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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!

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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.

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