# Bounding the exponential of a stochastic integral

Recently I found this question on bounding

$$$$\mathbb{E} \left[ \exp\left(\sup_{0\leq s\leq t} \left| \int_0^s f_u \mathrm{d}W_u \right| \right) \right],$$$$

where $$f_u$$ is adapted to the Wiener process $$W_u$$ and uniformly bounded by a constant $$C$$.

So far no answer was provided. However in the comments it was suggested to consider a random time change. That is $$$$\int_0^s f_u \mathrm{d}W_u = W_{_s}$$$$ for the quadratic variations process $$_s = \int_0^s f^2_u \mathrm{d}u.$$ Using the uniform bound on $$f$$, the quadratic variation satisfies $$$$_s \leq C^2s.$$$$ Moreover, by the reflection principle $$$$\mathbb{P} \left( \sup_{0\leq s\leq t} \left| \int_0^s f_u \mathrm{d}W_u \right| \geq \lambda \right) \leq 4 \mathbb{P} \left( W_{C^2 t} \geq \lambda \right).$$$$

How can I go on and obtain a bound on the above expectation?

Since $$\mathbb{E}(Y) = \int_{(0,\infty)} \mathbb{P}(|Y| \geq r) \, dr$$ for any non-negative random variable $$Y$$, we have \begin{align} \mathbb{E}\exp(|X|) = \int_{(0,\infty)} \mathbb{P}( \exp(|X|) \geq r) \, dr &= \int_{(0,\infty)} \mathbb{P}(|X| \geq \ln(r)) \, dr \\ &= \int_{(-\infty,\infty)} \exp(u) \mathbb{P}(|X| \geq u) \, du \tag{1} \end{align} for any random variable $$X$$. Using this estimate for $$X:=M_t := \sup_{s \leq t} \int_0^t f(s) \, dW_s$$ it follows from the probability estimate which you gave at the end of your question that \begin{align*} \mathbb{E}\exp(|M_t|) &\stackrel{(1)}{=} \int_{\mathbb{R}} \exp(u) \mathbb{P}(|M_t| \geq u) \, du \\ &\leq 2 \int_{\mathbb{R}} \exp(u) \mathbb{P}(|W_{C^2 t}| \geq u) \, du \\ &\stackrel{(1)}{=} 2 \mathbb{E}\exp(|W_{C^2 t}|).\end{align*} As $$W_{C^2 t}$$ is Gaussian with mean 0 and variance $$C^2 t$$, we obtain that \begin{align*} \mathbb{E}\exp(|M_t|) \leq 2 \mathbb{E}\exp(W_{C^2 t}) + 2 \mathbb{E}\exp(-W_{C^2 t}) &= 4 \mathbb{E}\exp(W_{C^2 t}) \\ &= 4 \exp \left( \frac{C^2 t}{2} \right). \end{align*}