Show that this process is not a martingale

Question

I am cross-posting this question from MSE since I did not received any answer, furthermore I tried asking some professors in my university but still we could not find an answer. The most surprising thing is that this exercise was taken from an introductory text on Stochastic Analysis. (Introduction to Stochastic Integration by Hui-Hsiung Kuo)

Assume we have the following stochastic process:

$$X_t=\int_0^t e^{B(s)^2}dB(s)\, ,0\leq t \leq 1$$

where $(B)_{t\geq 0}$ is a Brownian Motion.

I have to show that $X_t$ is not a martingale.

I know that if $t< \frac 1 4$ then $\int_0^t \mathbb E(e^{2B(s)^2})ds < \infty $ and then the process is a martingale, this makes me think that $X_t$ is actually a local martingale, but I don't see how to prove that it's not a proper martingale.

Thanks in advance.

Answer 1

Here's an approach that comes from Li, Xue-Mei, Strict local martingales: examples, Stat. Probab. Lett. 129, 65-68 (2017). ZBL1386.60159, https://arxiv.org/abs/1609.00935. Indeed, she mentions this very example after Corollary 4 (bottom of page 4 in the arXiv version).

Lemma. Suppose $X_t$ is a continuous martingale, and let $\langle X \rangle_t$ be its quadratic variation. Then for every $0 < \alpha < 1$ and every $t$, we have $E[\langle X \rangle_t^{\alpha/2}] < \infty$.

Proof. Let $M_t = \sup_{0 \le s \le t} |X_s|$. By Doob's maximal inequality, for any $x > 0$ we have $$P(M_t \ge x) \le \frac{1}{x} E|X_t|.$$ So $$\begin{align*} E[M_t^\alpha] &= \int_0^\infty P(M_t^\alpha \ge x)\,dx \\&\le 1 + \int_1^\infty P(M_t^\alpha \ge x)\,dx \\& \le 1 + E|X_t| \cdot \int_1^\infty x^{-1/\alpha}\,dx \\&< \infty.\end{align*}$$ Then by the Burkholder-Davis-Gundy inequality we have$$E[\langle X \rangle_t^{\alpha/2}] \le C_\alpha E[M_t^\alpha] < \infty$$ as desired. $\Box$

Now for the process at hand, we have $\langle X \rangle_t = \int_0^t e^{2 B_s^2}\,ds$. By Hölder's inequality or Jensen's we have $\langle X \rangle_t^\alpha \ge t^{\alpha-1} \int_0^t e^{2 \alpha B_s^2}\,ds$, so by Fubini/Tonelli $E[ \langle X \rangle_t^\alpha] \ge t^{\alpha - 1} \int_0^t E[e^{2 \alpha B_s^2}]\,ds$. The integrand is infinite for all $s > 1/4$, so by the lemma $X_t$ cannot be a martingale for any $t > 1/4$.

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I feel like perhaps there should be a more direct way to relate $E[\langle X\rangle_t^{\alpha/2}]$ to $E[X_t]$, perhaps by some clever application of Itô's formula and Hölder's inequality, but I don't quite see how.

Answer 2

Let $B_t:=B(t)$. By the Itô formula $$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$ with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have \begin{equation*} X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt. \tag{1} \end{equation*} From here, it is not hard to see that $EX_1$ does not exist.

Indeed, consider the event \begin{equation*} A:=\{B_u\in[b,b+1/b],B_1-B_u<-2/b,M_u>-b\}, \end{equation*} where $b\to\infty$, \begin{equation*} u:=1-1/b^2, \end{equation*} \begin{equation*} M_u:=\min_{t\in[u,1]}(B_t-B_u). \end{equation*} Note that on the event $A$ we have $B_u\ge b$, $B_1<b-1/b$ and $B_t>0$ for all $t\in[u,1]$. Therefore and because (by the l'Hospital rule) $f(b)\sim e^{b^2}/(2b)$, it follows from (1) that on $A$ \begin{align*} X_1&\le\frac{e^{(b-1/b)^2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &=\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &\le\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u (\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt, \end{align*} where $B_t^u:=B_t-\frac tu\,B_u$; for the latter, displayed inequality, we recall that $B_u\ge b$ on $A$ and use the fact that $se^{s^2}$ is increasing in real $s$.

The Brownian bridge $(B_t^u)_{t\in[0,u]}$ is a zero-mean (Gaussian) process independent of $(B_u,B_1-B_u,M_u)$; so, $(B_t^u)_{t\in[0,u]}$ is independent of the event $A$. Introduce also the event $$C_x:=\{\max_{t\in[0,u]}|B_t^u|\le x\}$$ for real $x>0$, which allows us to use the Fubini theorem to get \begin{align*} EX_1\,1_{A\cap C_x}\le\Big(&\frac{e^{b^2-2}}{(2+o(1))b}P(C_x) \\ &-\int_0^u E(\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}1_{C_x}\,dt\Big)\,P(A). \tag{2} \end{align*} Because $g_a(b):=(a+b) e^{(a+b)^2}+(a-b)e^{(a-b)^2}$ is convex and even in real $b$ for each real $a\ge0$, we have $g_a(b)\ge g_a(0)=2ae^{a^2}$. Therefore and because the distribution of the Brownian bridge $(B_t^u)_{t\in[0,u]}$ is symmetric,
\begin{align*} E(a+B_t^u) \exp\{(a+B_t^u)^2\}1_{C_x} =\tfrac12\,Eg_a(B_t^u)\,1_{C_x} \ge ae^{a^2}\,P(C_x) \end{align*} for $a\ge0$. So, by (2), \begin{align*} EX_1\,1_{A\cap C_x}&\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u \tfrac tu\,b \exp\{(\tfrac tu\,b)^2\}\,dt\Big)\,P(C_x)P(A) \\ &=\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\frac u{2b} \, (e^{b^2}-1)\Big)\,P(C_x)P(A) \\ &=-e^{b^2(1+o(1))}\,P(C_x)P(A). \end{align*}

On the other hand, letting now $x\to\infty$, we have $P(C_x)\to1$ and, still with $b\to\infty$, \begin{align*} P(A)&=P(B_u\in[b,b+1/b])P(B_1-B_u<-2/b,M_u>-b) \\ &\ge P(B_u\in[b,b+1/b])[P(B_1-B_u<-2/b)-P(M_u\le-b)] \\ &=e^{-b^2/(2+o(1))}[P(B_1<-2)-o(1)]=e^{-b^2/(2+o(1))}. \end{align*} Thus, \begin{align*} EX_1\,1_{A\cap C_x}&\le-e^{b^2(1+o(1))}\,(1-o(1))\,e^{-b^2/(2+o(1))}=-e^{b^2/(2+o(1))}\to-\infty, \end{align*} which shows that indeed $EX_1$ does not exist.