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?


1 Answer 1


When the measure is changed, the distribution of the same random variable is also changed. That is, under the new measure, you obtain a new distribution for the same random variable. However, while you hold the new distribution fixed and go back to the original measure, then the random variable has to be changed. If you do not take the new distribution, then you do not have to change the random variable. In other words, to have a new distribution, you can either change the measure or change the random variable.

  • $\begingroup$ 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 $\mathbb{R}$)? $\endgroup$
    – bcf
    Commented Nov 27, 2015 at 21:36
  • $\begingroup$ You have the distribution under the new measure, and you want to obtain the same distribution under the original measure, then you have to change the random variable. $\endgroup$
    – Gordon
    Commented Nov 27, 2015 at 21:43

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