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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?

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how did you create the simulation? i am working on something similar. specifically, once you change measures how do you simulate from it? do you simply use the standard density inversion, generate uniform rv, etc.? – Alex Oct 10 at 2:59

1 Answer 1

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, $\mathrm E(Z)=\mathrm e^{-\theta^2/(2\sigma^2)}\mathrm e^{\lambda^2/2}=1$.

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