Variance and Covariance computation

Question

We have a random walk defined as: $$m_t=m_{t-1}+u_t$$ with $u_t$ representing a noise i.i.d. with expectation 0 and variance $\sigma^2_u$. Then, we define: $$p_t=m_t+q_tc$$ where $q_t=+1$ or $-1$ and $c$ is a constant. Notice also that $q_t$ are assumed to be serially independent and independent form both $m_t$ and $p_t$. Could someone help me to understand, step by step, how to get that: $$\operatorname{Var}(p_t-p_{t-1})=2c^2+\sigma^2_u$$ $$\operatorname{Cov}((p_{t-1}-p_{t-2}),(p_t-p_{t-1}))=-c^2 \text{ ?}$$

Answer 1

\begin{align} & \operatorname{Var}(p_t-p_{t-1}) \\[8pt] = {} & \operatorname{Var} \big((m_t+q_tc) - (m_{t-1}+q_{t-1}c)\big) \\[8pt] = {} & \operatorname{Var} (m_t-m_{t-1}) + c^2\operatorname{Var}( q_t-q_{t-1}) \end{align} The last line is valid if $c$ is a constant (i.e. not random) and $q_t-q_{t-1}$ is not correlated with $m_t-m_{t-1}.$ And the first term in the last line above is $\operatorname{Var}(u_t).$

As above, we have \begin{align} p_t - p_{t-1} = {} & (m_t-m_{t-1}) + c(q_t-q_{t-1}) \\ p_{t-1} - p_{t-2} = {} & (m_{t-1} - m_{t-1}) + c(q_{t-1}-q_{t-2}) \end{align}

\begin{align} & \operatorname{Cov}((p_{t-1}-p_{t-2}),(p_t-p_{t-1})) \\[8pt] \text{and } & \operatorname{Cov}(m_t-m_{t-1}, m_{t-1}- m_{t-2}) \\ = {} & \operatorname{Cov}(u_t,u_{t-1}) \\ = {} & 0 \text{ ?} \\[8pt] \text{and } & \operatorname{Cov}(q_t-q_{t-1},q_{t-1}-q_{t-2}) \\ = {} & \operatorname{Cov}(-q_{t-1}, q_{t-1}) \\ & \text{if the other covariances here are 0.} \end{align}