I am cross-posting [this question from MSE](https://math.stackexchange.com/questions/3528142/how-to-show-this-is-not-a-martingale) 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.



Show that this process is not a martingale
Here's an approach that comes from 
*Li, Xue-Mei*, [**Strict local martingales: examples**](http://dx.doi.org/10.1016/j.spl.2017.04.025), Stat. Probab. Lett. 129, 65-68 (2017). [ZBL1386.60159](https://zbmath.org/?q=an:1386.60159), <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.


