The exercise is from Yuh-Dauh Lyuu's Financial Engineering and Computation: Principles, Mathematics, Algorithms (the parts of the text the exercise refers to are included):
Exercise 13.1.4. (1) Use Eq. (13.1) to characterize the random walk in Example 13.1.1. (2) Show that the variance of the symmetric random walk's position after $n$ moves is $n$.
EXAMPLE 13.1.1 Consider a particle on the integer line, $0, ±1, ±2, \dots$. In each time step, this particle can make one move to the right with probability $p$ or one move to the left with probability $1 − p$ (see Fig. 13.1). [...] This random walk is symmetric when $p = 1/2$.
EXAMPLE 13.1.2 The random walk with drift is the following discrete-time process:
$$ X_n = \mu + X_{n-1} + \xi_n\, , \qquad (13.1) $$
where $\xi_n$ are independent and identically distributed with zero mean. The drift $\mu$ is the expected change per period.
So what does "characterize" mean? I guess putting it in the form of that equation.
Then my solution for 13.1.4 (1) would be that for any $p \in [0, 1]$
- the $\xi_n$ are independent with probability distribution given by $\mathrm{Prob}[\xi_n = 2 - 2p] = p$ and $\mathrm{Prob}[\xi_n = - 2p] = 1-p$ and
- $\mu = -1 + 2p$.
This makes $\mathrm{E}[\xi_n] = p\cdot(2-2p) + (1-p)\cdot(-2p) = 0$, as was required.
With probability $p$ we get $\xi_n + \mu = 2 - 2p - 1 + 2p = 1$ and with probability $1-p$ we get $\xi_n + \mu = -2p -1 + 2p = -1$, therefore $X_n = \sum_{i=1}^n\xi_i + \mu$ is a random walk (right-stepping with probability p, starting at zero). By substituting the sum up to $n-1$ by $X_{n-1}$ we have written the stochastic process in the form (13.1).
Then for 13.1.4 (2) we note that $$\mathrm{Var}[\xi_i-\mu] = \mathrm{Var}[\xi_i] = p(2-2p)^2+(1-p)(-2p)^2 = 4p(1-p)$$ which is 1 in case of a symmetric ($p = 1/2$) walk. Then we get the result, because the $\xi_i$ are i.i.d: $\mathrm{Var}[X_n] = \mathrm{Var}[\sum_{i=1}^n\xi_i + \mu] = \mathrm{Var}[\sum_{i=1}^n\xi_i] = n\, \mathrm{Var}[\xi_1] = n $
Now my question is: The official solution to this exercise in the book does not agree with what I wrote above:
Official solution to 13.1.4: (1) In Eq. (13.1), use $\mu = 0$ and $$ξ_n =\begin{cases} 1& \text{ with probability }p\\ −1& \text{ with probability }q ≡ 1 − p \end{cases}.$$ (2) Because the mean is zero, the variance equals $n[ (1/2) × 1 + (1/2) × (−1)^2 ] = n$.
What then is the correct solution?