Constructing equivalent martingale measure (lognormal distribution)

Question

Please, help me with next problem. It's from Follmer's and Schied's Stochastic Finance. An introduction in discrete time.

First, I'm looking for conditions for $X_t$ to be a martingale:

$$ E[X_t|\mathcal{F}_{t-1}] = E[X_0 \prod_{i=1}^{t} e^{\sigma_i Z_i + m_i}|\mathcal{F}_{t-1}] = X_{t-1}\cdot E[e^{\sigma_t Z_t + m_t}] = X_{t-1}\cdot e^{m_t + \frac{\sigma_t^{2}}{2}} $$ where the last (unconditional) expectation was taken w.r.t. a standard normal density.

Under a new measure $ P^{*}$, $E^{*}[e^{\sigma_t Z_t + m_t}] = E[\frac{dP_{t}^*}{dP_t} \cdot e^{\sigma_t Z_t + m_t}]$ must be equal to 1 of $m_t + \frac{\sigma_t^2}{2} = 0 \implies m_t = -\frac{\sigma_t^2}{2}$. I want ot find $\frac{dP^*}{dP}$ using two densities (or measures), but I confused, which to take.

Am I on the right way and how to proceed?

I decided to look for a Radon-Nikodym derivative $\frac{dP_{t}^*}{dP_t}$ of a one step, and then take a product of such densities.

Thanks in advance!

Answer 1

I know this comes late, but here is a possible solution: As you already indicated, it might be reasonable guess, that $dP^*/dP$ has some kind of product structure; i.e. there are $\sigma(Z_t)$-m.b. random variables $Y_t$ s.t. $$dP^*_t/dP=\prod_{i=1}^tY_i\,.$$ With this ansatz we get by Prop. A.16: \begin{align*} E^*(X_t\mid X_{t-1}) &=\frac{1}{E(dP^*_t/dP\mid X_{t-1})}E(X_tdP^*_t/dP\mid X_{t-1})\\ &=\frac{1}{\left(\prod_{i=1}^{t-1}Y_i\right)E(Y_t\mid X_{t-1})}\left(\prod_{i=1}^{t-1}Y_i\right)X_{t-1}E(e^{\sigma_tZ_t+m_t}Y_t\mid X_{t-1})\\ &=\frac{1}{E(Y_t)}X_{t-1}E(e^{\sigma_tZ_t+m_t}Y_t)\,. \end{align*} Further we know that in order to be a density, we have by induction on $i$ and the independence of the $Y_i$ that $E(Y_i)\overset{!}{=}1$ for each $i$. Thus we have to choose $Y_i$ in such a way that $$ E(Y_i)=1 \quad \text{and}\quad E(e^{\sigma_iZ_i+m_i}Y_i)=1\,.$$ Moreover we have the restriction that $dP^*/dPX_i$ is supposed to be log-normal distributed. One can easily check that by the formula for the expected value of a log-normal r.v. the following does the job: $$ Y_i=\exp\left(-(\sigma_i/2+m_i/\sigma_i)Z_i-(\sigma_i/2+m_i/\sigma_i)^2/2\right)\,.$$