The convexity of a stochastic integral with respect to the initial condition

Question

Let $B(t)$ be the standard Brownian motion, $$dx(t) = \mu(x(t),t)\,dt+\sigma(x(t),t)\,dB(t),$$ where $\mu(x,t)$ and $\sigma(x,t)$ satisfy the usual conditions laid out e.g. on the Wikipedia page on the existence and almost sure uniqueness of the t-continuous sample path solution of a stochastic differential equation. $f(x)$ is convex in $x$. Is $\mathbf E[f(x(t))|x(0)]$ convex with respect to $x(0)$?

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1. I can show if $f$ increases, $\mathbf E[f]$ increases in $x(0)$ by looking at path $x(t,\omega)$ for each sample $\omega$. By the Markovness and the uniqueness of the t-continuous sample path solution $x(t,\omega)$ given the initial condition $x(0,\omega)$, $x_1(0,\omega)\le x_2(0,\omega) \Longleftrightarrow x_1(t,\omega)\le x_2(t,\omega) \Longleftrightarrow f(x_1(t,\omega))\le f(x_2(t,\omega))$. Taking the expectation on both sides generates the desired result. Can we do something similar to answer the question?

2. As an alternative, I applied Ito's lemma to $f(x)$ $$f(x(t)) = f(x(0)) +\int_0^t \Big(\mu\frac{\partial f}{\partial x}+\frac12\sigma^2\frac{\partial^2 f}{\partial x^2}\Big)\,d\tau+\int_0^t \sigma\frac{\partial f}{\partial x}\,dB_\tau,$$ and $$\mathbf E_{t=0}[f(x(t))]=f(x(0)) +\int_0^t \mathbf E\Big[\mu\frac{\partial f}{\partial x}+\frac12\sigma^2\frac{\partial^2 f}{\partial x^2}\Big]\,d\tau$$ The first term on the right hand side of the above equation is obviously convex with respect to $x(0)$. However, what can one say about the $x(0)$ dependency of the remaining terms?

Answer 1

The answer is affirmative if $\mu$ is independent of $x$.

Proof:

Define $x(t)$ as a stochastic path and $$u(x,t):=\mathbf E[f(x(T))|x(t)=x].$$ By the Feynman-Kac formula, which is derived from the tower theorem of the expectation $$u(x,t) = \mathbf E[\mathbf E[f(x(T))|x(s)]|x(t)=x]=\mathbf E[u(x(s),s)|x(t)=x].$$ $$\frac{\partial u}{\partial t}+\mu\frac{\partial u}{\partial x}+\frac12\sigma^2\frac{\partial^2 u}{\partial x^2}=0.$$ Differentiating PDE with respect to $x$, and using subscript $1$ on a function to denote the partial differentiation with respect to $x$ of that function, we obtain $$\frac{\partial u_1}{\partial t}+(\mu+\sigma\sigma_1)\frac{\partial u_1}{\partial x}+\frac12\sigma^2\frac{\partial^2 u_1}{\partial x^2}+\mu_1u_1=0.$$ Again by the Feynman-Kac formula, this is equivalent to $$u_1(x,t)=\mathbf E\big[e^{\int_t^s\mu_1}u_1(x(s),s)\big|x(t)\big]=E\big[e^{\int_t^T\mu_1}f'(x(T))\big|x(t)=x\big],$$ under the probability measure such that $x$ is an Ito process driven by $$dx = (\mu+\sigma\sigma_1)\,dt+\sigma\,dB_t.$$ Note that now the current $x$ is a different Ito process with the drift different from the one described in the question.

Now we use the condition that $\mu_1=0$. Then $u_1(x,t)=\mathbf E\big[f'(x(T))\big|x(t)=x\big]$.

We apply method 1 mentioned in the question. Since $f'$ increases, $\mathbf E\big[f'(x(t)\,\big|\,x(0)\big]$ increases in $x(0)$ by looking at path $x(t,\omega)$ for each sample $\omega$. By the Markovness of $x(t,\omega)$ stemming from it being a diffusion and the P-almost surely uniqueness of t-continuous sample path, $$x_1(0,\omega)\le x_2(0,\omega) \Longleftrightarrow x_1(t,\omega)\le x_2(t,\omega) \Longleftrightarrow f'(x_1(t,\omega))\le f'(x_2(t,\omega)).$$ Taking the expectation on both sides arrives at the convexity of $\mathbf E\big[f(x(T))\big|x(0)\big]$ with respect to $x(0)$.

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The answer is also affirmative if we can transform $x$ into something that makes the effective $\mu_1=0$ and $f$ transformed still convex.

Answer 2

Looks complicated in general. Writing \begin{align}\tag{1} \mathbf{E}[f(x(t))]=f(x(0))+\mathbf{E}\Bigg[\int_0^t\Big(\mu\frac{\partial f}{\partial x}\Big)(x(s))+\frac{1}{2}\Big(\sigma\frac{\partial^2 f}{\partial x^2}\Big)(x(s))\,ds\Bigg] \end{align} we can replace $x(s)$ by $x_0+\int_0^s\mu(x(r))\,dr+\int_0^s\sigma(x(r))\,dB(r)$ and repeat that with $x(r)$ and so on. We will never be able to reach a simple expression that shows how that second part depends on $x_0$.

A typical example where we know the convexity of the expectation on the LHS of (1) is the Black-Scholes formula where \begin{align} f(x)&=(x-K)^+\,,\quad\mu(x)=rx\,,\quad\sigma(x)=\sigma x\,,\\[3mm] \mathbf{E}[f(x(t))]&=e^{rt}x_0\Phi(d_1)-K\Phi(d_2)\,,\\[3mm] d_{1,2}&=\frac{\log(x_0/K)+rt\pm\sigma^2t/2}{\sigma\sqrt{t}}\,,\\[3mm] \frac{\partial \mathbf{E}[f(x(t))]}{\partial x_0}&=e^{rt}\Phi(d_1)>0\,. \end{align} To derive this we use the known lognormal distribution of the stock price $x(t)$.