Krdytkyu

I was looking at a proof of the Feynman-Kac theorem and I don't really understand a passage. The theorem is as follows

Let $k$ be a piecewise continuous function, $g in C^2(mathbb{R})$ and $B(t),t>0$ a standard Brownian motion. Then $$w(x,t) = Eleft{g(B(t))expleft{-int_0^t k(B(s))dsright}vert B(0)=x right}$$

is a unique solution to the following Cauchy problem $$left{begin{matrix}
frac{partial w}{partial t} = frac{1}{2}frac{partial^2 w}{partial t^2} -kw quad\
w(x,0)=g(x)
end{matrix}right.$$
for $xinmathbb{R}$ and $t>0$.

One of the proofs I've seen of this result is the following.

Consider the following identity

$$g(B(t))expleft{-int_0^t k(B(s))dsright} = \
g(B(t))expleft{-int_{Delta t}^t k(B(s))dsright} + \
g(B(t))expleft{-int_{Delta t}^t k(B(s))dsright}left(expleft{-int_0^{Delta t} k(B(s))dsright} -1right)$$

Now, keeping in mind the strong Markov property of Brownian motion and letting $mathcal{A}_{Delta t} = sigma{B(s), s leq Delta t }$, we have that

$$w(x,t) = \
Eleft{Eleft{g(B(t))expleft{-int_{Delta t}^t k(B(s))dsright} | mathcal{A}_{Delta t}right}right} + \
Eleft{Eleft{g(B(t))expleft{-int_{Delta t}^t k(B(s))dsright}left(expleft{-int_0^{Delta t} k(B(s))dsright} -1right) | mathcal{A}_{Delta t}right}right}$$

(in this step we just apply the law of conditional expectation)

Following on, the last expression equals

$(1)$ $$w(x,t) = \
Eleft{Eleft{ g(B(t-Delta t) e ^{-int_0^{t-Delta t}k(B(s))ds}right}| B(Delta t) + xright} + \
Eleft{ left( e^{-int_0^{Delta t}k(B(s))ds} -1right)Eleft{ g(B(t-Delta t))e^{-int_0^{t-Delta t}kB(s))ds}|B(Delta t) +x right}right}$$

which can be written as

$(2)$ $$E{ w(x+B(Delta t), t - Delta t)} + Eleft{left( e^{-int_0^{Delta t}k(B(s))ds} -1right)w(x+B(Delta t), t - Delta t) right} = \
Eleft{ left( w(x,t) + B(Delta t)frac{partial w}{partial x} + frac{1}{2}B^2(Delta t) frac{partial^2 w}{partial x^2}-Delta t frac{partial w}{partial t}right) (1-Delta tk(x))right} + o(Delta t)=\
w(x,t) + frac{1}{2}Delta tfrac{partial^2 w}{partial x^2} - Delta tfrac{partial w}{partial t} - Delta t k(x)w + o(Delta t)$$

dividing everything by $Delta t$ and letting $Delta t rightarrow 0$ we show that $w$ satisfies the starting Cauchy problem.

Excuse me for the clumpiness of the notation. Now, to the questions

$a)$ in $(1)$, how exactly was the strong Markov property of Brownian motion used? I know that it states that

$$ P(B(t) in A|mathcal{F}_s) = P(B(t) in A | B(s)) quad forall tgeq s quad A in mathcal{B}(mathbb{R})$$

so I can see how the conditioning to the $sigma$-algebra was dropped, but what justifies the time shift in the Brownian motion and in the integral bounds?

$b)$ in $(2)$, how is that series expansion made? I can see how it is done for the first summand, but how is it done for the second summand?