# Discounted price process in Black-Scholes model is a martingale with respect to Q.

I have been presented a proof that the discounted price process in the Black and Scholes formula is a martingale, but there is something important omitted, and I am not able to fill in the gap. I will present the presentation of the proof I am given here:

start of proof:

The price process is modelled by $dS_t = \mu dt+\sigma dB_t$, $\mu, \sigma$ constant. We also have a constant interest rate r, and we define $\tilde{S}_t = e^{-rt}S_t$.

Using the Ito's lemma on $e^{-rt}S_t$, we find that $d\tilde{S}_t =\tilde{S}_t(\mu-r)dt+\tilde{S}_t\sigma d B_t$

Then we have this version of Girsanovs theorem:

If $X_t$ is a stochastic process where: $E[\exp(\int_o^ tX_t^2/2)]<\infty, \forall t.$ And B is a Brownian motion under an original probability space $(\Omega, \mathcal{F},P)$. Then $\tilde{B}_t=B_t-\int_o^tX_s ds$ is a brownian motion on the space $(\Omega, \mathcal{F},Q)$, where $Q(A)=E_P[1_A \exp(\int_0^tX_t dB_t-1/2\int_0^tX_t^2ds)]$.

We then define $X_t = (r-\mu)/\sigma$. It satisfies the novikov condition because it is a constant, and hence. $\tilde{B}_t=B_t-(\mu-r)/\sigma \cdot t$ is a Brownian motion under Q.

But now comes by problem, in order to finish the proof they say that:

since we have $\tilde{S}_t=\tilde{S}_0+\int_0^t\tilde{S}_s(\mu-r)ds+\int_0^t\tilde{S}_s\sigma dB_S=\tilde{S}_0+\int_0^t\tilde{S}_s\sigma(\frac{\mu-r}{\sigma}ds+dB_s)=\tilde{S}_0+\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$.

And this is a martingale under Q.$\square$

My problems are these:

1. I agree that $\tilde{S}_0+\int_0^t\tilde{S}\sigma d\tilde{B}_t$. is a martingale under Q. But how do we have that $\tilde{S}_0+\int_0^t\tilde{S}_s\sigma(\frac{\mu-r}{\sigma}ds+dB_s)=\tilde{S}_0+\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$. I do see that it is logical, and where it comes from. But we can't just play around with differentials like that? I see where it comes from if we can manipulate differentials in the formula like this, but we can't just do algebra on differentials?

2. If we in some way are able to deduce that $\tilde{S}_0+\int_0^t\tilde{S}_s\sigma(\frac{\mu-r}{\sigma}ds+dB_s)=\tilde{S}_0+\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$ (from point 1), we still can't know that the integral considered as a r.v. will be the same under the measure Q, because in the definition of the stochastic integral when we defined it, it depended very much on what probability-space we started with(it realies heavily on cauchy and convergence of random variables). So how do we know that the integral $\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$ is the same considered in both $(\Omega, \mathcal{F},P)$ and $(\Omega, \mathcal{F},Q)$? Is $\int_0^t\tilde{S}_s\sigma d\tilde{B}_s[P]=\int_0^t\tilde{S}_s\sigma d\tilde{B}_s[Q]?$

Can you guys please help me? Do you see how to make the problem complete?

In summery the problem is this: How do we see that $\tilde{S}_0+\int_0^t\tilde{S}_s(\mu-r)ds+\int_0^t\tilde{S}_s\sigma dB_s=\tilde{S}_0+\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$. When the first stochastic integral is created using $(\Omega,\mathcal{F},P)$, and the second stochastic integral is created using $(\Omega, \mathcal{F},Q)$?

• You have to consider how a stochastic integral $\int f(s) dB_s$ is defined and confirm that it is linear with respect to the integrator. For an analogy just consider a non-stochastic Riemann-Stieltjes integral. If $h(x) = g_1(x) + g_2(x)$, then $\int f dh = \int f dg_1 + \int f dg_2$ – RRL May 10 '16 at 17:27
• You have $\tilde{B}_t = B_t - (\mu-r)/\sigma t$. To find $d\tilde{B}_t$, apply Ito's lemma. I agree, it is not a very clean way to write integrals down. – Cavents May 10 '16 at 17:33
• @Siron Thank you, I agree that Itos lemma tells us that $d\tilde{B}_t=dB_t-(\mu-r)/\sigma$. From what I read this is just a notational way of writing: $\tilde{B}_t=\tilde{B}_0+\int_{0}^tdB_s+\int_{0}^t-(\mu-r)/\sigma dt=\tilde{B}_0+B_t-(\mu-r)/\sigma t$. But are there results also saying that if I make an integral with respect to $\tilde{B}_t$ like $\int_0^tY_sd\tilde{B}_s$, I can also just plug in the $d\tilde{B}_s$ from Ito's lemma here? – user119615 May 10 '16 at 17:51
• @user119615: I'm trying to better understand your primary concern. Is it why you can substitute for $\frac{\mu-r}{\sigma}ds+dB_s$ with $d\tilde{B}_s$ in a stochastic integral? – RRL May 10 '16 at 18:55
• @RRL There are two primary concerns one is that yes, but I think you explained why that was allowed in your post. The other concern is after you have replaced it, how do you know that the integrals are the same in the sense that a construction of a stochastic integral also depends heavily on the underlying probability space so how do we know that $\int_0^t \tilde{S}_s d\tilde{B}_t$ is the same if we consider it as a stochastic integral in $(\Omega, \mathcal{F},P)$ or as a stochastic integral in $(\Omega, \mathcal{F},Q)$ . – user119615 May 11 '16 at 15:41

I'll rewrite the proof in terms of the process $$\tilde{S}_t$$, rather than using stochastic integrals. The idea behind Girsanov's theorem is to 'eliminate' the drift term of $$\tilde{S}_t$$ by changing the probability measure (to $$\mathbb{Q}$$). We have, $$d\tilde{S}_t = \tilde{S}_t(\mu-r)dt + \tilde{S}_t \sigma dB_t.$$ Since $$\tilde{B}_t = B_t - \frac{r-\mu}{\sigma}t$$, we obtain by Itô's lemma $$d\tilde{B}_t = \left(\frac{\mu-r}{\sigma}\right)dt+dB_t.$$ Hence, we obtain \begin{align} d\tilde{S}_t & = \tilde{S}_t(\mu-r)dt + \tilde{S}_t \sigma dB_t \\ &=\tilde{S}_t(\mu-r)dt + \tilde{S}_t\sigma\left[d\tilde{B}_t - \left(\frac{\mu-r}{\sigma}\right)dt\right] \\ &= \tilde{S}_t \sigma d\tilde{B}_t. \end{align} We have showed that $$\tilde{S}_t$$ is a martingale w.r.t $$\tilde{B}_t$$, hence a $$\mathbb{Q}$$-martingale (since $$\tilde{B}_t$$ is a Brownian motion under $$\mathbb{Q}$$ by Girsanov and a Brownian motion is a martingale).
To answer your first question, stochastic integrals of this type are Ito integrals. They are defined as a specialized limit of a Riemann-like sum with respect to a partition. Under appropriate conditions on processes $X_s$ and $B_s,$ we can define the integral as a limit of left-hand (non-anticipatory) sums:
$$\int_0^t X_s dB_s = \lim_{n \to \infty} \sum_{i=1}^n X_{s_{i-1}}[B_{s_i}- B_{s_{i-1}}].$$
If $B_s = Y_s + Z_s$ then it should be obvious how we can justify decomposing the integral as
$$\int_0^t X_s dB_s = \int_0^t X_s dY_s + \int_0^t X_s dZ_s$$