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Given the following Lemma: Let $A_{t}=\int_{0}^{t}a_{s}dB_{s}$ where $a$ is an adapted process satisfying $\mathbb{P}\Big(\int_{0}^{T}a^{2}_{u}du < \infty\Big) = 1$ and $B$ is a standard Brownian motion. $A$ is a supermartingale if $A_{t}(\omega)$ is bounded by a constant $K$ for all $t,\omega$.

For a given adapted process $\eta$ consider the following stochastic differential equation: $$dZ_{t}:=-Z_{t}\eta_{t}dB_{t},\;\;\;Z_{0}:=1.$$ -Explain why we have $$(Z_{t})^{\beta}=e^{-\beta\int_{0}^{t}\eta_{u}dB_{u}-\frac{1}{2}\beta\int_{0}^{t}\eta^{2}_{u}du}$$ where $\beta$ is a positive constant. (I've gotten this one.)

-Show that $$\mathbb{E}\Bigg[\Big(e^{-\beta\int_{0}^{t}\eta_{s}dB_{s}-\beta^{2}\int_{0}^{t}\eta^{2}_{s}ds}\Big)^{2}\Bigg ]\leq 1$$ where $\beta$ is a positive constant and $\eta$ is an adapted process, using the given lemma. (The given lemma only seems to apply to part of the exponent?)

-Use the Cauchy-Schwartz inequality to derive the expression $$\mathbb{E}[Z_{t}^{\beta}]\leq \mathbb{E}\Bigg[e^{\beta(2\beta - 1)\int_{0}^{t}\eta_{u}^{2}du}\Bigg]^{\frac{1}{2}}$$ where $\beta$ is a positive constant. (Where did the $\beta(2\beta-1)$ come from?)

-Use the previous question with $\beta := 2$, to show that for $\eta$ bounded, we have for all $T>0$ $$\mathbb{E}\Bigg[\int_{0}^{T}(Z_{t}\eta_{t})^{2}dt\Bigg]<\infty,$$ and conclude that for a bounded $\eta$, the process $Z$ is a martingale. (Totally confused)

-Define the process $\eta$ by $\eta_{t}:=\frac{1}{R_{t}}$ where $R$ is the Bessel process, defined as the solution to the following equation$$dR_{t}:=\frac{1}{R_{t}}dt+dB_{t},\;\;\;R_{0}:=1.$$ Compute the differential $d\Big(\frac{\eta_{t}}{Z_{t}}\Big)$ and conclude that $Z_{t}=\eta_{t}$. (I've worked on this but didn't get the right conclusion.)

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