Is $\mathbb{E}\exp \left( k \int_0^T B_t^2 \, dt \right)<\infty$ for small $k>0$?

Question

Suppose that $B$ is a Brownian motion. Does it hold that \begin{equation} \mathbb{E}\left[\exp\left(k\int_0^T[B(t)]^{2}\,dt\right)\right] <\infty\text{ ?} \end{equation} for some positive constant $k$?

my idea: I think that $\int_{0}^{T}[B(t)]^{2}dt$ is actually a Normal random variable $X\sim N(\frac{T^{2}}{2},\sigma(T))$ where $\sigma(T)$ is some function of $T$. Then, we know that $\mathbb{E}[\exp(\theta X)]$ is finite for such $X$. Am I correct?

Answer 1

Obviously,

$$\int_0^T B_t^2 \, dt \leq T \cdot \sup_{t \leq T} B_t^2$$

Since

$$\sup_{t \leq T} B_t^2 \leq \left(\sup_{t \leq T} B_t \right)^2 + \left( \inf_{t \leq T} B_t \right)^2$$

and

$$\sup_{t \leq T} B_t \sim - \inf_{t \leq T} B_t \sim |B_T|$$

(this is a direct consequence of the reflection principle), we get by the Cauchy-Schwarz inequality

$$\begin{align*} \mathbb{E}\exp \left(k \int_0^T B_t^2 \, dt \right) &\leq \sqrt{\mathbb{E}\exp \left(2kT \left[ \sup_{t \leq T} B_t \right]^2 \right)} \sqrt{\mathbb{E}\exp \left(2kT \left[ \inf_{t \leq T} B_t \right]^2 \right)}\\ &= \mathbb{E}\exp \left(2kT B_T^2 \right). \end{align*}$$

Finally, as $B_T \sim N(0,T)$, we observe that

$$\mathbb{E}\exp (2kT B_T^2) = \int_{\mathbb{R}} e^{2kT x^2} \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{x^2}{2T} \right) \, dx$$

is finite if $2k T< \frac{1}{2T}$, i.e. if $k < \frac{1}{4T^2}$.

Answer 2

I was confronted to a similar problem, here's one possible solution \begin{equation} \mathbb{E}\left[\exp\left(k\int_0^T[B(t)]^{2}\,dt\right)\right] <\infty\text{ ?} \end{equation}

First of all, we introduce the following sequence of process

\begin{equation} \ B_{t}^{n}= \sum_{i=0}^{n}\sqrt\lambda_{i}*\phi_{i}*\eta_{i} \end{equation}

where

\begin{equation} \ \eta_{i} \sim N(0,1) \\ \ \phi_{i}=\sqrt2\sin((i+\frac{1}{2})\pi*t) \\ \ \lambda_{i} = \frac{4}{\pi^2}\frac{1}{(2i+1)^{2}} \end{equation}

We will show that this sequence of process is a representation of the brownian motion on [0;1]. Then thanks to a change in variables and the invariance property of the brownian motion it will become "easier" to calcul the expectation.

This sequence of process is converging, to show that you can show that it is a Cauchy sequence in . We denote X_t that limit.

You can easily show that X_t is a gaussian process. Or a gaussian process is characterized by its covariance function and its expected function. Thanks to the Theorem of Mercer you can calculate the covariance function and the calculation of the expected function is very simple because it is the constante 0. \begin{equation} K(s,t)=cov(X_{t},X_{s}) = min(s,t) \\ E(t)= \mathbb{E}[X_{t}]=0 \end{equation}

\begin{equation} \ B_{t}^{n}= \sum_{i=0}^{\infty}\sqrt\lambda_{i}*\phi_{i}*\eta_{i}\text{ is a representation of the brownian motion on [0;1]} \end{equation}

We can use that representation to calculate

\begin{equation} \int_{0}^{T}[B(t)]^{2}dt= T^{2}*\int_{0}^{1} [\frac{B_{T*t}}{\sqrt{T}}]^{2}dt \\ \frac{B_{T*t}}{\sqrt{T}}\text{is a brownian motion, it is just the invariance by expansion of time} \end{equation}

\begin{equation} \int_{0}^{T}[B(t)]^{2}dt= \sum_{i=0}^{\infty}\lambda_{i}\eta_{i}^2\ \end{equation}

Now you "just" have to calculate the Laplace transformation of a square normal law which is painful (it is basically the calculation of a Laplace transformation for a random variable following a chi2 law) \begin{equation} \mathbb{E}\left[\exp\left(k\int_0^T[B(t)]^{2}\,dt\right)\right] = \prod_{i=0}^{\infty}\mathbb{E}[exp(T^{2}*k*\lambda_{i}*\eta_{i}^2)] \end{equation} If I have made any mistake or if you need more explanations on some particular points don't hesitate