Model-wise uncertainty decomposition in multi-model ensemble hydrological projections

Abstract

There has been a growing interest in model-wise uncertainty decomposition, which quantifies contribution of individual models such as emission scenarios, global circulation models, bias correction techniques and hydrological models, to the total uncertainty of a hydrological projection. However, little research has been conducted for model-wise uncertainty decomposition in spite of its usefulness. In this paper, we propose a novel method for decomposing the total uncertainties into model-wise uncertainties. The proposed model-wise uncertainty decomposition method can be applied with general uncertainty measures, which include mean absolute deviation and variance measures. Moreover, the proposed method provides an intuitive interpretation of the quantified model-wise uncertainties. The results of analyzing real data by the proposed method are presented.

Appendix 1: Proofs

1.1 Proofs for Sect. 3

1.1.1 Proof of Eq. (3.3)

By the identifiability condition (3.2), we have that for any $\mathbf {x}\in \mathcal {X}$,

$$\begin{aligned} \sum _{\mathbf {v}\in \mathcal {X}:\mathbf {v}_A=\mathbf {x}_A}Y_\mathbf {v}= n_{A^c}\sum _{B:B\subset A}\beta ^B(\mathbf {x}) \end{aligned}$$

(5)

for any $A\subset \{1,\dots , K\}$. By forward substitution where the equations are ordered by the number of elements in A, we have

$$\begin{aligned}&\beta ^\emptyset (\mathbf {x})={\bar{Y}}\\&\beta ^{\{k\}}(\mathbf {x})={\bar{Y}}_{\{k\}, \mathbf {x}}- \beta ^\emptyset (\mathbf {x}) = {\bar{Y}}_{\{k\}, \mathbf {x}}-{\bar{Y}} \\&\beta ^{\{k,h\}}(\mathbf {x})={\bar{Y}}_{\{k,h\}, \mathbf {x}}- \beta ^{\{k\}}(\mathbf {x})-\beta ^{\{h\}}(\mathbf {x})- \beta ^\emptyset (\mathbf {x}) \\&\quad = {\bar{Y}}_{\{k,h\}, \mathbf {x}}-{\bar{Y}}_{\{k\}, \mathbf {x}}-{\bar{Y}}_{\{h\}, \mathbf {x}}+{\bar{Y}} \end{aligned}$$

and in general,

$$\begin{aligned} \beta ^A(\mathbf {x})=\sum _{B:B\subset A}(-1)^{|A|-|B|}{\bar{Y}}_{B, \mathbf {x}}, \end{aligned}$$

(6)

which completes the proof. $\square$

1.1.2 Proof of Theorem 1

Let two different subsets A and $A'$ of $\{1,\dots , K\}$ be given. Without loss of generality we assume that there is $k\in \{1,\dots , K\}$ such that $k\in A'$ but $k\notin A.$ Let $\mathcal {X}_{-k}:=\mathcal {X}_{\{1,\dots ,K\}\setminus \{k\}}$ and let $\mathbf {x}_{-k}:=\mathbf {x}_{\{1,\dots ,K\}\setminus \{k\}}$ for $\mathbf {x}\in \mathcal {X}$. Then by the identifiability condition (3.2),

$$\begin{aligned} \textsc {cov}(\beta ^A(\mathbf {X}), \beta ^{A'}(\mathbf {X}))= n^{-1}\sum _{\mathbf {x}_{-k}\in \mathcal {X}_{-k}}\sum _{x_k\in \mathcal {X}_k}\beta ^A(\mathbf {x})\beta ^{A'}(\mathbf {x})\\=\sum _{\mathbf {x}_{-k}\in \mathcal {X}_{-k}}\beta ^A(\mathbf {x})\sum _{x_k\in \mathcal {X}_k}\beta ^{A'}(\mathbf {x})=0. \end{aligned}$$

This shows that all the covariance terms are zero, and hence proved the proof is done. $\square$

