In part (v) of Exercise 5.12, we saw that when we change measures and change Brownian motions, correlations can change if the instantaneous correlations are random. This exercise shows that a change of measure without a change of Brownian motions can change correlations if the market prices of risk are random
Let $W_1(t)$ and $W_2(t)$ be independent Brownian motions under a probability measure $\widetilde{\mathbb{P}}$. Take $\theta_1(t)=0$ and $\theta_2(t)=W_1(t)$ in the multidimensional Girsanov Theorem, Theorem 5.4.1. Then $\widetilde{W}_1(t)=W_1(t)$ and $\widetilde{W}_2(t)=W_2(t)+\int_0^t W_1(u) d u$.
(i) Because $\widetilde{W}_1(t)$ and $\widetilde{W}_2(t)$ are Brownian motions under $\widetilde{\mathbb{P}}$, the equation $\widetilde{\mathbb{E}} \widetilde{W}_1(t)=\widetilde{\mathbb{E}} W_2(t)=0$ must hold for all $t \in[0, T]$. Use this equation to conclude that
$$
\tilde{\mathbb{E}} W_1(t)=\tilde{\mathbb{E}} W_2(t)=0 \text { for all } t \in[0, T] .
$$
(ii) From Itô's product rule, we have
$$
d\left(W_1(t) W_2(t)\right)=W_1(t) d W_2(t)+W_2(t) d W_1(t) .
$$
Use this equation to show that
$$
\widetilde{\operatorname{Cov}}\left[W_1(T), W_2(T)\right]=\widetilde{\mathbb{E}}\left[W_1(T) W_2(T)\right]=-\frac{1}{2} T^2 .
$$
This is different from
$$
\operatorname{Cov}\left[W_1(T), W_2(T)\right]=\mathbb{E}\left[W_1(T) W_2(T)\right]=0 .
$$