Consider the following SDE:
$$\mathrm{d} X_t = – \lambda X_t + \mathrm{d} B_t$$
with initial condition $X_0 = x$, and where $B_t$ is a standard Brownian motion.
An application of Ito's formula gives us that the solution can be expressed as:
$$X_t = e^{- \lambda t} x + \int_0^t e^{- \lambda (t – s)} \mathrm{d} W_s$$
Is there an obvious way to check, for a given $p$, whether or not $\mathrm{sup}_t\, \mathbb{E} |X_t|^p < \infty$ ?
Best Answer
Since $$\int_0^t e^{-\lambda (t-s)} \, dW_s$$ is Gaussian (as a stochastic integral of a deterministic function with respect to Brownian motion), it follows that $X_t$ is Gaussian for each $t$. Its mean is given by
$$m_t := \mathbb{E}(X_t)= e^{-\lambda t} x + \underbrace{\mathbb{E} \left( \int_0^t e^{-\lambda(t-s)} \, dW_s \right)}_{0} = e^{-\lambda t} x$$
and the variance equals, by Itô's isometry,
\begin{align*} \sigma_t^2 := \mathbb{E}((X_t-m_t)^2) &= \mathbb{E} \left( \left| \int_0^te^{-\lambda (t-s)} \, dW_s \right|^2 \right) \\&= \int_0^t e^{-2\lambda (t-s)} \, ds \\ &= \int_0^t e^{-2 \lambda s} \, ds \\ &= \frac{1-e^{-2\lambda t}}{2\lambda}. \end{align*}
Consequently, we have $X_t \stackrel{d}{=} m_t + \sigma_t U$ for $U \sim N(0,1)$. Using that $$|x+y|^p \leq c_p |x|^p + c_p |y|^p$$ for some constant $c_p>0$, it follows that
$$\mathbb{E}(|X_t|^p) \leq c_p |m_t|^p + c_p \sigma_t^p \mathbb{E}(|U|^p).$$
This means that $\sup_{t \geq 0} \mathbb{E}(|X_t|^p) < \infty$ if $$ \sup_{t \geq 0} |m_t|^p + \sup_{t \geq 0} \sigma_t^p < \infty. \tag{1}$$
As $\sigma_t \leq 1/\sqrt{2\lambda}$ and $|m_t| \leq |x|$ for any $t>0$, we find that $(1)$ holds for any $p>0$. Hence, $$\sup_{t \geq 0} \mathbb{E}(|X_t|^p)< \infty \quad \text{for all $p>0$.}$$