If $W_t$ is a Wiener Process/Brownian motion standard, what is the expectation of:
$H_t=exp(W_t/(1+t))$
I know the answer is:
$E(H_t)=exp(t/(2(1+t)^2))$
But the solution I find is $1/2*exp(t/(2(1+t)^2))*erf((t(x-1)+x)/(sqrt(2)*sqrt(t)*(t+1))$
How can I do this? I can't do the passages!!! Thank you so much.
For $t\geq 0$, we have : \begin{align} \mathbb{E}[H_t] &= \mathbb{E}\left[\exp\left(\frac{W_t}{1+t}\right)\right] \\ &= \mathbb{E}\left[\exp\left(\underbrace{\frac{\sqrt{t}}{1+t}}_\alpha X\right)\right] \quad X\sim\mathcal{N}(0,1)\\ &=\int_\mathbb{R} \frac{1}{\sqrt{2\pi}}\exp\left(\alpha x -\frac12x^2\right)dx \\ &=\underbrace{\int_\mathbb{R} \frac{1}{\sqrt{2\pi}}\exp\left(-\frac12(x-\alpha)^2\right)dx}_A\exp(\frac12\alpha^2) \\ &= \exp\left(\frac12\alpha^2\right) \end{align} The last equality follows from the fact that the quantity $A$ is the p.d.f of gaussian variable.