I am trying to set up utilities for an infinitely repeated game and I am having some trouble figuring out how to write the correct functional form. This game has a stochastic component where a realization for variable p is drawn in each period from some distribution $g(\cdot)$. An agent is choosing between two actions, A and B. When choosing A, the agent gets utility
$A = x*p$
When choosing B, utility is
$B = w$
I need to figure out an expression for the probability that $ B>A $. In a single period game this is simply $Pr(B>A) = Pr(w>xp) = Pr(w/x > p)$ with CDF $G(w/x)$. I'm having trouble getting this into a repeated game context. With discounted average utilities, the long-run payoff for A is
$A = xp(1-\delta) + \delta(Pr(B>A)*B + (1 - Pr(B>A))*A)$
Where $0< \delta <1$ is the discount parameter. Payoff for B
$B = w(1-\delta) + \delta(Pr(B>A)*B + (1 - Pr(B>A))*A)$
In the long run context, the tricky part is that the $Pr(B>A)$ in a given period is a function of $Pr(B>A)$ in future periods. Writing out $Pr(B>A)$ with long-run utilities out gives me
$Pr(B>A) = Pr[w(1-\delta) + \delta(Pr(B>A)*B + (1 - Pr(B>A))*A) > xp(1-\delta) + \delta(Pr(B>A)*B + (1 - Pr(B>A))*A)]$
Is there some way to derive a clean expression for $Pr(B>A)$ that doesn't have this kind of recursion?
If you look at $B-A$ from your long expressions you seem to get this equal to $(w-xp)(1-\delta)$ though I am not sure where the $(1-\delta)$ comes from, even if it does not seem to matter.
You could then say $\Pr(B \gt A)= \Pr(B -A \gt 0) = \Pr(p \lt w/x)$ as you had before. This seems intuitively sensible: the infinite repeats with discounting may affect the total value, but should not affect which of $A$ or $B$ gives the higher expected benefit.
You could have found this from your final expression: $\Pr[w(1-\delta) + \delta(Pr(B>A)\times B + (1 - Pr(B>A))\times A) > \\ \quad xp(1-\delta) + \delta(Pr(B>A)\times B + (1 - Pr(B>A))\times A)] \\= \Pr[w(1-\delta) > xp(1-\delta) ] \\ = \Pr[w > xp ]= \Pr[p < w/x ] $