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Properties of filtration space

J

Johnson Adeleke

Member
Hey,
I have an understanding issue. Why is the E[X(t)|F(t)] = X(t) true? All I see here is that given a adapted-filtration space X then the expected value of X is still a random variable X. The only reason this makes sense to me perhaps would be through the tower of expectations i.e. E[E[X(t)|F(t)]] = E[X(t)]. If I try to dive deeper into how filtrations work I get confused. My understanding of the tower of expectations goes as follows: If E[X] = E_X[E_Y[X|Y]] but in this scenario why would the E_F(t)[X(t)|F(t)] = X(t)?
 
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Are you sure that E[X|F(t)] = X is true? Where have you seen this written down? Maybe you meant E[X(t)|F(t)] = X(t)?
 
Oh right, I see what you mean. So the statement E[X|F(t)] = X is only true if X is F(t)-measurable.
Being F(t)-measurable means that X is known at time t, so there's no uncertainty about its value because it's actually embedded within the filtration F(t). At time t, the variable X has been fully resolved and so is no longer random.
Hope that helps.
 
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