Application of the Burkholder Davis Gundy inequality

Question

The proof of the Feynman-Kac formula uses a lemma which I need to prove, but I can not figure it out.

The lemma is the folllowing:

Let $X$ be a weak solution of $$dX_t=b(t,X_t)dt+\sigma(t,X_t)dW_t$$ With $b$ and $\sigma$ continuous and satisfying the linear growth condition: $$|b(t,x)|+|\sigma(t,x)|\leq K(1+|x|)$$ Then for any finite time $T$ and $p\geq 2$ there is a constant $C$ such that $$\mathbb{E}\sup_{t\leq T}|X_t|^p\leq Ce^{CT}(1+\mathbb{E}|X_0|^p)$$

This proof of this lemma should use the Burkholder Davis Gundy inequality:

Let $p \geq 2$. There exists a constant $C_p$ such that for all continuous local martingales $M$ with $M_0 = 0$ and all finite stopping times $T$ one has $$\mathbb{E} \sup_{t \leq T} |M_t|^p \leq C_p \mathbb{E}\langle M \rangle_T^{p/2}$$

And maybe should also use Doobs inequality.

Can anyone help me with the details?

Thanks

Answer 1

Let $(X_t)_{t \geq 0}$ be a weak solution of the given SDE, i.e. there exists a Brownian motion $(B_t)_{t \geq 0}$ such that

$$X_t = X_0 + \int_0^t b(s,X_s) \, ds + \int_0^t \sigma(s,X_s) \, dB_s.$$

Using the elementary estimate

$$(a+b+c)^p \leq 3^p (a^p+b^p+c^p), \qquad a,b,c \geq 0,$$

we find

$$|X_t|^p \leq 3^p |X_0|^p + 3^p \left| \int_0^t b(s,X_s) \, ds \right|^p + 3^p \left| \int_0^t \sigma(s,X_s) \, dB_s \right|^p. \tag{1}$$

We estimate the terms separately. By Jensen's inequality (applied with the probability measure $\frac{ds}{t}$), we get

$$\begin{align*} \left| \int_0^t b(s,X_s) \, ds \right|^p &= t^p \left| \int_0^t b(s,X_s) \, \frac{ds}{t} \right|^p\\ &\leq t^{p-1} \int_0^t |b(s,X_s)|^p \, ds \\ &\leq K^p T^{p-1} \int_0^t (1+|X_s|^p) \,ds \\ &\leq K^p T^{p-1} \int_0^t \left( 1+ \sup_{r \leq s} |X_r|^p \right) \, ds\end{align*}$$

for any $t \in [0,T]$. In the penultimate line, we have used the linear growth condition. In order to estimate the third term in $(1)$, we apply the Burkholder-Davis-Gundy inequality to obtain

$$\begin{align*} \mathbb{E} \left( \sup_{r \leq t} \left| \int_0^r \sigma(s,X_s) \, dB_s \right|^p \right) &\leq C_p \mathbb{E} \left( \left| \int_0^t |\sigma(s,X_s)|^2 \, ds \right)^{p/2} \right). \end{align*}$$

Using Jensen's inequality as above yields

$$\begin{align*} \mathbb{E} \left( \sup_{r \leq t} \left| \int_0^r \sigma(s,X_s) \, dB_s \right|^p \right) &\leq C_p t^{p/2-1} \mathbb{E} \left( \int_0^t |\sigma(s,X_s)|^p \, ds \right) \\ &\leq C_p T^{p/2-1} K^p \mathbb{E}\left( \int_0^t (1+|X_s|^p) \, ds \right) \\ &\leq C_p T^{p/2-1} K^p \int_0^t \left(1+ \mathbb{E}\left[ \sup_{r \leq s} |X_r|^p \right] \right) \, ds. \tag{3} \end{align*}$$

Adding all up, we get

$$\mathbb{E} \left( \sup_{r \leq t} |X_r|^p \right) \leq 3^p \mathbb{E}|X_0|^p + \underbrace{3^p K^p \left( T^{p-1} + T^{p/2-1} C_p \right)}_{=: \tilde{C}} \int_0^t \left(1+ \mathbb{E}\left[ \sup_{r \leq s} |X_r|^p \right] \right) \, ds.$$

This means that

$$u(t) := \mathbb{E} \left( \sup_{r \leq t} |X_r|^p \right)$$

satisfies

$$u(t) \leq (3^p \mathbb{E}|X_0|^p + \tilde{C} T) + \tilde{C} \int_0^t u(s) \, ds$$

for any $t \in [0,T]$. Now the claim follows from Gronwall's lemma.

Reference: René Schilling, Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 19 (2nd edition).