Hi Sunil,
The derivation has since been added to the notes. I've reproduced it below for you (note that \( t_n \) has been replaced by \( n \)):
When we make forecasts using the random walk model, the errors accumulate over time, meaning that we are increasingly uncertain about our forecasts further into the future. In a linear regression model, the error (between actual and predicted values) is assumed to be stationary, ie it has the same distribution (with constant variance) over time.
Under the random walk model, \( \mu=E(k_t-k_{t-1}) \), ie \( \mu \) is the mean change in the time trend factor. \( \hat{\mu} \) is calculated from observed historical data, and so is itself also subject to uncertainty. If the estimator of \( \mu \) is based on \( n \) years of data, then it is the average of \( n-1 \) values of \( k_t-k_{t-1} \).
If we denote the estimator of \( \mu \) as \( \tilde{\mu} \), then since \( var(k_t-k_{t-1})=\sigma^2 \) for all \( t \) and the increments are independent:
\[ var(\tilde{\mu})=var\left[\frac{\left(k_n-k_{n-1}\right)+\left(k_{n-1}-k_{n-2}\right)+\cdots+\left(k_2-k_1\right)}{n-1}\right] \]
\[ =\sum_{i=2}^{n}var\left[\frac{(k_i-k_{i-1})}{n-1}\right] \]
\[ =\sum_{i=2}^{n}\frac{\sigma^2}{(n-1)^2} \]
\[ =\frac{\sigma^2}{n-1} \]
Hope this clears things up for you.
Dave
