Large Deviation Properties of a function of a geometric random variable Suppose I have a variable $s$ that has a geometric distribution with success parameter x. So the probability of success on  trial $s$ is
$p_s = (1 - x)^{s - 1} x$,
Consider the following function of $s$  
$V(s) = \delta^s$,  
where $\delta < 1 $.  Then the expected value of this function is given by
$\sum_ 1^{\infty} V(s) p_s = \frac{\delta x} {1 - \delta (1 - x)}$.
Consider now the deviation of V(s) from its mean :
$(V(s) - \frac{\delta x} {1 - \delta (1 - x)}) $
What distribution does this have?  
What I am really interested in is the behavior of the follow ordinary stochastic differential equation:
$\dot{V} =  \delta^s  - V$,
stochastic because the law of motion depends on the random variable $s$.  The stochastic approximation technique allows me to focus on the mean dynamics, which are given by
$\dot{V} =  \frac{\delta x} {1 - \delta (1 - x)}  - V$,
whose equilibrium is simply $\hat{V} = \frac{\delta x} {1 - \delta (1 - x)}$.  
I would like to characterize the large deviation properties of the original ODE, IE calculate the exponential likelihood $V$ crosses a particular boundary $c$. 
For example, if $\delta = .95$ and $x= .01$, $\hat{V} = 0.159664$.  If we set $V_0 = 0.159664$, and let $V_t = V_{t-1} + .2 (\delta^s - V_{t-1})$, and $s$ has the above distribution, how do I calculate the expected time to $V_t$ crossing, say, $c=.4$?  What is the associated rate function?

Editing to give the background to the problem:
I am interested in the following stochastic dynamic system:

$V_t = V_{t-1} + \gamma (\delta^{S_{t-1}} - V_{t-1})$,

where $\delta <1$, $\gamma<1$, both positive, and each $S_t$ is an i.i.d. geometric random variable with success parameter $x$.  What this models is agents who have to wait a geometric length of time to get a reward of value 1, and they have time discount factor $\delta$, and they are learning  the expected value of their reward using a constant gain adaptive  learning procedure, with gain $\gamma$.  
So considering $S$ as following a Poisson, rather than Geometric, makes little difference to the sense of the problem.  The goal is


*

*Find the equilibria of the learning dynamics

*Characterize the mean time to escape from this equilibrium, the large deviation properties.

So, for large deviation properties, I should be able to use something like Cramer's theorem, I think.  As Mike says below, iterating the dynamics gives
$V_t = (1−γ)^tV_0+γ\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$
So it all depends on $\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$, the discounted sum of some random variables.
The Probability that $V_t >c$ is then given by
$Pr(V_t > c ) = Pr(\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k} > \frac{c-(1−γ)^tV_0}{\gamma})$
Let $Z:=\frac{c-(1−γ)^tV_0}{\gamma}$.  so we need the distribution of $\sum_{k=0}^{t−1}(1−γ)^{t−1−k}δ^{S_k}$, which as Mike says is a discounted sum of independent random variables, and so we have that this is approximately normal, for large $t$.
Turning now to the question of large deviations, if things were not discounted by powers of $(1-\gamma)$, we could appeal directly to Cramer's Theorem; 
$ Pr(\sum \delta^{S_k} > n a) \leq Exp[-n(r^* a - Log(E[Exp(r^* \delta^S)])] $
where $r^* $ is chosen to maximize $r a - Log(E[Exp(r \delta^S)]$.  The problem would be to simply calculate the moment generating function of  $\delta^S$.  But, I don't exactly want $Pr(\sum \delta^{S_k} > n a)$, I need $ Pr(\sum (1-\gamma)^{t-1-k} \delta^{S_k} > a)$; so not an average of $\delta^{S_k}$s, but a discounted sum.  so it is not quite a direct application.
 A: If $\delta > 0$, then the transformations you are applying to the random variable $S$ to get the quantity $T = \delta^S - \frac{\delta x}{1 - \delta(1-x)}$ are all one-to-one.  Since $S$ is discrete, this means you can construct the probability distribution of $T$ by inverting those transformations.  This would yield
$$P(T = t) = P\left(\delta^S - \frac{\delta x}{1 - \delta(1-x)} = t\right) = P\left(S = \frac{\ln\left(\frac{\delta x}{1 - \delta(1-x)} + t\right)}{\ln \delta}\right) = (1 - x)^{y-1} x,$$
where $$y = \frac{\ln\left(\frac{\delta x}{1 - \delta(1-x)} + t\right)}{\ln \delta},$$
for any value of $t$ such that $y$ is in the set $\{1, 2, \ldots\}$.
Maybe someone else can help you with the other questions.

I've thought some more about your problem.  The recursion in your dynamical system yields the expression
$$V_t = (1- \gamma)^t V_0 + \gamma \sum_{k=0}^{t-1} (1 - \gamma)^{t-1-k} \delta^{S_k}.$$
So you have a sum of discounted random variables.  If they were not discounted the distribution would be approximately normal for large $t$.  It turns out, though, that there is an analogous result for the distribution of a sum of discounted random variables -- a discounted central limit theorem, as it were.  You might have some success using this discounted central limit theorem as an approximation.
Here are the specifics.  Let $Y_v = \sum_{k=0}^{\infty} v^k X_k$, where the $X_k$'s are independent and have a common distribution function.  Let
$$Z_v = \frac{\sqrt{1-v}}{\sigma} \left(Y_v - \frac{\mu}{1-v}\right).$$
Then $Z_v$ is asymptotically normal with mean $0$ and variance $\frac{1}{1+v}$, for $v \to 1$, where $\mu$ and $\sigma$ are the common mean and standard deviation of the $X_k$'s.  (The expected value of $|X_k - \mu|^3$ must exist, too.)
Moreover, there is a bound on the error in this approximation.  Let $F_v(x)$ and $N_v(x)$ be the cdf's of the $Z_v$ random variable and of a normal with mean $0$ and variance $\frac{1}{1+v}$, respectively.  Then
$$|F_v(x) - N_v(x)| \leq \frac{C \rho \sqrt{1-v}}{\sigma^3},$$
where $\rho = E[|X_k - \mu|^3]$ and $C = 5.4$.
The reference is Hans Gerber, "The Discounted Central Limit Theorem and Its Berry-Esseen Analogue," Annals of Mathematical Statistics 42(1), pp. 389-392, 1971.
