In stochastic calculus, I use the notation $E_t\left[X_T\right]$ more than I do $E \left[ X_T | \mathcal{F}_t \right]$. Sometimes, for convenience, I use the former freely and interchangeably with the latter. But, in the back of my mind, I always wonder and hence my question:
Are the two expressions, $E_t\left[X_T\right]$ and $E \left[ X_T | \mathcal{F}_t \right]$, really equivalent?
I understand that the full expression $E \left[ X_T | \mathcal{F}_t \right]$ denotes the time-$T$expectation of the stochastic variable $X(t)$ given its filtration, or all information, up to time $t$. I thought that $E_t\left[X_T\right]$ denotes just as much. Then, again, people use the long-hand $E \left[ X_T | \mathcal{F}_t \right]$ religiously, which makes me pause and post the question above.
I would like to know If they are different in anyway, however subtle it may be. Or, otherwise, what is the stigma to use the more economical $E_t[X_T]$?
Yes, they are explicitly equivalent. It is in fact not that unusual for people to write $E_{\mathcal{F}} X = E[X|\mathcal{F}]$ by emphasizing the projective nature of conditional expectations; See Thm 5.1. in Kallenberg's Foundations of Modern Probability. So formally denoting $E_{\mathcal{F}_t} =: E_t$ is perfectly formal. Now, suppose we are interested in different filtrations, say one minimally augmented and one not. Then $E_t$ must be contrasted with another conditional expectation operator (say $\tilde{E}_t$). At this stage I believe the notation complicates things. Another situation as mentioned in the comments is Markov processes, where the law of a Markov process is typically denoted $P_x$ or $P^x$. At to that the dependence of the filtration then is the suggested notation $E_x^t,E_{x,t}$ or $E_t^x$?