# Probability that a stochastic process is below a special random level

Given a stochastic process $x(t)$ over time $t \in [0,T]$, and a given (deterministic) $\tau$, where $0<\tau<T$, define a random variable $x^{*}$ as $$x^{*} \triangleq \inf\bigg\{y: \int_{0}^{T} \mathbf{1}_{\{x(t) \leq y\}}\:dt = \tau\bigg\}$$ How to compute the probability $\mathbb{P}\left(x(t) \leq x^{*}\right)$? In words, this asks to compute the probability that a stochastic process is below the lowest (random) level such that the time measure corresponding to the part of the process that is below the level mentioned, is equal to $\tau$. Is there a name for such a thing? I will appreciate any reference or suggestions on how to compute the probability. I am also interested in special cases such as $x(t)$ being a (continuous time countable state) Markov chain.

• Typically, a value of $y$ that satisfies $\int_0^T 1_{\{x(t)\leq y\}} dt = \tau$ will not exist (particularly when $x(t)$ takes discrete values). You might want to change your definition to: $$x^* = \inf\left\{ y : \int_0^T 1_{\{x(t)\leq y\}} dt \geq \tau\right\}$$ Commented Aug 27, 2015 at 3:51

Let's redefine $x^*$ according to my comment above: $$x^* = \inf\left\{ y : \int_0^T 1_{\{x(t)\leq y\}} dt \geq \tau\right\}$$ Assume $x(t)$ takes values in a state space of nonnegative integers, so $x(t) \in \{0, 1, 2, \ldots\}$ for all $t$. Assume that $x(0)=0$. Then the infimum value is achieved by a particular integer $x^* \in \{0, 1, 2, \ldots\}$ and we have: $$\frac{1}{T}\int_0^T 1_{\{x(t)\leq x^*\}} dt \geq \frac{\tau}{T}$$ The above holds for all realizations of the random process $x(t)$ (and the corresponding random variable $x^*$). Taking expectations of both sides gives: $$\frac{1}{T}\int_0^T Pr[x(t)\leq x^*] dt \geq \frac{\tau}{T}$$ This does not tell you what the value of $Pr[x(t) \leq x^*]$ is for a particular time $t$, but says that if we average these probabilities over the duration of the interval, the result is at least $\tau/T$.
If you assume the Markov chain has a steady state distribution, you can say a lot as $T\rightarrow\infty$. In particular, let $X$ be a random variable with distribution equal to the steady state distribution. Then with prob 1, and for all integers $y$: $$\lim_{T\rightarrow\infty} \frac{1}{T}\int_0^T 1_{\{x(t)\leq y\}}dt = \lim_{t\rightarrow\infty} Pr[x(t) \leq y] = Pr[X\leq y]$$ So then you choose the smallest integer $y$ such that $Pr[X\leq y] \geq \tau/T$. Call this $y^*$. Since we have a discrete collection of probabilities $Pr[X \leq y]$ for $y \in \{0, 1, 2, \ldots\}$, we typically have: $$Pr[X \leq y] > \tau/T > Pr[X \leq y-1]$$ In this case you can show that $x^*$ converges to $y^*$ in probability as $T\rightarrow\infty$, and: $$\lim_{t\rightarrow\infty} Pr[x(t)\leq x^*] = \lim_{t\rightarrow\infty}Pr[x(t)\leq y^*] = Pr[X\leq y^*]$$