**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](https://math.stackexchange.com/questions/755261/is-there-a-standard-proof-for-mathbb-psx-n-text-hits-a-text-before-b)).



> 
> **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 $(X,Y)$.
> 
> 
> 


**Added:**(12.05.2014)


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 $\partial F/\partial G$.


**Added:**(14.05.2014)


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](https://math.stackexchange.com/questions/775498/comparing-the-expected-stopping-times-of-two-stochastically-ordered-random-proce). 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.



Comparing the expected stopping times of two stochastically ordered random processes (Added:(14.05.2014))
I found a counter example for the claim with mean shifted Gaussian distributions. So the claim is not true even with the condition in $d$.


