Home

Downscaled Climate Projections

In a deterministic approach, the variance of a downscaled variable is in general smaller than the observed variance of the same variable. This is the case simply because the large-scale does not "explain" all the variance of the small-scale. For example, consider classical linear regression with a single large-scale predictor, $L(t)$, and a small-scale predictand, $s(t)$, where both are a function of time, $t$. Using ordinary least squares, we can obtain the constants $a$ and $b$ that "best" predict $s$ from $L$: \begin{equation*} \tilde{s}(t) = a L(t) + b, \end{equation*} where $\tilde{s}$ is the $s$ predicted from the large-scale. The variance of $\tilde{s}$, however, is too small: \begin{equation*} \mathrm{var}(\tilde{s}) = r^2 \mathrm{var}(s), \end{equation*} where $r$ is the correlation between $s$ and $L$.
The most consistent solution is too explicitly consider the fact that some variance is unexplained: $$\label{noise} \tilde{s}(t) = a L(t) + b + \sqrt{1-r^2} \sigma_s n(t),$$ where $\sigma_s$ is the standard deviation of $s$ and $n(t)$ is random noise with zero mean and unit standard deviation. When classical linear regression applies, $n(t)$, is normally distributed.
The above equation \eqref{noise} can also be expressed as: \begin{align*} \Pr(\tilde{s} = s_0) &= \frac{1}{\sigma_s \sqrt{2 \pi}} \exp \left( -\frac{(s_0 - \mu)^2}{2 \sigma^2} \right)\\ \mu &= a L(t) + b \\ \sigma &= \sigma_s \sqrt{1-r^2} \end{align*} Hence, we are predicting the Probability Density Function (PDF) of $s(t)$ as a function of the large-scale, $L(t)$. When we downscale precipitation, the form of the parametric PDF will be something other than the normal distribution used in the example above.