Change of measure for a normal random variable). A nonrigorous but informative derivation of the formula for the Radon-Nikodým derivative $Z(\omega)$ in Example 1.6.6 is provided by this exercise. As in that example, let $X$ be a standard normal random variable on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$, and let $Y=X+\theta$. Our goal is to define a strictly positive random variable $Z(\omega)$ so that when we set
$$
\widetilde{\mathbb{P}}(A)=\int_A Z(\omega) d \mathbb{P}(\omega) \text { for all } A \in \mathcal{F},
$$
the random variable $Y$ under $\widetilde{\mathbb{P}}$ is standard normal. If we fix $\bar{\omega} \in \Omega$ and choose a set $A$ that contains $\bar{\omega}$ and is "small," then (1.9.4) gives
$$
\tilde{\mathbb{P}}(A) \approx Z(\bar{\omega}) \mathbb{P}(A),
$$
where the symbol $\approx$ means "is approximately equal to." Dividing by $\mathbb{P}(A)$, we see that
$$
\frac{\widetilde{\mathbb{P}}(A)}{\mathbb{P}(A)} \approx Z(\bar{\omega})
$$
for "small" sets $A$ containing $\bar{\omega}$. We use this observation to identify $Z(\bar{\omega})$.
With $\bar{\omega}$ fixed, let $x=X(\bar{\omega})$. For $\epsilon>0$, we define $B(x, \epsilon)=\left[x-\frac{\epsilon}{2}, x+\frac{\epsilon}{2}\right]$ to be the closed interval centered at $x$ and having length $\epsilon$. Let $y=x+\theta$ and $B(y, \epsilon)=\left[y-\frac{\epsilon}{2}, y+\frac{\epsilon}{2}\right]$.
(i) Show that
$$
\frac{1}{\epsilon} \mathbb{P}\{X \in B(x, \epsilon)\} \approx \frac{1}{\sqrt{2 \pi}} \exp \left\{-\frac{X^2(\bar{\omega})}{2}\right\} .
$$
(ii) In order for $Y$ to be a standard normal random variable under $\widetilde{\mathbb{P}}$, show that we must have
$$
\frac{1}{\epsilon} \widetilde{\mathbb{P}}\{Y \in B(y, \epsilon)\} \approx \frac{1}{\sqrt{2 \pi}} \exp \left\{-\frac{Y^2(\bar{\omega})}{2}\right\} .
$$
1.9 Exercises
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(iii) Show that $\{X \in B(x, \epsilon)\}$ and $\{Y \in B(y, \epsilon)\}$ are the same set, which we call $A(\bar{\omega}, \epsilon)$. This set contains $\bar{\omega}$ and is "small" when $\epsilon>0$ is small.
(iv) Show that
$$
\frac{\tilde{\mathbb{P}}(A)}{\mathbb{P}(A)} \approx \exp \left\{-\theta X(\bar{\omega})-\frac{1}{2} \theta^2\right\} .
$$
The right-hand side is the value we obtained for $Z(\bar{\omega})$ in Example 1.6.6.