In Shreve's book on pp. 37-39 I read that given a standard normal random variable $X \sim N(0, 1)$ and another random variable $Y = X+ \theta$, we can define a measure change
$$\frac{d \widetilde{\mathbb{P}}}{d \mathbb{P}} = Z = \exp\left(-\theta X – \tfrac12\theta^2\right)$$
so that under $\widetilde{\mathbb{P}}$ the random variable $Y$ has the same distribution as $X$ under the original probability measure.
I am trying to extend it to the case when $X \sim N(\mu, \sigma^2)$. I guessed that in this case:
$$Z = \exp\left(\frac{-\theta (X – \mu) – \tfrac12\theta^2}{\sigma^2}\right)$$
and the integration seem to confirm (if I haven't made a mistake that is) that it indeed works as expected.
However, when I used it in a simulation then for small $\sigma$ the average of the simulated $Z_i$ was much lower than 1 (whereas for the measure change we should have $\mathbb{E}[Z] = 1$). For $\sigma=1$ the simulation works perfectly, which makes me think that I made a mistake somewhere.
What I am doing wrong? What is the correct measure change?
Best Answer
Yours is the correct change of measure.
The random variable $U=(X-\mu)/\sigma$ is standard normal hence its Laplace transform is such that $\mathrm E(\mathrm e^{-\lambda U})=\mathrm e^{\lambda^2/2}$ for every $\lambda$. Furthermore, $Z=\mathrm e^{-\theta^2/(2\sigma^2)}\mathrm e^{-\theta U/\sigma}$ hence $$\mathrm E(Z)=\mathrm e^{-\theta^2/(2\sigma^2)}\mathrm E(\mathrm e^{-\lambda U}),$$ for $\lambda=\theta/\sigma$, that is, $\mathrm E(Z)=\mathrm e^{-\theta^2/(2\sigma^2)}\mathrm e^{\lambda^2/2}=1$.