Expected position of 1D random walk
This is problem taken from Chap. 03 from Larry Wasserman's All of Statistics:
A particle starts at the origin of the real line and moves along the line in jumps of one unit. For each jump the probability is $p$ that the particle will jump one unit to the left and the probability is $1-p$ that the particle will jump one unit to the right. Let $X_n$ be the position of the particle after $n$ units. Find $\mathbb{E}[X_n]$.
So simply applying the definition of expectation, after 1 jump we have:
$\mathbb{E}[X_1] = -1 \cdot (1-p) + 1 \cdot p = 2p-1$
At this point my intuition tells me that the expected position after $n$ jumps will be $\mathbb{E}[X_n] = n(2p-1)$, however if I take the expectation after 2 jumps I have:
$\mathbb{E}[X_2] = -2 \cdot (1-p)^2 -1 \cdot (1-p) + 1 \cdot p + 2 \cdot p^2 = -2(1 -2p + p^2) -1 + p + p + 2p^2 = -2 + 4p - 2p^2 -1 + 2p + 2p^2 = 6p - 3 = 3(2p - 1)$
which would mean $\mathbb{E}[X_n] = (n+1)(2p-1)$. So I'm not sure which is correct? I can't spot something problematic in my algebra so is it a case of naive intuition or a case naive extrapolation to further jumps?
$\endgroup$ 51 Answer
$\begingroup$Revisit your second equation (In particular the second and third terms -1 (1-p) +1 p ) and you can figure it out.
$\endgroup$
