Kolmogorov's backward equation with initial value

Question

I am refreshing basic financial mathematics concepts and self-learning from the text, A first course in Stochastic Calculus, by Louis Pierre Arguin.

I understand that, the transition probability density function $p(x,t|y,s)$ of a diffusion satisfies the Kolomogorov backward equation with a certain initial condition. It can be represented as a specific type of space average.

However, I do not quite follow the reasoning behind a particular step in the proof. I would like to ask for some help in clearly understanding how this step comes about and the justification for it.

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Theorem. Let $(X_t,t\geq 0)$ be a diffusion in $\mathbb{R}$ with the SDE:

$$ dX_t = \sigma(X_t)dB_t + \mu(X_t) dt $$

Let $g\in C^2(\mathbb{R})$ be such that $g$ is $0$ outside an interval. Then, the solution of the PDE with initial value

$$ \begin{align*} \frac{\partial f}{\partial t}(t,x) &= \frac{\sigma(x)^2}{2}\frac{\partial^2 f}{\partial x^2} + \mu(x)\frac{\partial f}{\partial x}\\ f(0,x) &= g(x) \end{align*} $$

has the representation:

$$ f(t,x) = \mathbb{E}[g(X_t)|X_0 = x] $$

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Proof.

Step 1. Let's fix $t$ and consider the function of space $h(x)=f(t,x)=\mathbb{E}[g(X_t)|X_0=x]$. Applying Ito's formula to $h$, we have:

\begin{align} dh(X_s) &= h'(X_s) dX_s + \frac{1}{2}h''(X_s) (dX_s)^2\\ &= h'(X_s) (\sigma(X_s)dB_s + \mu(X_s) ds) + \frac{\sigma(X_s)^2}{2}h''(X_s)ds\\ &= \sigma(X_s)h'(X_s)dB_s + \left(\frac{\sigma(X_s)^2}{2}h''(X_s) + \mu(X_s)h'(X_s)\right)ds \end{align}

In the integral form this is:

\begin{align*} h(X_s) - h(X_0) &= \int_0^s \sigma(X_u)h'(X_u)dB_u \\ &+ \int_0^u \left(\frac{\sigma(X_u)^2}{2}h''(X_u) + \mu(X_u)h'(X_u)\right)du \tag{1} \end{align*}

Step 2. Take expectations on both sides, divide by $s$ and let $s \to 0$.

(a) The expectation of the first term on the right hand side is zero, by the properties of the Ito integral.

(b) The second term on the RHS becomes,

\begin{align} &\lim_{s \to 0} \frac{1}{s} \int_0^s \mathbb{E}[ \left(\frac{\sigma(X_u)^2}{2}h''(X_u) + \mu(X_u)h'(X_u)\right) \vert X_0 = x] du \\ &= \frac{\sigma(x)^2}{2}h''(x) + \mu(x)h'(x) \tag{2} \end{align}

by the Fundamental Theorem of Calculus(FTC) (and continuity of $\sigma,\mu,h',h''$).

How does this step come about? From FTC, I know that $\int_a^b f'(u)du = f(b) - f(a)$. But, I don't follow the above step.

Step 3. As for the left-hand side, we have:

$$ \lim_{s \to 0} \frac{\mathbb{E}[h(X_s)|X_0 = x] - h(X_0)}{s} = \lim_{s \to 0} \frac{\mathbb{E}[h(X_s)|X_0 = x] - f(t,x)}{s} $$

To prove that this limit is $\frac{\partial f}{\partial t}(t,x)$, it remains to show that $\mathbb{E}[h(X_s)|X_0 = x]=\mathbb{E}[g(X_{t+s})|X_0 = x]=f(t+s,x)$.

To see this, note that $h(X_s) = \mathbb{E}[g(X_{t+s})|X_s]$. We deduce:

\begin{align*} \mathbb{E}[h(X_s)|X_0 = x] &= \mathbb{E}[\mathbb{E}[g(X_{t+s})|X_s]|X_0 = x]\\ &= \mathbb{E}[\mathbb{E}[g(X_{t+s})|\mathcal{F}_s]|X_0 = x]\\ & \{ (X_t,t\geq 0) \text{ is Markov }\} \\ &= \mathbb{E}[g(X_{t+s})|X_0 = x]\\ & \{ \text{ Tower property }\} \\ &= f(t+s,x) \end{align*}

This closes the proof. $\blacksquare$

Answer 1

The integrand of the second term (RHS) is a conditional expectation $\mathbb{E}[\xi(X_u)|X_0 = x]$, it is an average at time $u$, of the paths of the process starting at initial position $X_0 = x$, so it is a function of $u$ and $x$. So, $\mathbb{E}[\xi(X_u)|X_0 = x] = p(u,x)$. Suppressing the argument $x$, we have the representation:

\begin{align} \int_0^s p(u) du \end{align}

Recall that, if $p$ is a continuous function, then it is Riemann integrable. Further, since integration and differentiation are inverse operations, there exists a unique antiderivative $P$ given by

$$ P(s) = \int_{0}^{s}p(u)du $$

satisfying $P'(0) = p(0)$.

By the definition of the derivative:

$$P'(0) = \lim_{s \to 0} \frac{P(s) - P(0)}{s} = \lim_{s\to 0} \frac{P(s)}{s} = p(0) \quad \{ P(0)=0 \text{ by definition }\}$$

Thus, we have:

$$ p(0,x) = \mathbb{E}[\xi(X_0)|X_0 = x] = \frac{\sigma(x)^2}{2} h''(x) + \mu(x)h'(x) $$