Question about Conditional Expectation and Brownian Motion I have been struggling with this problem for quite some time. Suppose $B_t$ is the standard Brownian motion, then let $X_t= e^{B_t}$. I calculated the expectation as $$E[X_t]=e^{t^2/2}$$ and the variance as $$Var[X_t]=e^{t^2}[e^{t^2}-1]$$ I believe these are the correct solutions, but I am not absolutely sure and if they are incorrect, I would like to go into more detail about how I found those solutions to get some feedback, but in the interest of time, the more pressing part is that I am meant to calculate $$\lim_{h\rightarrow 0^+}E[\frac{f(X_{t+h})-f(X_t)}{h}|X_t=x]$$ With f assumed to be nice enough (I assume nice enough to swap the integral and the limit). I have attempted this problem several times. I used the form of $X_t$ and Taylor Thm. to write: $$X_{t+h}=e^{B_t}+e^{B_t}(B_{t+h}-B_t)+\frac{1}{2}e^{B_t}(B_{t+h}-B_t)^2+...$$
So we have $$f(X_{t+h})-f(X_t)=f'(X_t)(X_{t+h}-X_t)+\frac{1}{2}f''(X_t)(X_{t+h}-X_t)^2+...$$
$$=f'(X_t)(e^{B_t}(B_{t+h}-B_t)+\frac{1}{2}e^{B_t}(B_{t+h}-B_t)^2+...)+\frac{1}{2}f''(X_t)(e^{B_t}(B_{t+h}-B_t)+\frac{1}{2}e^{B_t}(B_{t+h}-B_t)^2+...)^2+...$$
By assumption $B_{t+h}-B_t \sim N(0,h).$ So to me it seems like the first term should get wiped out in the average. $E[(B_{t+h}-B_t)^2]=Var[B_{t+h}-B_t]+E[B_{t+h}-B_t]^2=h.$ So this term should survive in the average as $\frac{1}{2}xf'(x).$ 
I have a couple questions at this point. 
1) Is what I've done correct so far?
2) Are there more surviving terms?
3) What did I just compute. What does it mean? It's clearly some kind of derivative, but why use this particular process?
 A: 
1) Is what I've done correct so far?

Your results for expectation and variance seem incorrect. Provided that $B_t\sim\mathcal{N}(0,t)$ with $t$ being its variance, the correct results should be
\begin{align}
\mathbb{E}X_t&=\mathbb{E}e^{B_t}=\int_{\mathbb{R}}e^x\frac{1}{\sqrt{2\pi t}}e^{-x^2/\left(2t\right)}{\rm d}x=e^{t/2},\\
\mathbb{D}X_t&=\mathbb{E}X_t^2-\left(\mathbb{E}X_t\right)^2=\mathbb{E}e^{2B_t}-\left(\mathbb{E}e^{B_t}\right)^2=\int_{\mathbb{R}}e^{2x}\frac{1}{\sqrt{2\pi t}}e^{-x^2/\left(2t\right)}{\rm d}x-e^t=e^t\left(e^t-1\right).
\end{align}
I am afraid you have taken $t$ as $t^2$ by mistake.
Your result for limit happens to be incomplete. The correct limit should be
$$
\frac{1}{2}\left(xf'(x)+x^2f''(x)\right).
$$
I am afraid you have left out the $f''(x)$ term, which has been included in the second step from the last in your derivation.

2) Are there more surviving terms?

Your expansion to the quadratic term is indeed a wise choice. I am not sure if you have in mind Ito calculus, especially Ito's lemma, but keeping the quadratic terms coincides with the concept of quadratic variation. Therefore, no more term would survive with mathematical rigor (here I assume that you did not leave out the $f''(x)$ term by mistake).

What did I just compute. What does it mean?

