Assume you have past consumption levels $c_1, \dots c_n$ at times $t_1, \dots t_n$ and cumulated consumption levels $y_1=c_1, y_2 = c_1 + c_2, \dots y_n=\sum_{k=1}^{n} c_k$.
(I use the quadratic term to indicate an expected positive / negative change in the consumption $c$.)

The question is, how can we estimate the predicted consumption $E\left[c_{n+1}\right]$ at $t_{n+1}$ and its variance $\text{Var}\left(c_{n+1}\right)$ for a $t_{n+1} > t_n$?

My thoughts are as follows, formulas are in line with the great book by Kutner et. al., "Applied Linear Statistical Models" (2004):
I use a quadratic model $y \sim 1 + t + t^2$ to obtain the estimated coefficients $\hat{\beta}$ (5.60) and their estimated variance $\hat{s}^2$ (5.93). Using these estimates I get the model

$$y\left(t\right) = \hat{\beta_0} + \hat{\beta_1}\cdot t + \hat{\beta_2}\cdot t^2$$

and I can calculate $E\left[c_{n+1}\right] = \hat{y}\left(t_{n+1}\right) - \hat{y}\left(t_n\right)$. Then I would estimate the variance $\text{Var}\left(c_{n+1}\right) = \text{Var}\left(y_{n+1}\right) + \text{Var}\left(y_n\right)$ as $y_n$ and $y_{n+1}$ are uncorrelated in case of uncorrelated error terms $\epsilon_i$. Then I use similar to (2.33)

$$\left[E\left[c_{n+1}\right] - t_{n-3}\left(1-\frac{\alpha}{2}\right) \cdot \sqrt{\text{Var}\left(c_{n+1}\right)} ; E\left[c_{n+1}\right] + t_{n-3}\left(1-\frac{\alpha}{2}\right) \cdot \sqrt{\text{Var}\left(c_{n+1}\right)} \right]$$

to obtain the $1- \alpha = 95$ % confidence interval, where according to (5.100) and (5.98)

$$\text{Var}\left(y_{n+1}\right) = \text{MSE}\cdot \left( 1 + X_{n+1}'\left(X'X\right)^{-1}X_{n+1}\right)$$ $$\text{Var}\left(y_{n}\right) = \text{MSE}\cdot X_n'\left(X'X\right)^{-1}X_n$$

and as (1.22 with 3 parameters)

$$MSE = \frac{\sum \left(y_k - \hat{y}_k\right)^2}{n - 3}, X_k = \left(1, t_k, t_k^2\right)$$

Is this procedure correct?

PS: There are for sure other ways (timeseries, ...) to do the job, but I would be interested, in how far I can get with this approach.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.