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
$$\frac{d \mathbb{\tilde{P}}}{d \mathbb{P}} = Z = \exp(-\theta X - 0.5\theta^2)$$
so that under $\mathbb{\tilde{P}}$ 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: $$Z = \exp(\frac{-\theta (X - \mu) - 0.5\theta^2}{\sigma^2})$$
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?
