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Assume that $X_t$ is a process with dynamics $dX_t = \sigma X_t dW_t$ is where $W_t$ is a standard Brownian motion. Given two deterministic functions $p(t)$ and $q(t)$, compute $\mathbb{E}[p(t)X(t)+q(t)]$.

Solution

Since $p$ and $q$ are deterministic we have $$\mathbb{E}[p(t)X(t)+q(t)] = p(t) \mathbb{E}[X(t)]+q(t)=p(t)X_0+q(t)$$ where the last equation from the fact that $X_t$ is a martingale. To be more precise, $\mathbb{E}[X(t)|X_0]=X_0$.

SDE dynamics

I want to derive the SDE for $p(t)X(t)+q(t)$ using Ito's lemma. First, I am considering the case $V(t)=p(t)X(t)$, then according to Ito's lemma: \begin{align} dV_t&=\frac{\partial p}{\partial t}X_tdt+p(t)\sigma X_tdW_t \\ &=\frac{1}{p(t)}\frac{\partial p}{\partial t}V_tdt+\sigma V_tdW_t \end{align} The SDE that remains is quite straightforward to solve then with solution: $$V(t) = V(0)\exp \left \{ \left(\frac{1}{p(t)}\frac{\partial p}{\partial t} - \frac{\sigma^2}{2}\right)T + \sigma W_t\right \} $$ This will give a different answer for $\mathbb{E}[p(t)X(t)]$, so I am wondering where my mistake is.

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Let's just write $p'(t)$ for $\frac{\partial p}{\partial t}$. Then your last expression for $V$ should rather be $$ V(t)=V(0)\exp\Bigg\{\int_0^t\frac{p'(s)}{p(s)}\,ds-\frac{\sigma^2}{2}\color{red}{t}+\sigma W_t\Bigg\}. $$ This gives \begin{align} E[V(t)]&=V(0)\exp\Bigg\{\int_0^t\frac{p'(s)}{p(s)}\,ds\Bigg\}\\[3mm] &=V(0)\exp\Bigg\{\int_0^t\frac{d}{ds}\log p(s)\,ds\Bigg\}\\[3mm] &=V(0)\exp\{\log p(t)-\log p(0)\}\\[3mm] &=\frac{V(0)p(t)}{p(0)}=X(0)p(t) \end{align} as it should.

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