What exactly means the notation $\mathbb{E}_xX_t$

Question

I can't find a rigorous introduction of the notation $\mathbb{E}_xX_t$ for a stochastic process $X$. Where I have seen it so far the authors normally just say that it means that the process starts at $x$. But just with this intuition the following formula would allways make sense to me: $$ \mathbb{E}_xX_t=\mathbb{E}(X_t+x). $$ Nevertheless it seems that this is true for Levy processes but not for general Markov processes. Could someone give a rigorous introduction of the notation $\mathbb{E}_x$ that clears this confusion?

Answer 1

A Markov process is a tuple $$(X_t,t \geq 0, \mathbb{P}^x, x \in \mathbb{R}^d, \mathcal{F}_t, t \geq 0)$$ consisting of a stochastic process $(X_t)_{t \geq 0}$ which is adapted to the filtration $(\mathcal{F}_t)_{t \geq 0}$ and a family of probability measures $(\mathbb{P}^x)_{x \in \mathbb{R}^d}$. This tuple is called a Markov process if $\mathbb{P}^x(X_0 = x)=1$ and $(X_t)_{t \geq 0}$ satisfies the Markov property, i.e. $$\mathbb{P}^x(X_t \in B \mid \mathcal{F}_s) = \mathbb{P}^{X_s}(X_{t-s} \in B) \quad \mathbb{P}^x\text{-a.s.} \tag{1}$$ for all $s \leq t$ and $x \in \mathbb{R}^d$. For each $x \in \mathbb{R}^d$, we write $$\mathbb{E}^x f(X_t) := \int_{\Omega} f(X_t) \, d\mathbb{P}^x.$$

In general, $$\mathbb{E}^x f(X_t) = \mathbb{E}^0 f(x+X_t) \tag{1}$$ does not hold true. To see this, consider for instance the process $$X_t(\omega) := \omega \cdot e^t$$ on $\Omega := \mathbb{R}$. If we define $\mathbb{P}^x := \delta_x$ and $\mathcal{F}_t := \{\emptyset,\Omega\}$, then it is not difficult to see that $(X_t)_{t \geq 0}$ is a Markov process. However,

$$\mathbb{E}^x f(X_t)= f(x e^t) \neq f(x) = f(x+0 \cdot e^t) = \mathbb{E}^0 f(x+X_t),$$

i.e. $(1)$ is not satisfied.

Lévy processes are, essentially, the only processes satisfying $(1)$ for a large class of functions $f$ (e.g. all bounded and continuous functions). Recall that a Lévy process $(L_t)_{t \geq 0}$ on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ has càdlàg sample paths, independent and stationary increments and satisfies $L_0 = 0$ almost surely. In order to show that $(L_t)_{t \geq 0}$ is a Markov process, we have to find a suitable family of probability measures $\mathbb{P}^x$. One possible approach is to define a larger space $\tilde{\Omega} := \mathbb{R}^d \times \Omega$ endowed with the product $\sigma$-algebra $\tilde{A} := \mathcal{B}(\mathbb{R}^d) \otimes \mathcal{A}$. Then

$$\mathbb{P}^x := \delta_x \otimes \mathbb{P}$$

defines a measure on $(\tilde{\Omega},\tilde{\mathcal{A}})$. Moreover, we set $\tilde{L}(t,(x,\omega)) = x+L_t(\omega)$. The process $(\tilde{L}_t)_{t \geq 0}$ has still independent and stationary increments and $\mathbb{P}^x(\tilde{L}_0 = x)=1$. Moreover, because of the independence and stationarity of the increments, it is not difficult to see that $(\tilde{L}_t)_{t \geq 0}$ is a Markov process. (Here, it is crucial that $(L_t)_{t \geq 0}$ is a Lévy process; for an arbitrary stochastic process $(L_t)_{t \geq 0}$ this construction does not yield a Markov process!) Finally, it follows from the very definition of $\mathbb{P}^x$ that $$\mathbb{E}^x f(\tilde{L}_t) = \mathbb{E}f(x+L_t),$$ i.e. $(1)$ holds.

As mentioned above, it is possible to show under certain additional assumptions on the associated semigroup $T_t f(x) := \mathbb{E}^x f(X_t)$ that Lévy processes are the only stochastic processes satisfying $(1)$ for a large class of functions. This is, however, not easy to prove.

Answer 2

A stochastic process $X$ is simply a random variable taking values in a space of functions, for example the set of all continuous functions from $[0,\infty)$ to $\mathbb R$. Since $X$ is a random variable, it has a law, which is a probability measure on the space of functions.

Oftentimes, one considers stochastic processes as solutions to stochastic differential equations, in which case there is a freedom to choose the starting point $X_0$. If we start $X_0$ at some deterministic value $x$, then the law of the resulting process started $x$ is denoted by $\mathbb P_x$. As mentioned above, this is a probability measure on the space of functions (let's call this space $\Omega$ from now on, for example $\Omega=C([0,\infty),\mathbb R)$.)

Now $\mathbb E_x$ denotes expectation with respect to the measure $\mathbb P_x$. In other words, for an arbitrary stochastic process $Y$, we define $\mathbb E_x Y=\int_{\Omega}Y\ d\mathbb P_x$.

Thus we have a collection of different measures $\{\mathbb P_x\colon x\in\mathbb R\}$ on the same space $\Omega$, one for each value of $x\in\mathbb R$. Note that these measures are mutually singular, and $\mathbb P_x(X_0=x)=1$ for all $x$.

Lastly, you are asking about whether $\mathbb E_x X_t=\mathbb E(X_t+x)$. This will be the case if the distribution $\mathbb P_x$ is equal to the distribution of $\mathbb P_0$ after shifting by $x$. Formally, there is a translation operator $\tau_x\colon \Omega\to\Omega$ sending a (deterministic) process $(y_t)_{t\geq 0}$ to $(y_t+x)_{t\geq 0}$. Then if $\mathbb P_x$ is the pushforward of $\mathbb P_0$ under the map $\tau_x$, denoted $\mathbb P_x=(\tau_x)_*\mathbb P_0$, then $\mathbb E_x X_t=\mathbb E_0(X_t+x)$ for all $x$ and for all $t$.

Answer 3

It's $$\mathbb E[X_t\mid X_0=x]$$