# Proving $\frac{\Pi(t)}{B(t)}$ is a martingale

Consider the stanadard Black-Scholes model and a T-claim $$\mathcal{X}$$ of the form $$\mathcal{X}=\Phi(S(T))$$.

Denote the corresponding arbitrage free price process by $$\Pi(t)$$.

Show that, under the martingale measure $$Q$$,$$\Pi(t)$$ has a local return equal to the short rate of interest r. In other words show that $$\Pi(t)$$ has a differential of the form:

$$d\Pi(t)=r.\Pi(t)dt+g(t)dW_t$$

Show that under the martingale measure $$Q$$, the process $$Z(t)=\frac{\Pi(t)}{B(t)}$$ is a martingale. More precisely, show that the stochastic differential for $$Z$$ has zero drift term,i.e, it is of the form:

$$dZ(t)=Z(t)\sigma_{Z}(t)dW_t$$

Determine also the diffusion process $$\sigma_Z(t)$$ (in terms of the pricing function F and its derivatives).

The first question is easily solved by recognizing f(t,S(t))=\frac{\pi(t)}{B(t)}, so that f(t,S(t)) satisfies the BS pde.

$$df(t,S(t))=f_t dt+f_S S(t)dS(t)+\frac{1}{2} f_{SS} S^2(t) dS^2(t)\\=f_t dt+f_S S(t)r dt+f_S S(t)\sigma dW_t+\frac{1}{2}f_{SS}\sigma^2 dt=f_S S(t)\sigma dW_t +r f(t,S(t))$$ as I wanted to prove.

My problem is with the first part of the second question. I do not want to use the Black-Scholes pde to prove that $$\Pi(t)$$ is a martingale, since I want to prove the risk-neutral valuation formula independently of the BS pde.

So the risk neutral valuation formula is given by $$\Pi(t)=B(t)E^Q_t\left(\frac{\Pi(T)}{B(T)}\right)\implies\frac{\Pi(t)}{B(t)}=E^Q_t\left(\frac{\Pi(T)}{B(T)}\right)$$, therefore by the tower property of the conditional expected values the right side of the expression is a martingale which implies $$\Pi(t)$$ to be a martingale.

Question:

Is this reasoning correct? If not, how would I proceed to prove $$\Pi(t)$$ is a martingale?