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, $N=\min\{n\geq 1:S_n>A \,\,\mbox{or}\,\, S_n<B\}$.

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 $P(Z_i<y|Y\sim f_0)>P(Z_i<y|Y\sim g_0)$ 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 thecdfof $Z_i^{f_0}$ is larger than thecdfof $Z_i^{g_0}$ 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.