This may be a duplicate but I cannot find the corresponding question. I have been asked to show:
$\mathbb{E}[\exp(\sigma(B_t-B_s))] = \exp\left(-\dfrac{\sigma^2}{2}(s-t)\right)$
As a side note I have seen that $\mathbb{E}[B_t^{2k}] = \dfrac{(2k)!t^k}{2^k k!}$ with $\mathbb{E}[B_t^k]=0$ if k is odd, from the generator function being real.
So if this means that if $\mathbb{E}[(B_t-B_s)^{2k}] = \dfrac{(2k)!(t-s)^k}{2^k k!} $ expanding the LHS of the first equation arrives at the RHS quite easily.
I have a problem is finding how, for example, $G_{t,s} = B_t-B_s$ is a Brownian motion $~\mathscr{N}(0,t-s)$. Also, my solution really does not seem very rigorous at all and it seems that I have only managed to solve it since I have seen an expression for $\mathbb{E}[B_t^{2k}]$. In fact, the question asks to arrive at a proof from this:
$$\displaystyle P^x(B_{t_1} \in F_1, \cdots, B_{t_k} \in F_k ) = \int_{F_1\times \cdots \times F_k} p(t_1, x, x_1) \cdots p(t_k-t_{k-1}, x_{k-1}, x_k ) dx_1 \cdots dx_k $$
So what I would like, if possible, is a technique for getting to answers to questions like this. Specifically, in this question I would like to know how I can get to $\mathbb{E}[\exp(\sigma(B_t-B_s))]$ from the above. I thought I would begin with:
$$\mathbb{E}[f(X(\omega)] = \int_{\mathbb{R}^n}f(x)d\mu_x(x)$$
I am unsure how to proceed from here.
Write that $B_t - B_s \sim N(0,t-s)$ and the bilinear form as a sum of squares:
$$ E\exp (a(B_t - B_s))= \int \frac{dx}{\sqrt{2\pi}} \exp \left(a\sqrt{t-s} x - \frac{x^2}2\right) \\= \int \frac{dx}{\sqrt{2\pi}} \exp \left(-\frac 12\left(x-a\sqrt{t-s}\right)^2\right) \exp\left(\frac 12 a^2\left(t-s\right)\right) = \exp\left(\frac 12 a^2\left(t-s\right)\right) $$