# Comparing the stopping times of two stochastic processes

Let $$f_0$$, $$g_1$$, $$g_0$$ be $$3$$ distinct density functions on the real numbers $$\mathbb{R}$$ with the corresponding distribution functions $$F_0$$, $$G_1$$, and $$G_0$$, respectively. The following relation is known to hold: $$F_0(y)>G_0(y)>G_1(y)\quad \forall y$$ Define a random variable $$Z_i:=\log\left(\frac{g_1(Y_i)}{g_0(Y_i)}\right)$$ $$\large(Y_i$$ are independent replicas of random variable $$Y\large)$$

$$\star$$

The stochastic process, $$S_n=\sum_{i=1}^{n}Z_i$$ is terminated whenever $$S_n$$ exceeds either $$A>0$$ or $$B<0$$ for the first time, $$N$$. Here, $$N$$ is random and it is the stopping time of the stochastic process, $$N=\min\{n\geq 1:S_n>A \,\,\mbox{or}\,\, S_n.

Consider the following two cases:

Case one $$\rightarrow$$ $$Y\sim g_0$$

Case two $$\rightarrow$$ $$Y\sim f_0$$

Question: Is it true that $$E_{g_0}[1_{S_N>A}]>E_{f_0}[1_{S_N>A}]$$? where $$1$$ is the indicator function and, $$E_{h}[\cdot]$$ is the expectation of $$[\cdot]$$ when $$Y$$ follows the density $$h$$.

Clarification: To remove the confusion I suggest to simplify the problem a bit from its original version. We won't loose too much generality anyways. The transition from the distribution of $$Y$$ under $$f_0$$ or $$g_0$$ to the distribution of $$Z_i$$ under $$f_0$$ or $$g_0$$ is a bit tricky. Therefore I suggest to take $$Z_i$$ as a random variable with $$P(Z_iP(Z_i for all $$y$$. Then the problem starts right after the $$\star$$ above and which is equivalent to the comparision of the test between two random variables $$Z_i^{f_0}$$ and $$Z_i^{g_0}$$ which are stochastically ordered, namely the cdf of $$Z_i^{f_0}$$ is larger than the cdf of $$Z_i^{g_0}$$ everywhere and we compare the average hitting times of $$\sum_{i=1}^n Z_i^{f_0}$$ and $$\sum_{i=1}^n Z_i^{g_0}$$ at the upper threshold $$A$$.

Wordy explanation: For the simplified case, lets assume we are running Monte-Carlo simulations in the range $$(B,A)$$. Assume we have 1 million sequences under $$Z_i^{g_0}$$ and under $$Z_i^{f_0}$$. Due to stochastical ordering we also know that the expected value of $$Z_i^{g_0}$$ is larger than the expected value of $$Z_i^{f_0}$$ (as well as its higher moments). Then on average one should expect that the sequences under $$Z_i^{g_0}$$ are more likely to terminate at $$A$$ than the sequences under $$Z_i^{f_0}$$ because $$A>0$$ and the expected value of $$Z_i^{g_0}$$ is larger. As an example if $$E[Z_i^{g_0}]=-0.5$$, $$E[Z_i^{f_0}]=-1$$, and $$(A=2,B=-2)$$ then I would guess like for $$Z_i^{f_0}$$ may be $$50.000$$ sequences would terminate at $$A$$ and $$950.000$$ at $$B$$, whereas for $$Z_i^{g_0}$$ may be $$100.000$$ at $$A$$ and $$900.000$$ at $$B$$.

Note: This question was asked in math.stackexchange before, with no comments or answers here. I want to learn at least the way which I could follow.

Thank you very much.

• @UwF They are related to improve the understandability of the question. I updated the question with those remarks. What I mean as a way or a hint or anything related to the solution but not the question. Commented Jan 29, 2014 at 9:00
• @UwF yes thats true. $f_1$ doesn't play any role. I better delete it. I already added the link into the question. I believe that the assertion is true. I did examples for the mean shifted Gaussian densities and it was true. Actually I've never seen any single case where it doesn't hold, even with non-Gaussian densities. I will explain in a wordy way why I believe that it should hold in the question. Updating... Commented Jan 29, 2014 at 11:51
• @UwF I updated gave an example and simplified the question in the direction of your confusion. Does it seem nicer now? Commented Jan 29, 2014 at 13:29

## 2 Answers

Suppose you have two random variables $X$ and $Y$. Let $X_i$, $Y_i$, $i\ge 1$ be independent copies of $X$, $Y$, respectively.

Suppose $\mathbb P(X<a) < \mathbb P(Y<a)$ for all $a$, $S^X_n=\sum_{i=1}^n X_i$, $S^Y_n=\sum_{i=1}^n Y_i$, $A>0$, and $B<0$. Then I claim that $$\mathbb P(S^X_n\text{ hits }A\text{ before }B) > \mathbb P(S^Y_n\text{ hits }A\text{ before }B).$$

Indeed, the event that $S^X_n$ hits $A$ before $B$ is a union of countably many finite conjunctions of independent events of the form $X_i\ge x_i$, $x_i\in\mathbb Q$.

