Change of Measure
Question:
Given \( Z \tilde{} N(0, 1) \) on probability space \( (\Sigma, \mathcal{F}, \mathbb{P}) \), define a new probability measure \( \mathbb{Q} \), by taking
\[ G(z) = \frac{d\mathbb{Q}}{d\mathbb{P}} = e^{-\gamma z - \frac{\gamma^2}{2}} \]
where \( \gamma \in \mathbb{R} \) is a free parameter. Prove that
\[ \tilde{Z} = \gamma + Z \]
is \( \tilde{Z}\tilde{}N(\gamma, 1) \) w.r.t \( \mathbb{P} \) \( \tilde{Z}\tilde{}N(0, 1) \) w.r.t \( \mathbb{Q} \)
Solution:
Part 1: \( \tilde{Z} \tilde{} N(\gamma, 1) \) w.r.t \( \mathbb{P} \)
\( d\mathbb{P}(z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz \implies d\mathbb{P}(\tilde{z}) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(\tilde{z} - \gamma)^2}{2}}d\tilde{z} \)
This represents \( \tilde{Z} \tilde{} N(\gamma, 1) \)
This can also be obtained from expectation algebra.
\( E[\tilde{Z}] = E[\gamma + Z] = \gamma + E[Z] = \gamma \)
\( E[(\tilde{Z}-\gamma)^2] = E[(Z + \gamma - \gamma)^2] = E[Z^2] = 1 \)
\( \)
Part 2: \( \tilde{Z} \tilde{} N(0,1) \) w.r.t \( \mathbb{Q} \)
\( d\mathbb{Q}(z) = G(z)d\mathbb{P}(z) \)
\( = \frac{1}{\sqrt{2\pi}}e^{-\gamma z - \frac{\gamma^2}{2}}e^{-\frac{z^2}{2}}dz \)
\( = \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2 + 2\gamma z + \gamma^2}{2}}dz \)
\( = \frac{1}{\sqrt{2\pi}}e^{-\frac{(z + \gamma)^2}{2}}dz \)
\( \implies d\mathbb{Q}(\tilde{z}) = \frac{1}{\sqrt{2\pi}}e^{-\frac{\tilde{z}^2}{2}}d\tilde{z} \)
This represents \( \tilde{Z} \tilde{} N(0,1) \)
