Can someone help me with the following?
We have $B(t),t\geq 0$ which is a standard Brownian Motion with $B(0)=0$. I have to calculate the mean of $e^{B(2)}$. I am not sure if this ($Y(t)=e^{B(t)}$) is a geometric Brownian Motion, since my notes say this $Y(t)$ is a geometric BM for $X(t)$ a BM with drift.
But how do I calculate this $\mathbb{E}(e^{B(2)})$? I read something on using characteristic functions, however I do not see how since this involves $e^{itX}$, so a complex power.
By the way, the answer should be $e$ :)
Edit: I might also use the moment generating function, so: $\mathbb{E}(e^tX)$. For a normal distribution this is $e^{t\mu + 1/2 \sigma^2 t^2} = e^{t^3/2}$ since $B(t)$ is a $N(0,t)$. But then is the $t$ in the power of the exponent the same t as in the expectation (so 2)? Because then we would have $\mathbb{E}(e^2B(2))$ and then I am again stuck because what about the 2?
Edit: I also computed it directly: so $\int_{-\infty}^{\infty} e^{x}\frac{1}{\sqrt{2\pi t}}e^{-x^2/2t} dx$ (where I used that $B(t)$ is distributed $N(0,t)$) and then $t=2$, which gives indeed $e$. However I would also like to do it via some other way that does not involve integrals since I do not have any calculators or Wolfram Alpha during my exam ;) (Also, let me know if this correct)
Since you want to compute the expectation of $e^{B(2)}$ which is a distributional property, you can use the fact that $e^{B(2)}\sim e^{\sqrt{2}Z}$ where $Z$ is a standard normally distributed random variable. Hence, $$\mathbb{E}[e^{B(2)}] = \int_{-\infty}^{\infty} e^{\sqrt{2}z} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}z^2} dz.$$ The latter integral can be computed by mergeing together the exponential functions, completing the squares. That is, use $-\frac{1}{2}z^2 + \sqrt{2}z = -\frac{1}{2}(z-2)^2 +1$. At the end you get, $$\mathbb{E}[e^{B(2)}] = e\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(z-2)^2} dz = e.$$
Another way without using integrals could be using the Taylor series of the exponential $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$: \begin{align*} \mathbb{E}\left[e^{B(2)}\right] = \sum_{n=0}^\infty \frac{2^{n/2}}{n!} \mathbb{E}[Z^n], \end{align*} where $Z\sim N(0,1)$, but it is well known that $$\mathbb{E}[Z^n] =\begin{cases} (n-1)!!, n \mbox{ even}\\ 0, \mbox{ otherwise},\end{cases}$$ where $(n-1)!!$ is the double factorial. By properties of the double factorial (see here) it holds that $$(n-1)!! = \frac{n!}{n!!}.$$
Further, \begin{align*} \mathbb{E}\left[e^{B(2)}\right] &= 1+\sum_{\substack{n=2\\ n\mbox{ even}}}^\infty \frac{2^{n/2}}{n!!} \\ &=1+\sum_{k=1}^\infty \frac{2^{k}}{(2k)!!} \end{align*} Another crucial property of the double factorial is that $(2k)!!=k!2^k$ for all $k\geq 1$ (see here).
Hence, \begin{align*} \mathbb{E}\left[e^{B(2)}\right] &= 1+\sum_{k=1}^\infty \frac{2^{k}}{(2k)!!}\\ &=1+\sum_{k=1}^\infty \frac{1}{k!}\\ &=e. \end{align*}