depending on how you interpret it, yes, because ##S_J = \sum_{i=1}^J Y_i##
... it belatedly occurs to me that I should have been using ##S_J## or given the lack of homogeneity, ##S_J^{(n)}## or something like that to denote the ##S_J## on the nth "day".
I think I merged notations in a not so great way.
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so, remember what lurks in the background for anything Wald Equality related is (a) a valid stopping rule and (b) ##E\big[J\big] \lt \infty##. These are both satisfied here though the latter takes some work to show. (If you're interested in this sort of stuff, may I suggest
https://ocw.mit.edu/courses/electri...62-discrete-stochastic-processes-spring-2011/ ? It's a lot of work, but not a lot of wasted work...)
Ignoring nuisances around boundary conditions with minimum bet size, for the time being, we stop when our cumulative bankroll = 0 or equivalently, when accumulated losses are equal to the bankroll we walked into the casino with on that day. Otherwise the game never stops. Hence we can say ##E\big[S_J\big] =\text{bankroll at start of the day}##
The equation of Wald tells us ##E\big[S_J\big] = E\big[J\big]\bar{Y}## and the same game is being played over and over (and independently) so ##\bar{Y}## which is expected payoff per game, is constant.
Now let's have some fun with scaling. Choose some ##\alpha \gt 0## and scale
##\alpha E\big[S_J\big] = E\big[\alpha S_J\big] = E\big[(\alpha J)\big]\bar{Y} = \big(\alpha E\big[J\big]\big)\bar{Y} = \alpha E\big[J\big]\bar{Y}##
And then repeatedly choosing the right scalars along the way to homogenize your starting bankroll and hence homogenize expectation of J as well. Then back into expected value of Y after
a lot of trials. That's really it.