Note that $a$ has a mean of 0.
My approach:
$$X_t=X_{t-1}+a_t$$
$$E[X_{t+1}mid X_1 + dots+X_{t-1}]$$
$$=E[X_{t-1}+2amid X_1 + dots+X_{t-1}]$$
$$=E[X_{t-1}mid X_1 + dots+X_{t-1}]+E[2amid X_1 + dots+X_{t-1}]$$
$$=E[X_{t-1}mid X_1 + dots+X_{t-1}]+0$$
$$=E[X_{t-1}mid X_1 + dots+X_{t-1}]$$
$$=X_{t-1}$$
Am I doing something wrong here? shouldn't the end product be $X_t$?
Contents
hide
Best Answer
begin{align} E[X_{t+1} mid X_1, ldots, X_t] &= E[X_t + a_{t+1} mid X_1, ldots, X_t] \ &= X_t + E[a_{t+1} mid X_1, ldots, X_t] \ &= X_t end{align}
Similar Posts:
- Solved – Prove Cov(X, Y) = Cov(X , E(Y|X) )
- Solved – P(A|B) and P(A|C) known, what is P(A|BC)
- Solved – P(A|B) and P(A|C) known, what is P(A|BC)
- Solved – Question about deriving posterior distribution from normal prior and likelihood
- Solved – The distribution of the product of a Bernoulli & an exponential random variable