This exercise provides an example where there exists a solution although the Lipschitz condition is not satisfied.

Let \(a\) and \(b\) be two constants and \(\zeta\) a bounded \(\mathcal{F}_{T}\)-measurable r.v.. Prove that the solution of \(-d X_{t}=\left(a Y_{t}^{2}+b Y_{t}\right) d t-Y_{t} d W_{t}, X_{T}=\zeta\) is

\[X_{t}=\frac{1}{2 a}\left(\frac{1}{2} b^{2}(t-T)-b W_{t}+\ln \mathbb{E}\left(e^{b W_{T}+2 a \zeta} \mid \mathcal{F}_{t}\right)\right)\]

\[-d X_{t}=a Y_{t}^{2} d t-Y_{t} d W_{t}, X_{T}=\zeta\]

is \(X_{t}=\frac{1}{2 a} \ln \mathbb{E}\left(e^{2 a \zeta} \mid \mathcal{F}_{t}\right)\). Then, using Girsanov's theorem, the solution of

\[-d X_{t}=\left(a Y_{t}^{2}+b Y_{t}\right) d t-Y_{t} d W_{t}, X_{T}=\zeta\]

is given by

\[X_{t}=\frac{1}{2 a} \ln \widehat{\mathbb{E}}\left(e^{2 a \zeta} \mid \mathcal{F}_{t}\right)\]

where \(\widehat{\mathbb{Q}}_{\mid \mathcal{F}_{t}}=e^{b W_{t}-\frac{1}{2} b^{2} t} \mathbb{P}_{\mid \mathcal{F}_{t}}\). Therefore,

\[\begin{aligned}
X_{t} & =\frac{1}{2 a} \ln \left(\mathbb{E}\left(\left.e^{b W_{T}-\frac{1}{2} b^{2} T} e^{2 a \zeta} \rightvert\, \mathcal{F}_{t}\right) e^{-b W_{t}+\frac{1}{2} b^{2} t}\right) \\
& =\frac{1}{2 a}\left(\ln \mathbb{E}\left(\left.e^{b W_{T}-\frac{1}{2} b^{2} T} e^{2 a \zeta} \rightvert\, \mathcal{F}_{t}\right)-b W_{t}+\frac{1}{2} b^{2} t\right)
\end{aligned}\]