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 $\mathrm{sup}_t\, \mathbb{E} |X_t|^p < \infty$ ?


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, $$\sup_{t \geq 0} \mathbb{E}(|X_t|^p)< \infty \quad \text{for all $p>0$.}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.