Example. If $X_1=−\pi$, $X_2=2e$ (where $e=2.7182...$), and $A=1$ and $B=−4$, then the conjunction of the two events $X_1\ge−3.5$, $X_2\ge 5$ implies that $S^X$ hits $A$ before $B$: so we can take $x_1=−3.5$ and $x_2=5$.

And if say $A$, $B$, $C$ $D$ are events such that each pair is either independent according to two distributions $\mathbb P_1$, $\mathbb P_2$, or one part of the part is contained in the other, and moreover each event is more likely according to $\mathbb P_2$ than according to $\mathbb P_1$, then $(A\cap B)\cup (C\cap D)$ is more likely according to $\mathbb P_2$. Say $A\subseteq C$ and apart from that they are independent. Then $$\mathbb P_2((A\cap B)\cup(C\cap D)) = \mathbb P_2(A\cap B) + \mathbb P_2(C\cap D) - \mathbb P_2(A\cap B\cap D)$$ $$= \mathbb P_2(A)\mathbb P_2(B) + \mathbb P_2(C)\mathbb P_2(D) - \mathbb P_2(A)\mathbb P_2(B)\mathbb P_2(D)$$ is bigger than the corresponding expression for $\mathbb P_1$ since the quantity $ab+cd-adb$ is increasing in each variable for variable values in $[0,1]$ with $a\le c$. This should generalize to infinite unions of finite intersections of such events...

• First of all thanks for the answer. I couldnt understand this part very well "the event that $S^X_n$ hits $A$ before $B$ is a union of countably many finite conjunctions of independent events of the form $X_i\ge x_i$, $x_i\in\mathbb Q$." could you please give a bit more details? Commented Jan 29, 2014 at 22:30
• ohhh I see your point. It was only when $N=2$ and the conjunction is an increasing function for all $N$ so that it applies to all $N$ and as a result the assertion is complete. I must only check if I could find a proof for the countably many conjunctions. Thanks again. Where should I search for that? Commented Jan 29, 2014 at 23:34
• I have still a problem about it. $X_1\geq -3.5$ is necessary but seems not sufficient. Because there are $X_1$ which are greater than $-3.5$ but also greater than $4$. Those hit already $A$ before $B$ at the previous step. I have difficulties to link the probability that we are after $P(X_1+X_2>A,B<X_1<A)$ to $P(X_1>-3.5,X_2>5)$ as in the given example. Commented Jan 30, 2014 at 14:14
• Okay we take $N=2$, $N=3$... and for all $N$, we show that the probability that $S_n^X$ reaches $A$ before it reaches $B$ is greater under $X$ than under $Y$, then we conclude that $\mathbb P(S^X_n\text{ hits }A\text{ before }B) > \mathbb P(S^Y_n\text{ hits }A\text{ before }B)$. This is also the same with $E_{g_0}[1_{S_N>A}]>E_{f_0}[1_{S_N>A}]$. In your previous answer you had $A$ and $B$ and then $A$, $B$ and $C$. why is it different with $4$ variables now? We are still missing the monotonicity condition for all $N$ right? Commented Jan 30, 2014 at 14:43
• No no I didnt mix them with the former $A$ and $B$. I only didnt understand why you changed it to $4$ events. Before it was $2$ and $3$, respectively. At any case we will still need that the difference of the conjunction under $\mathbb{P}_2$ and $\mathbb{P}_1$ is greater than $0$ or monotone increasing. Commented Jan 30, 2014 at 15:32

Here's a discrete example to show its not true. Stochastic monotonicity says little about likelihood ratios. In this example you can replace point masses by uniforms with short supports to get continuous. Let $g_0$ and $g_1$ be supported on $\{0,1,2,3\}$, $g_0$ assigning probabilities $1/4$, $3/8$, $1/8$, and $1/4$, $g_1$ assigning $1/4$, $1/8$, $3/8$, and $1/4$. So $G_0 \geq G_1$. Fudging with epsilon adjustments to the probabilities can make it strict. Consider two possibilities for $f_0$: $f_{01}$ assigns mass $1-\epsilon$ to 0 and $\epsilon$ to $1$, where $\epsilon$ is small. $f_{02}$ assigns mass $1-\epsilon$ to 0 and $\epsilon$ to $2$.

The $f_{01}$ distribution of $Z$ puts mass on $0$ and $\log(1/3)$, while $f_{02}$ has $Z$ with mass on $0$ and $\log(3)$. Thus the two $Z$ walks go in opposite directions, so one must be a counter example.

• Very interesting. Thx for the answer. So $f_{02}$ is a counter example. It gets benefit of the fact that from $f_0,g_0,g_1$ we can have density of $Z$ which is not ordered as mentioned in the continuation of the question. I think such a counter example doesnt exist if $\log(g_1/g_0)(y)$ is an increasing function do you think so? Commented Jan 30, 2014 at 11:41