexample would look like. I did not work through the details. Let f be a density concentrated just to the right of A. Let g be a density concentrated around B/N for some large N, with a small mass eps at M*B, for some large M and small eps. Let X have unnormalized density f(x)+g(x)/sqrt(1-x). Let Y have unnormalized density g(x)+f(x)/sqrt(1+x). Add in eps times a normal(0,1) if you want fully support. Then normalize. (continued...) – guest May 12 '14 at 22:50
• With a suitable choice of constants (sufficiently big M) the conditions will be satisfied. Y will take O(N) steps to cross the lower boundary with high probability, while X will cross the upper boundary in O(1) steps with high probability. The expectations should behave similarly. Again, this isn't a complete answer, but I hope it is a help. – guest May 12 '14 at 22:50
• How can $f$ be concentrated to the right of $A$? you mean the expected value? if so it cannot be. – Seyhmus Güngören May 12 '14 at 23:15

I found a counter example for the claim with mean shifted Gaussian distributions. So the claim is not true even with the condition in $d$.