Girsanov’s Theorem


We can change the probability measure, and then make a random variable follows a certain probability measure.

  • Radon-Nikodym Derivative:

$$Z(\omega) = \frac{\tilde{P}(\omega)}{P(\omega)}$$

  • $\tilde{P}(\omega)$ is the risk-neutral probability measure.
  • ${P}(\omega)$ is the actual probability measure.
  • Properties:
    • $Z(\omega)>0$
    • $\mathbb{E}(Z)=1$
    • As \tilde{P}(\omega) = Z(\omega) P(\omega), so if Z(\omega), then \tilde{P}(\omega)>P(\omega). vice versa.

We can calculate that,

$$ \underbrace{\tilde{\mathbb{E}}(X)}_{\text{Expectation under Risk-neutral Probability Measure}} = \underbrace{\mathbb{E}(ZX)}_{\text{Expectation under Actual Probability Measure}} $$

Proof & Example

Under (\Omega,\mathcal{F},P), A\in \mathcal{F}, let X be a random variable X\sim N(0,1). \mathbb{E}(X)=0, and \mathbb{Var}(X)=1.

$Y=X+\theta$, $\mathbb{E}(Y)=\theta$, and $\mathbb{Var}(Y)=1$.

$X$ here is s.d. normal under the actual probability measure.

However, Y here is not standard normal under the current probability P(.), because \mathbb{E}(Y)\neq0.

What do we do?

We change the probability measure from P(.)\to\tilde{P}(.) to let Y be standard normal under the new probability measure!

We set the Radon-Nikodym Derivative,

$$Z(\omega) = exp\{ -\theta\ X(\omega) – \frac{1}{2}\theta^2 \}$$

Now, we can create the probability measure \tilde{P}(A), A={ \omega;Y(\omega)\leq b) }

$$\tilde{P}(A) = \int_A Z(\omega)\ dP(\omega)$$

such that Y=X+\theta would be standard normal distributed under the new probability measure \tilde{P}(A).

$$\tilde{P}(A) = \tilde{P}(Y(\omega \leq b)$$

$$ = \int_{{ Y(\omega)\leq b } } exp{ -\theta\ X(\omega) – \frac{1}{2}\theta^2 } \ dP(\omega)$$

, then change the integral range from the set A to \Omega by multiplying that indicator.

$$ = \int_{\Omega }\mathbb{1}_{ Y(\omega)\leq b }\ exp{ -\theta\ X(\omega) – \frac{1}{2}\theta^2 } \ dP(\omega)$$

, change from dP to dX,

$$ = \int_{-\infty }^{\infty }\mathbb{1}_{ b-\theta}\ exp{ -\theta\ X(\omega) – \frac{1}{2}\theta^2 } \ \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}X^2(\omega)} \ dX(\omega)$$

$$ =\frac{1}{\sqrt{2\pi}} \int_{-\infty }^{b-\theta}\ exp{ -\theta\ X(\omega) – \frac{1}{2}\theta^2- \frac{1}{2}X^2(\omega)} \ dX(\omega)$$

$$ =\frac{1}{\sqrt{2\pi}} \int_{-\infty }^{b-\theta}\ exp\Bigg\{ -\frac{1}{2}\bigg(\theta+ X(\omega)\bigg)^2\Bigg\} \ dX(\omega)$$

, as Y=X+\theta, dY = dX, we now change dX to dY,

$$ =\frac{1}{\sqrt{2\pi}} \int_{-\infty }^{b}\ exp\big\{ -\frac{1}{2}Y(\omega)^2\big\} \ dY(\omega)$$

, the above is now a standard normal distribution for Y(\omega).