Let B = (Bt)t¸0 be a standard Brownian motion started at zero, let $X_t$ be a non negative stochastic process solving:
$dX_t=1/X_tdt+dB_t$
Compute $E[\sigma]$ when $\sigma=\inf \{ t\ge 0 : X_t= 1 \}$
I have tried using itos formula to derive an expression $E[X_\sigma]=E[\sigma]$. Is this correct and if so how can I go about evaluating this? Would $E[X_\sigma]=1$ which implies $E[\sigma]=1$?
We apply Itô's formula to $f(x) :=x^2$ and the Itô process $(X_t)_{t \geq 0}$:
$$\begin{align*} X_t^2 -X_0^2 &= 2\int_0^t X_s \, dX_s + \int_0^t 1 \langle X \rangle_s \\ &= 2 \int_0^t X_s \, dB_s + 2 \int_0^t \frac{X_s}{X_s} \, ds + \int_0^t ds \\ &= 2 \int_0^t X_s \, dB_s + 3t \end{align*}$$
Using that $\int_0^t X_s \, dB_s$ is a martingale, we find by the optional stopping theorem
$$3 \mathbb{E}(t \wedge \sigma) = \mathbb{E}(X_{t \wedge \sigma}^2) - \mathbb{E}(X_0^2)$$
By definition of $\sigma$, $|X_{t \wedge \sigma}| \leq 1$, therefore, the monotone convergence theorem yields
$$3 \mathbb{E}\sigma = \mathbb{E}(X_{\sigma}^2)-\mathbb{E}(X_0^2)=1-\mathbb{E}(X_0^2)$$
i.e. $\mathbb{E}\sigma= (1-\mathbb{E}(X_0^2))/3$.
Remark Note that this calculation is sloppy: Obviously, $\frac{1}{X_t} \, dt$ is not well-defined if $X_t=0$, i.e. you either have to assume that $X_t$ is a (strictly) positive solution to the given SDE or you have to take care of the indicator function $1_{\mathbb{R} \backslash \{0\}}$. In the latter case, you will get a different result, I guess.