As is mentioned above, you were re-deriving the quadratic variation of standard Brownian motion.
To make your derivations mathematically rigorous, I will be following the schematic way to restate your result. First of all,
\begin{align}
\frac{f(X_{t+h})-f(X_t)}{h}&=\frac{1}{h}\int_{s\in\left[0,h\right]}{\rm d}f(X_{t+s})\\
&=\frac{1}{h}\int_{s\in\left[0,h\right]}\left(f'(X_{t+s}){\rm d}X_{t+s}+\frac{1}{2}f''(X_{t+s}){\rm d}\left<X\right>_{t+s}\right),
\end{align}
where we have used Ito's lemma
$$
{\rm d}g(Y_t)=g'(Y_t){\rm d}Y_t+\frac{1}{2}g''(Y_t){\rm d}\left<Y\right>_t
$$
with $\left<Y\right>_t$ being the quadratic variation of $Y_t$.
Thereafter, since $X_t=e^{B_t}$, Ito's lemma yields
$$
{\rm d}X_t={\rm d}e^{B_t}=e^{B_t}{\rm d}B_t+\frac{1}{2}e^{B_t}{\rm d}\left<B\right>_t=e^{B_t}{\rm d}B_t+\frac{1}{2}e^{B_t}{\rm d}t,
$$
where we have used $\left<B\right>_t=t$. This relation, as per the fundamental identity, reveals that
$$
{\rm d}\left<X\right>_t=\left(e^{B_t}\right)^2{\rm d}\left<B\right>_t=e^{2B_t}{\rm d}t.
$$
With these results, we eventually obtain
\begin{align}
\frac{f(X_{t+h})-f(X_t)}{h}&=\frac{1}{h}\int_{s\in\left[0,h\right]}\left[f'(X_{t+s})\left(e^{B_{t+s}}{\rm d}B_{t+s}+\frac{1}{2}e^{B_{t+s}}{\rm d}s\right)+\frac{1}{2}f''(X_{t+s})e^{2B_{t+s}}{\rm d}s\right]\\
&=\frac{1}{h}\int_{s\in\left[0,h\right]}\left[H_{t+s}{\rm d}B_{t+s}+\frac{1}{2}\left(f'(X_{t+s})e^{B_{t+s}}+f''(X_{t+s})e^{2B_{t+s}}\right){\rm d}s\right],
\end{align}
where $H_{t+s}=f'(X_{t+s})e^{B_{t+s}}$.
Define
$$
Z_t=\int_0^tH_s{\rm d}B_s.
$$
Since $B_t$ is a martingale, according to Ito calculus, $Z_t$, as a stochastic integral of a martingale, is also a martingale. Note that
$$
\int_{s\in\left[0,h\right]}H_{t+s}{\rm d}B_{t+s}=Z_{t+h}-Z_t.
$$
As per the definition of martingale, we have
$$
\mathbb{E}\left(Z_{t+h}-Z_t|\mathcal{F}_t\right)=0,
$$
where $\mathcal{F}_t$ means the given information no later than the moment $t$ (provided that $X_t=x$, $B_t$ is known as well; so are $H_t$ and $Z_t$).
Consequently,
$$
\mathbb{E}\left(\frac{f(X_{t+h})-f(X_t)}{h}\Bigg|X_t\right)=\mathbb{E}\left[\frac{1}{2h}\int_{s\in\left[0,h\right]}\left(f'(X_{t+s})e^{B_{t+s}}+f''(X_{t+s})e^{2B_{t+s}}\right){\rm d}s\Bigg|X_t\right].
$$
For simplicity, define
$$
g(x)=xf'(x)+x^2f''(x).
$$
With this definition, the last result witnesses further simplification
$$
\mathbb{E}\left(\frac{f(X_{t+h})-f(X_t)}{h}\Bigg|X_t\right)=\mathbb{E}\left[\frac{1}{2h}\int_{s\in\left[0,h\right]}g(X_{t+s}){\rm d}s\Bigg|X_t\right]=\frac{1}{2h}\int_0^h\mathbb{E}\left(g(X_{t+s})|X_t\right){\rm d}s.
$$
Since $\mathbb{E}\left(g(X_{t+s})|X_t\right)$ is no more than a function of $s$, denote
$$
\alpha(s)=\mathbb{E}\left(g(X_{t+s})|X_t\right).
$$
Thus
$$
\mathbb{E}\left(\frac{f(X_{t+h})-f(X_t)}{h}\Bigg|X_t\right)=\frac{1}{2h}\int_0^h\alpha(s){\rm d}s.
$$
Therefore,
\begin{align}
\lim_{h\to 0^+}\mathbb{E}\left(\frac{f(X_{t+h})-f(X_t)}{h}\Bigg|X_t\right)&=\lim_{h\to 0^+}\frac{1}{2h}\int_0^h\alpha(s){\rm d}s\\
&=\frac{1}{2}\alpha(0)\\
&=\frac{1}{2}\mathbb{E}\left(g(X_t)|X_t\right)\\
&=\frac{1}{2}g(x)\\
&=\frac{1}{2}\left(xf'(x)+x^2f''(x)\right).
\end{align}
I hope these could be helpful for you.
