This is the excercise:
Let $B_t$ be 1d Brownian motion with $B_0=0$. Define
$$ X_t=X_t^x=x\cdot\exp\left(ct+\alpha B_t\right) $$
where $c,\alpha$ are constants and $x$ is non-random. Prove directly from the definition that $X_t$ is a Markov process, i.e., that
$$ \mathbb{E}\left[f\left(X_{t+h}\right)|\mathcal{F}_t\right]=\mathbb{E}^{X_{t}}\left[f\left(X_{h}\right)\right] $$
for bounded Borel-measurable $f$.
Here is my attempt
Let $\mathcal{F}_t^X:=\sigma\left(X_s:s\le t\right)$ and $\mathcal{F}_t^B:=\sigma\left(B_s:s\le t\right)$. Since $X_s$ is $\mathcal{F}_t^B$-measurable then $\mathcal{F}_t^X\subset\mathcal{F}_t^B$ we have, for any bounded Borel function $f(x)$,
$$ \begin{aligned} \mathbb{E}^x\left[f\left(X_{t+h}\right)\bigg|\mathcal{F}_t^X\right]&=\mathbb{E}^x\left[f\left(x\cdot\exp\left(c(t+h)+\alpha B_{t+h}\right)\right)\bigg|\mathcal{F}_t^X\right]\\ &=\mathbb{E}^x\left[f\left(x\cdot\exp\left(c(t+h)+\alpha B_{t+h}\right)\right)\bigg|\mathcal{F}_t^B\right]\\ &=\mathbb{E}^{B_t}\left[f\left(x\cdot\exp\left(c(t+h)+\alpha B_{t+h}\right)\right)\right]\in\sigma(B_t)=\sigma(X_t). \end{aligned} $$
So $\mathbb{E}^x\left[f\left(X_{t+h}\right)\bigg|\mathcal{F}_t^X\right]=\mathbb{E}\left[f\left(X_{t+h}\right)|X_t\right]$
Let $\mathcal{M}_t=\mathcal{F}_t^X:=\sigma\left(X_s:s\le t\right)$ and $\mathcal{F}_t=\mathcal{F}_t^B:=\sigma\left(B_s:s\le t\right)$. Since $X_s$ is $\mathcal{F}_t^B$-measurable then $\mathcal{F}_t^X\subset\mathcal{F}_t^B$ we have, for any bounded Borel function $f(x)$,
$$ \begin{aligned} \mathbb{E}\left[f\left(X_{t+h}\right)|\mathcal{F}_t\right]&=\mathbb{E}\left[f\left(x\cdot\exp\left(c(t+h)+\alpha B_{t+h}\right)\right)\bigg|\mathcal{F}_t\right]\\ &=\mathbb{E}\left[f\left(x\cdot\exp\left(ct+ch+\alpha B_{t+h}\pm\alpha B_{h}\right)\right)\bigg|\mathcal{F}_t\right]\\ &=\mathbb{E}\left[f\left(x\cdot\exp\left(ct+\alpha\left( B_{t+h}-B_{h}\right)+ch+\alpha B_{h}\right)\right)\bigg|\mathcal{F}_t\right]\\ &=\mathbb{E}\left[f\left(x\cdot\exp\left(ct+\alpha\left(B_{t+h}-B_{h}\right)\right)\exp\left(ch+\alpha B_{h}\right)\right)\bigg|\mathcal{F}_t\right] \end{aligned} $$
Notice that $\hat{B_{t}}=B_{t+h}-B_{h}$ is a Brownian Motion and it is equal in distribution to $B_{t}$, so
$$ \begin{aligned} \mathbb{E}\left[f\left(X_{t+h}\right)|\mathcal{F}_t\right]&=\mathbb{E}\left[f\left(x\cdot\exp\left(ct+\alpha\left(B_{t+h}-B_{h}\right)\right)\exp\left(ch+\alpha B_{h}\right)\right)\bigg|\mathcal{F}_t\right]\\ &=\mathbb{E}\left[f\left(\exp\left(ct+\alpha\hat{B_{t}} \right)\cdot x\cdot\exp\left(ch+\alpha B_{h}\right)\right)\bigg|\sigma\left(B_s:s\le t\right)\right]\\ &=\mathbb{E}\left[f\left(\exp\left(ct+\alpha\hat{B_{t}} \right)X_h\right)\bigg|\sigma\left(B_s:s\le t\right)\right]\\ &=\mathbb{E}\left[f\left(\exp\left(ct+\alpha B_t \right)X_h\right)\bigg|\sigma\left(B_s:s\le t\right)\right]\\ &=\mathbb{E}\left[f\left(x\cdot\exp\left(ct+\alpha B_t \right)\cdot\exp\left(ch+\alpha B_{h}\right)\right)\bigg|\sigma\left(B_s:s\le t\right)\right] \end{aligned} $$
And at the same time, if we where to start at $y=X_t=x\cdot\exp(ct+\alpha B_t)$, we would have that
$$ \begin{aligned} \mathbb{E}^{X_{t}}\left[f\left(X_{h}\right)\right]&=\mathbb{E}\left[f\left(y\cdot\exp\left(ch+\alpha B_{h}\right)\right)\right]\bigg|_{y=X_t}\\ &=\mathbb{E}\left[f\left(x\cdot\exp(ct+\alpha B_t)\cdot\exp\left(ch+\alpha B_{h}\right)\right)\bigg|X_t\right]\\ &=\mathbb{E}\left[f\left(x\cdot\exp(ct+\alpha B_t)\cdot\exp\left(ch+\alpha B_{h}\right)\right)\bigg|\sigma\left(X_s:s\le t\right)\right]. \end{aligned} $$
Since $\sigma(X_t)=\sigma(B_t)$, we have that
$$ \begin{aligned} \mathbb{E}^{X_{t}}\left[f\left(X_{h}\right)\right]&=\mathbb{E}\left[f\left(y\cdot\exp\left(ch+\alpha B_{h}\right)\right)\right]\bigg|_{y=X_t}\\ &=\mathbb{E}\left[f\left(x\cdot\exp(ct+\alpha B_t)\cdot\exp\left(ch+\alpha B_{h}\right)\right)\bigg|\sigma\left(X_s:s\le t\right)\right]\\ &=\mathbb{E}\left[f\left(x\cdot\exp(ct+\alpha B_t)\cdot\exp\left(ch+\alpha B_{h}\right)\right)\bigg|\sigma\left(B_s:s\le t\right)\right]. \end{aligned} $$
In other words, we have just prove that
$$ \mathbb{E}\left[f\left(X_{t+h}\right)|\mathcal{F}_t\right]=\mathbb{E}^{X_{t}}\left[f\left(X_{h}\right)\right]. $$
Could anyone please check if my reasoning is correct? Thanks