# If $X$ is an Ito process, is $\mathbb E(\int X \mathrm d X)$ convex?

Consider the functional $$F$$, which is defined for each Ito process

$$X(t) = \int_0^t \mu(s) \mathrm d s + \int_0^t \sigma(s) \mathrm d W(s)$$

as

$$F(X) := \mathbb E\bigg(\int_0^T X(s) \mathrm dX(s)\bigg)$$

Now I would like to prove that $$F$$ is convex, which seems to be intuitive because it is true for absolutely continuous processes.

Due to the Ito formula / product rule,

$$\int_0^T X(s) \mathrm dX(s) = \frac 1 2 X(T)^2 - \int_0^T \sigma(s)^2 \mathrm ds$$

If we restrict ourselves on Ito processes for which we have $$\sigma = 0$$, the proof is thus very easy. However, the quadratic variation term for $$\sigma \ne 0$$ seems to mess everything up. On the other hand, the $$\sigma$$ is also included in the $$X(T)^2$$ term, so we don't immediately get a counterexample.

Is there another proof for the claim? Or is the claim wrong and there is a counterexample?

I believe that $$F$$ fails to be convex.

First of all, if $$\sigma$$ is a "nice" function, then the stochastic integral $$M_t := \int_0^{t} \sigma(s) \, dW_s$$ is a martingale which implies that $$\mathbb{E} \left( \int_0^T X_s \, dM_s \right)=0,$$ i.e.

$$F(X) = \mathbb{E} \left( \int_0^T X_s \mu(s) \, ds \right). \tag{1}$$

Now consider $$X_t := \int_0^t W_s \, ds.$$ It follows from the very definition of $$F$$ that $$F(W) = \mathbb{E} \left( \int_0^T W_s \, dW_s \right)=0,$$ and $$(1)$$ yields \begin{align*} F(X) = \mathbb{E} \left( \int_0^T X_s W_s \, ds \right) &= \mathbb{E} \left( \int_0^T \int_0^s W_s W_r \, dr \, ds \right) \\ &= \int_0^T \int_0^s \underbrace{\mathbb{E}(W_s W_r)}_{=r} \, dr \, ds \\ &= \frac{T^3}{6}. \end{align*}For any $$\lambda \in [0,1]$$ we have by (1) that

\begin{align*} F(\lambda X + (1-\lambda) W) &= \mathbb{E} \bigg( \int_0^T (\lambda X_s + (1-\lambda) W_s) (\lambda W_s+0) \, ds \bigg) \\ &=\lambda^2 \underbrace{\mathbb{E}\left( \int_0^T X_s W_s \, ds \right)}_{\stackrel{(2)}{=} T^3/6} + \lambda (1-\lambda) \mathbb{E} \left( \int_0^T W_s^2 \, ds \right) \\ &= \lambda^2 \frac{T^3}{6} + \lambda (1-\lambda) \frac{T^2}{2}. \end{align*}

If we choose $$\lambda=1/2$$ and $$T$$ small enough (e.g. $$T=1$$), then

$$F(\tfrac{1}{2} X + \tfrac{1}{2} W) = \frac{T^3}{24} + \frac{T^2}{8}$$

is strictly larger than

$$\frac{1}{2} F(X) + \frac{1}{2} F(W) = \frac{T^3}{12}.$$

This means that $$F$$ is not convex.