How to find expected value for this financial expression I am reading a research paper, and there is such condition:
$$
E_t\Big[\frac{(F_t - S_{t+1})P_t}{S_tP_{t+1}}X\Big]=0
$$
where $E_t$ is expected value. I suspect that $E_t[F_t] = F_t$ because this variable (forward exchange rate) is observable at time $t$. Variable $S_t$ is exchange rate today, $P_t$ is price level in the economy and $X$ is, simply put, another random variable.
Then, it is said that under assumption that all variables are conditionally log-normally distributed, this equation can be rewritten as:
$$
E_t(s_{t+1}) = f_t -0.5Var_t(s_{t+1})+Cov_t(s_{t+1}, p_{t+1}) - Cov_t(s_{t+1},x)
$$
where $f$, $s$, $p$ and $x$ are logarithms of $S$, $F$, $P$ and $X$.
I don't understand how can the first equation be written this way. How can I represent $E_t(s_{t+1})$ the way it is written here starting with the first equation? Could anyone direct me towards the theory that is used here?
 A: The question changed substantially just as this answer was completed.  I am retaining the answer because it demonstrates the techniques needed to address the new question.

We have to make assumptions and add conditions in order to justify this calculation: $f_t$, $s_{t+1}$, and $p_{t+1}$ must have a multivariate normal distribution.  (This is stronger than assuming they are separately normal.)
In this case, the starting condition $\mathbb{E}_t\left[\frac{(F_t - S_{t+1})}{P_{t+1}}\right]=0$ is equivalent to
$$\mathbb{E}_t\left(\frac{F_t}{P_{t+1}}\right) = \mathbb{E}_t\left(\frac{S_{t+1}}{P_{t+1}}\right).$$
Both sides have lognormal distributions, because
$$\log\left(\frac{F_t}{P_{t+1}}\right) = \log(F_t) - \log(P_{t+1}) = f_t - p_{t+1}$$
is a linear combination of jointly normal variables and likewise for the right hand side.  Let's compute expectations (relative to the Sigma algebra at time $t$) using basic information about multivariate lognormal distributions and see where this goes:
$$\mathbb{E}_t\left(\frac{F_t}{P_{t+1}}\right) = \mathbb{E}_t\left(\exp(f_t - p_{t+1})\right) = \exp\left(\mathbb{E}_t(f_t - p_{t+1}) + \frac{1}{2}\text{Var}_t(f_t - p_{t+1})\right); \\
\mathbb{E}_t\left(\frac{S_{t+1}}{P_{t+1}}\right) = \mathbb{E}_t\left(\exp(s_{t+1} - p_{t+1})\right) = \exp\left(\mathbb{E}_t(s_{t+1} - p_{t+1}) + \frac{1}{2}\text{Var}_t(s_{t+1} - p_{t+1})\right).$$
Equating the right hand sides and taking logarithms yields
$$\mathbb{E}_t(f_t - p_{t+1}) + \frac{1}{2}\text{Var}_t(f_t - p_{t+1}) = \mathbb{E}_t(s_{t+1} - p_{t+1}) + \frac{1}{2}\text{Var}_t(s_{t+1} - p_{t+1}).$$
Exploiting the linearity of expectation and the bilinearity of covariance converts this into a sum of various expectations, variances, and covariances.  Many terms appear on both sides and can be cleared out.  Here's what is left:
$$\mathbb{E}_t(f_t) + \frac{1}{2}\text{Var}_t(f_t)- \text{Cov}(f_t, p_{t+1})\\ = \mathbb{E}_t(s_{t+1}) + \frac{1}{2}\text{Var}_t(s_{t+1})  - \text{Cov}(s_{t+1}, p_{t+1}).$$
At time $t$, $f_t$ is constant (almost surely) and therefore can be identified with its expectation.  The covariances with random variables are zero and so is its variance.  These observations simplify the preceding to
$$f_t = \mathbb{E}_t(s_{t+1}) + \frac{1}{2}\text{Var}_t(s_{t+1}) - \text{Cov}(s_{t+1}, p_{t+1}),$$
which is the desired formula.
