The convexity of a stochastic integral with respect to the initial condition Let $B(t)$ be the standard Brownian motion,
$$dx(t) = \mu(t,x(t))\,dt+\sigma(t,x(t))\,dB(t),$$
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 $x(0)$?


*

*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?


*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?
 A: 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 $x(t)$.
A: 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 $x(0)$.

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