Conditional expection of a lognormal process Suppose I have the following process
$$y_{t+1} = y_t \exp (\varepsilon_{t+1})$$
where $\varepsilon_{t+1}\sim _{iid} N(\frac{\sigma v^2 }{2},v^2 )$, $v> 0$. I would like to calculate $E[y_{t+1}|y_t]$
For this, I apply conditional expectation in the equation above. So
$$E[y_{t+1} | y_t ] =E [y_t \exp (\varepsilon_{t+1})| y_t] = y_t E [ \exp (\varepsilon_{t+1})| y_t].$$
Now I’m unsure of saying that $\exp (\varepsilon_{t+1})$ and $y_t$ are independent. I suspect this to be true, since $y_t = y_{t-1} \varepsilon_{t} $ and $\{\varepsilon_t\}_{t=0}^{\infty}$ is a iid process. If the independence of  $\exp (\varepsilon_{t+1})$ and $y_t$ is true, I can conclude that
$$E[y_{t+1} | y_t ] =  y_t E [ \exp (\varepsilon_{t+1})| y_t] = y_t E [ \exp (\varepsilon_{t+1})] = y_t \exp( \frac{\sigma v^2 }{2} + \frac{v^2}{2}).$$
Is my argument correct?
 A: Your assumption of $y_t \perp \!\!\! \perp  \exp(\varepsilon_{t+1})$ is correct by definition of $\varepsilon_t$, i.e:
$X \perp \!\!\! \perp Y \Leftrightarrow X \perp \!\!\! \perp f(Y) \ \forall \ f(\cdot) \tag{1}$
where $\perp \!\!\! \perp$ is the independence symbol. In words: if $X$ is independent of $Y$, then $X$ is also independent of any deterministic function of $Y$.
Here: $\varepsilon_t \sim iid(\mu,\sigma^2)$, i.e. non-zero mean iid noise. Then $\varepsilon_t$ is per definition an innovation at time $t$, i.e. independent of all variables at $t$, i.e. $y_t, \textbf{X}_t \perp \!\!\! \perp \varepsilon_t$ and therefore also any deterministic function of it is also independent of all variables at $t$, i.e. $y_t, \textbf{X}_t \perp \!\!\! \perp f(\varepsilon_t)$ where $f(x) =exp(x)$.
For solving $E[y_{t+1}|y_t]$ given $y_{t+1}=y_t \exp(\varepsilon_{t+1})$:
Option 1:
Take $E[y_{t+1}|y_t]$ directly, as you did. The rest is correct.
Option 2:
In cases with exponential functions, it might be more attractive to work with the $\log$ of the variables, due to additive properties:
Take $\log(y_{t+1})$ to get $\log(y_{t+1})=\log(y_t) + \varepsilon_{t+1}.$ Then take
$$ E[\log(y_t)+\varepsilon_{t+1}|y_t]=\log(y_t)+E[\varepsilon_t]=\log(y_t)+\mu$$
Here, independence may be more obvious. However, I can't think of a direct way to get from this to your variable of interest $E[y_{t+1}|y_t]$. Note that $E[\log(y_{t+1})|y_t] \ne \log(E[y_{t+1}|y_t])$, therefore one can't simply take $\exp()$ of $\log(y_t)+\mu$.
