# Exponential Martingale in the BS-model with time depending parameters

Suppose we have the BS model with time depending coefficient. This means we have to processes $dX_t=rX_tdt$ and $dU_t=U_t(\mu(t) dt +\sigma(t) dW_t)$, where $W$ is a Brwonian Motion. Moreover we define $Y_t:=\frac{U_t}{X_t}$. Both, $\mu,\sigma$ are continuous function from $[0,T]$ to the reals and $\sigma(t)>0,\forall t\in[0,T]$. Then I should find an equivalent probability measure $Q$ on $\mathcal{F}_T$ such that $Y$ is $Q$ martingale and satisfies the SDE $$dY_t=Y_t\sigma(t)dB_t$$ for a Brownian Motion $B$ under the measure $Q$.

Motivatet from the non time-depedning case, I defined $L_t:=-\int_0^t \lambda(s) dW_s$ and defined the measure $Z_T:=\frac{dQ}{dP}=\mathcal{E}(L)_T$. We have proved in the lecture that $\mathcal{E}(L)$ is a martingale, hence we have $E[Z_T]=E[Z_0]=1$. Since $Z_T$ is also positive, we can define a probability measure $Q$ equivalent to $P$ through this density. I was also able to prove, that $Y_t$ satisfies the SDE above under the new measure $Q$ with this Brownian Motion $B$. It is just applying Girsanov. However, I do not see why $Y$ should be a martingale under $Q$. The problem is, that $\sigma(t)$ appears in the SDE under $Q$. So I am not sure how $Y$ looks like under the measure $Q$. In the non-time depending case, we have that $Y$ has under $Q$ the form

$$Y_t=e^{(\sigma B_t-\frac{1}{2}\sigma^2 t)}$$

which is obviously a martingale. However, in this time depending I am not sure how to prove that $Y$ is a martingale.

Moreover, I want to prove the following inequality: Let $H=((U_T)^2-K)^+$. Then it should be true that

$$E_Q[\frac{H}{X_T}]\ge \big(s^2e^{(\int_0^T(\sigma(u)^2+r)du)}-Ke^{(-rT)}\big)^+$$

I am not sure if $s$ is a mistake in this inequality, since it does not appear so far in this exercise. However, I thought just proving the inequality, will show me what $s$ has to be. So far, I did the following:

$$E_Q[\frac{H}{X_T}]=e^{-rT}E_Q[((U_T)^2-K)^+]\ge e^{-rT}\big(e^{2rT}E_Q[Y_T^2]-K\big)^+=\big(e^{rT}E_Q[Y_T^2]-e^{-rT}K\big)^+$$

Hence, all I have to show is $e^{rT}E_Q[Y^2_T]=s^2e^{\int_0^T(\sigma(u)^2+r)du}$. Even more it is enough to prove $E_Q[Y^2_T]=e^{\int_0^T\sigma(u)^2du}$

It would be very helpful, if someone could explain the things I did not get.

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Isn't $\sigma$ a deterministic function? Consult Section 4 of these notes –  Ilya Jan 21 '13 at 12:57
@Ilya Ah nice, thanks! Now everything is working. So the $s$ should be $1$, right? If you want, you can just give a short answer. I will accept it. –  user20869 Jan 21 '13 at 13:20

For the change of measure of a diffusion with time-dependent drift and volatility you shall check Section 4 of these notes. There you also have a formula for $Y_T$ through $\displaystyle{\int_0^t}\sigma_s\mathrm dB_s$ which quickly leads you to the formula of the expectation.