Consider a linear forecasting problem where all shocks $\{\epsilon_i\}_1^n$ are independently distributed with $\epsilon_i\sim N(0,\sigma_i^2)$ for all $i$. Suppose you want to forecast $\theta = \sum_{i=1}^m a_i \epsilon_i$ for $m<n$, while observing signals $\mathbf x = \{x_j\}_1^p$ with each $x_j=\sum_{i=1}^n b_{ij} \epsilon_i$.

Let $\mathbb E[\theta|\mathbf x]$ denote the optimal forecast, that minimizes the mean squared error. I want to show that

$$\frac{\partial\operatorname{Cov}(\mathbb E[\theta|\mathbf x],\theta)}{\partial \sigma_i}<0, \quad\text{for }i\in\{m+1,\dots,n\}.$$

I think this is true because an increase in the variance of $\sigma_i$ for $i\in\{m+1,\dots,n\}$ makes the signals more noisy, therefore, the forecast less accurate.

By a similar argument it should also be possible to prove that

$$\frac{\partial\operatorname{Var}(\mathbb E[\theta|\mathbf x]|\epsilon_i)}{\partial \sigma_i}<0, \quad\text{for }i\in\{m+1,\dots,n\}.$$

For simple examples of this problem I've been able to prove this, but I haven't been able to generalize it. I suspect these results might exist, so just a reference of where I could find them would be of great help.

  • $\begingroup$ I am having problems with the statement. I presume a "shock" is a disturbance/noise in some discrete readings and the noise is affecting the reading not the timing? "forecast": Are you accumulating past readings in an attempt to compensate out a characteristic of the noise level; say 1/F noise? What are your "optimal forecast" adjustable's; a or b? Did you mean the derivative to really be the noise variance or did you mean adding more readings to an estimate? $\endgroup$ – rrogers Feb 9 '17 at 14:20

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