# Radon-Nikodym: Changing Distribution vs Changing Random Variable

Let $X \sim \mathcal{N}(\mu,\sigma^2)$ under the probability measure $P$ on the measurable space $(\Omega, \mathcal{F})$. We may define a Radon-Nikodym derivative $Z$, also defined on $(\Omega, \mathcal{F})$, by $$Z(\omega) := \frac{e^{\alpha X(\omega)}}{M_X(\alpha)} = \exp\left(\alpha X(\omega) - \alpha\mu - \frac{1}{2}\alpha^2\sigma^2\right)$$ for $\alpha \in \mathbb{R}$, where $M_X(\alpha) = \exp(\alpha\mu + \frac{1}{2}\alpha^2\sigma^2)$ is the MGF of $X$. The random variable $Z$ indeed qualifies as a Radon-Nikodym derivative, so let's use it to define a new probability measure $\tilde{P}$ on $(\Omega, \mathcal{F})$ by $$\tilde{P}(A) := \int_A Z(\omega) \, dP(\omega) \qquad \text{for all } A \in \mathcal{F}.$$ It can be shown that, under $\tilde{P}$, $X \sim \mathcal{N}(\mu + \sigma^2\alpha, \sigma^2)$. In words, I've often seen this described (e.g. Shreve II, p. 37) as

We changed the distribution of the random variable without changing the random variable itself.

However, computing probabilities of $X$ under $\tilde{P}$ is then equivalent to computing probabilities of $X + \sigma^2\alpha$ under $P$, in which we do change the random variable; we add $\sigma^2\alpha$ to it.

Even more, for option pricing, Shreve gives Grisanov's theorem on the bottom of p. 212, and defines a new random variable $$\tilde{W}(t) = W(t) + \int_0^t \theta(u) \, du,$$ which is then substituted into the stock price model. So here, in its most applied context in quant finance, we are blatantly changing the random variable!

So, although the above quote about changing only the distribution is technically valid in one sense, in the other sense we really are changing the random variable. Am I missing something?

• How can I hold a distribution fixed but change the measure? Isn't the distribution determined by the measure? I assume by distribution, you mean distribution measure (on $\mathbb{R}$)? – bcf Nov 27 '15 at 21:36