# Showing that a process is a supermartingale using Ito's formula

Consider a stock with price dynamics $$dS_t=S_t\sigma_tdW_t$$ where $$(W_t)_{t\geq0}$$ is a Brownian motion and $$(\sigma_t)_{t\geq0}$$ a bounded and continuous process adapted to the filtration $$(\mathcal{F}_t)_{t\geq0}$$. Show that the process $$(\sqrt{S_t})_{t\geq0}$$ is a supermartingale.

Using Ito's formula yields $$\sqrt{S_t}=\sqrt{S_0}+\frac{1}{2}\int_0^t\sigma_s\sqrt{S_s}dW_s-\frac{1}{8}\int_0^t\sigma_s^2\sqrt{S_s}ds.$$ I suppose my intuition is that the drift term here is negative and so the expectation will be decreasing. However, I am not sure how to show that the stochastic integral term is a martingale. This I think is a gap in my knowledge as I am usually not aware of how to prove a stochastic integral is a true martingale. Any advice would be greatly appreciated!

It is well known that the only solution to the SDE $$dS_t=S_t\sigma_t\,dW_t$$ is $$S_t=S_0\exp(\int_0^t\sigma_s\,dW_s-\frac12\int_0^t\sigma^2_s\,ds)$$ (see Doléans-Dade exponential), which is a (uniformly integrable) martingale as soon as $$\mathbb E[\exp(\frac12\int_0^t\sigma^2_s\,ds)]<+\infty$$ (see for instance Novikov criteria). This is the case here because $$(\sigma_t)_{t\ge0}$$ is bounded.

Therefore $$(S_t)_{t\ge0}$$ is a martingale, and since $$x\mapsto\sqrt x$$ is concave, $$(\sqrt{S_t})_{t\ge0}$$ is a supermartingale.

You can also use Itô's formula as you did. The result you need is then the following (which is an important proposition to have in mind in stochastic calculus):

Let $$(X_t)_{t\ge0}$$ be a continuous and adapted process (or more generally a progressively measurable process) such that $$\int_0^{+\infty}X_t^2\,dt<+\infty$$ almost surely. Then $$(\int_0^tX_s\,dW_s)_{t\ge0}$$ is a martingale bounded in $$L^2$$ iff $$\mathbb \int_0^{+\infty}E[X_t^2]\,dt<+\infty$$.

You can use the latter proposition to see that $$\int_0^t\sigma_s\sqrt{S_s}dW_s$$ is a martingale. Since $$(-\int_0^t\sigma_s^2\sqrt{S_s}ds)_{t\ge0}$$ is an adapted and nonincreasing process, summing it to a martingale yields a supermartingale, hence $$(S_t)_{t\ge0}$$ is a supermartingale.