For simplicity, consider the sample space {-2,-1,0,1,2}. We have two distributions, say, $f$ and $g$ that are defined on that sample space. Let $x_i$ and $y_i$, $i = 1,2,\dots$, be the i.i.d. random variables according to $f$ and $g$ respectively. Let also $E(x_i)>E(y_i)$, e.g., $f$ has probabilities {0.2, 0.2. 0.2, 0.2, 0.2} while $g$ may have {0.3,0.3,0,0.2,0.2}.
Now suppose we are given the random variables $x_i$'s sequentially and we test if the running sum is greater than 0. In particular, initially we have $h=0$ and we are given $x_1$. We get to stop and set $h=1$ if $x_1>0$; otherwise $h=0$ and we continue observing and have $x_2$. If $x_1+x_2>0$ we stop and set $h=1$ otherwise we continue. The same experiment is done on $y_i$'s.
My question is to test if $x_i$'s stop earlier than $y_i$'s, or if $P_f(h=1)>P_g(h=1)$?
My intuition tells me it should be correct, but haven't got any proof yet. Simply summing the probabilities that the experiment stops at the first observation, second observation... seems inapplicable.
Let $F$ be the cumulative distribution for the probability function $f$ for random variable $X.$ Similarly $G$ is the cdf of $g$ and r.v. $Y.$ Now suppose $F(k)\le G(k)$ for $k=-2,-1,0,1,2.$ This implies $E(X) \ge E(Y)$ but it is a stronger condition, called first-order stochastic dominance. Now let $\{X_i\}$ be iid with cdf $F$ and $\{Y_i\}$ be iid cdf $G.$ Then it can be shown that their convolutions will also be stochastically ordered: $$P(X_1+X_2+...+X_n \le k) \le P(Y_1+Y_2+...+Y_n \le k) , \text{for all } k \text{ and all } n. $$ and therefore $$P(X_1+X_2+...+X_n \gt 0) \ge P(Y_1+Y_2+...+Y_n \gt 0) , \text{for all } n. $$
See the book: Muller & Stoyan "Comparison Methods for Stochastic Models and Risks", 2002.