Q:

Intuitive view of conditional expectation

I would like someone to give me an intuitive view of conditional expectation. I mean: I have always understood it through formulas but I don't "see" what it is yet.
Thank you

A:

as you have an open question, let me try to give you an example in finance.
Let $(M_t)$ be a martingale, i.e. a fair game. Such processes, on average (their expectation), do not increase or decrease, they remain constant. As a consequence, $\mathbb{E}[M_t]=M_0$, i.e. if you ask me right now what value I expect $M$ will have at time $t$, I expect on average it to be at the starting point $M_0$. An example may be the wealth process of a fair game or the price process of a discounted stock price (with respect to an appropriate measure).
What happens now if some time passes. We are not at time $t=0$ anymore but midway through? You could ask me tomorrow what I expect where $M$ is going to be in one week. More formally, this translates to $\mathbb{E}[M_t\mid\mathcal{F}_s]$ if $s<t$. Since $M$ is a martingale, of course, $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_0$ and I will still answer that I expect to be, on average, where it starts, at $M_0$. 
Here, $\mathcal{F}_s$ represents all the information I have gathered until time $s$. This may improve my expectation. Suppose I ask you today where the apple stock trades in one year. Either you have a crystal ball or guess a number, this number is $\mathbb{E}[M_t]$. I could ask you in 6 months again and depending what will happen over the next half a year, your expectation may have changed. After all, the uncertainty has decreased slightly and we know a bit better what may happen. If I ask you in 11 months, you can give a reasonable guess and so on. So, with an increase of available information, your expectation changes. The conditional expectation captures this notion. $\mathbb{E}[M_t\mid\mathcal{F}_s]$ really only means what do you expect $M_t$ to be given the information until time $s$.
The rules and formuale involving conditional expectation are thus very intuitive.

Let $s=0$. We typicall assume that $\mathcal{F}_0=\{\emptyset,\Omega\}$. This is the same as having no information at all. After all, at time zero, you could not have gathered any specifing further information. Thus, $$\mathbb{E}[M_t\mid \{\emptyset,\Omega\}] = \mathbb{E}[M_t].$$ This means that conditioning on $F_0$ does not help at all to improve your expectation. This generalises as follows. Let $M$ be independent of $\mathcal{G}$. Then, $\mathbb{E}[M_t\mid \mathcal{G}] = \mathbb{E}[M_t]$. Imagine if $\mathcal{G}$ simply does not contain any information about $M$. Financially, if I ask you again about the Apple stock and you gather some information about Japanese penny stocks, this may not improve your prediction about Apple and the price of Apple may be independent of Japanese penny stocks. Thus, if I ask you predict Apple after observing Japanese penny stocks for a month, your prediction will not have improved. If you however observe the Apple price for the last 4 weeks, this may help you to give a better expectation of the future Apple price.
Suppose you have a finite process $(M_t)_{t\in[0,T]}$ ending at some point $T$. Then, $\mathbb{E}[M_T\mid \mathcal{F}_T] = M_T$. Financially, if I ask you Apples stock price on 01/03/2019 and you know all prices from Apple until 01/03/2019, then of course you can tell me with absolute certainty which value you expect. This generalises as follows. If $M_t$ is $\mathcal{F}_t$-measurable, then $$\mathbb{E}[M_t\mid \mathcal{F}_t]= M_t.$$

This answer is already far too long and I would like to go on but let me conclude as follows. It is crucial to understand that $\mathbb{E}[X]$ is a real number, your standard expectation. The object $\mathbb{E}[X\mid \mathcal{F}_t]$ is a random variable. It does depend on $\mathcal{F}_t$ which, in turn, depends on the (random) realisation of a stochastic process.

