I have a chain $C_t$. At every instant $t$ an exponential random variable $X_t$ with parameter $\lambda$ is added to the chain or if the chain has a value greater than $Q$ then a value $Q$ is subtracted and $X_t$ is added so that the value of the chain at instant $t+1$ is $C_{t+1}=C_t+X_t-Q$ if $C_t\geq Q$ and it is $C_{t+1}=C_t+X_t$ if $C_t<Q$. I want to know if there is any steady state distribution that shows $Pr(C_t>Q)$.
I already know that the value of chain at any instant after the $Q$ is subtracted follows a steady state distribution. But I do not know if this can help in finding the above steady state distribution.
Any help in this regard will be much appreciated.
Assuming a steady state, the long term rate of increase from the arrival variables must be equal to the rate of decrease from subtracting by $Q$, so: $$ E[X] = QP[C_t \geq Q] \implies P[C_t \geq Q] = \frac{E[X]}{Q} $$ You will get such a steady state if $E[X] < Q$.
More formally, define the indicator function $1\{C_t\geq Q\}= 1$ if $C_t\geq Q$, and 0 else. You have $$C_{t+1} - C_t = X_t -Q1\{C_t\geq Q\}$$ Summing over $t \in \{0, 1, ..., n-1\}$ and dividing by $n$ gives: $$ \frac{C_n - C_0}{n} = \frac{1}{n}\sum_{t=0}^{n-1}X_t - Q\frac{1}{n}\sum_{t=0}^{n-1} 1\{C_t\geq Q\}$$ Suppose that $C_n/n\rightarrow 0$ with prob 1. Then: $$ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{t=0}^{n-1} 1\{C_t \geq Q\} = \frac{E[X]}{Q} \quad (w.p.1) $$ where I have used the law of large numbers to claim that $\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=0}^{n-1}X_i=E[X]$ with prob 1. Finally, it can be shown (but is not obvious) $C_n/n\rightarrow 0$ with prob 1 if and only if $E[X] \leq Q$. See below for proof.
The proof of my $C_n/n$ claim is related to my Theorem 2.4 here ("Rate stability theorem").
http://www.morganclaypool.com/doi/abs/10.2200/S00271ED1V01Y201006CNT007
It is not exactly the same system, but the same trick can be used: Add a virtual process that makes the arrival rate $\epsilon$ more than the service rate.
Claim:
Consider the iteration $C_{t+1}=C_t + X_t - Q1\{X_t\geq Q\}$, with $\{X_t\}_{t=0}^{\infty}$ i.i.d. with mean $E[X]$, and with $C_0=0$. Then $C_t/t\rightarrow 0$ with prob 1 if and only if $E[X]\leq Q$.
Proof part 1:
Suppose $E[X]>Q$. The analysis above gives (with $C_0=0$): \begin{align} \frac{C_n}{n} &= \frac{1}{n}\sum_{t=0}^{n-1}X_t - Q\frac{1}{n}\sum_{t=0}^{n-1}1\{C_t \geq Q\} \\ &\geq \frac{1}{n}\sum_{t=0}^{n-1}X_t - Q \end{align} Taking a limit and using LLN gives (with prob 1): $$ \liminf_{n\rightarrow\infty} \frac{C_n}{n} \geq E[X] - Q > 0 $$ and so we conclude $C_n/n$ does not converge to zero with prob 1. Further, we conclude that $C_n\rightarrow\infty$ with prob 1, meaning that after some finite time it will always be greater than $Q$, and so the time average service rate is exactly $Q$. As in the argument from the book above, we get an exact limit: If $E[X]>Q$ then with prob 1: $$ \lim_{n\rightarrow\infty} \frac{C_n}{n} = E[X] - Q $$
Proof part 2:
Now suppose $E[X]\leq Q$. Fix $\epsilon>0$ and define $a = Q-E[X]+\epsilon$. Note that $a> 0$. Consider the virtual process $\tilde{C}_t$ with $\tilde{C}_0=0$ and: $$ \tilde{C}_{t+1}=\tilde{C}_t + X_t + a - Q1\{\tilde{C}_t\geq Q\} $$ It can be shown (by induction) that $0 \leq C_t \leq \tilde{C}_t + Q$ for all $t$. Further, by the same argument as in the proof part 1, we get with prob 1: $$\lim_{n\rightarrow\infty} \frac{\tilde{C}_n}{n} = E[X]+a-Q=\epsilon $$ Thus: $$ 0 \leq \limsup_{n\rightarrow\infty} \frac{C_n}{n} \leq \limsup_{n\rightarrow\infty} \frac{\tilde{C}_n+Q}{n} = \epsilon $$ This holds for all $\epsilon>0$ and so $\lim_{n\rightarrow\infty} C_n/n=0$ with prob 1.