1.2 Proofs for Sect. 4

1.2.1 Proof of Eq. (4.2)

The assertion follows from that

$$\begin{aligned}&U_{A}^{\text {cumul}}\\&\quad = \frac{1}{n_{A^c}} \sum _{\mathbf {x}_{A^c}\in \mathcal {X}_{A^c}}\frac{1}{n_A} \sum _{\mathbf {v}\in \mathcal {X}:\mathbf {v}_{A^c}=\mathbf {x}_{A^c}}({Y_{\mathbf {v}}-\bar{Y}_{A^c, \mathbf {x}}})^2\\&\quad = \frac{1}{n} \sum _{\mathbf {x}\in \mathcal {X}}({Y_{\mathbf {x}}-\bar{Y}_{A^c, \mathbf {x}}})^2\\&\quad = \frac{1}{n} \sum _{\mathbf {x}\in \mathcal {X}} \left( {\sum _{B\subset \{1,\dots , K\}}\beta ^B(\mathbf {x})-\sum _{B\subset \{1,\dots , K\}:B\subset A^c}\beta ^B(\mathbf {x})}\right) ^2 \\&\quad = \frac{1}{n} \sum _{\mathbf {x}\in \mathcal {X}} \left( {\sum _{B\subset \{1,\dots , K\}:B \nsubseteq A^c}\beta ^B(\mathbf {x})}\right) ^2\\&\quad = \frac{1}{n} \sum _{\mathbf {x}\in \mathcal {X}} \sum _{B\subset \{1,\dots , K\}:B \nsubseteq A^c}({\beta ^B(\mathbf {x})})^2\\&\quad = \sum _{B\subset \{1,\dots , K\}:B \nsubseteq A^c}VAR(\beta ^B(\mathbf {X})) \\&\quad = \sum _{B\subset \{1,\dots , K\}:B\cap A\ne \emptyset }VAR(\beta ^B(\mathbf {X})) , \end{aligned}$$

where the third equality follows from Eq. (5) and the fifth equality follows from the identifiability condition (3.2). $\square$

1.2.2 On the properness of the stage uncertainties by the cumulative uncertainty approach

We have claimed the stage-wise uncertainties $\{U_k^{cumul,stage }: k\in \{1,\dots , K\}\}$ are properly decomposed. Here we give the proof of the claim. First, the sum-to-total condition can be verified as

$$\begin{aligned} \sum _{k=1}^K U_k^{\text {cumul,stage}}&= \frac{1}{K} \sum _{r=1}^K \frac{1}{\left( {\begin{array}{c}K-1\\ r-1\end{array}}\right) } \sum _{A\subset \{1,\ldots ,K\}: A\ni k, |A|=r} U^{\text {cumul}}_{A,k} \\&=\frac{1}{K} \sum _{k=1}^K \sum _{A\subset \{1,\ldots ,K\}: A\ni k}\frac{1}{\left( {\begin{array}{c}K-1\\ |A|-1\end{array}}\right) }\left[ {U_{A}^{\text {cumul}}-U_{A\setminus \{k\}}^{\text {cumul}}}\right] \\&= \frac{1}{K} \sum _{k=1}^K \left[ {\sum _{A\subset \{1,\ldots ,K\}: A\ni k}\frac{1}{\left( {\begin{array}{c}K-1\\ |A|-1\end{array}}\right) }U_{A}^{\text {cumul}}-\sum _{A\subset \{1,\dots ,K\}\setminus \{k\}}\frac{1}{\left( {\begin{array}{c}K-1\\ |A|\end{array}}\right) }U_{A}^{\text {cumul}}}\right] \\&=\frac{1}{K} \sum _{A\subset \{1,\dots ,K\}}\left[ {\sum _{k\in A}\frac{1}{\left( {\begin{array}{c}K-1\\ |A|-1\end{array}}\right) }U_{A}^{\text {cumul}} - \sum _{h\in A^c}\frac{1}{\left( {\begin{array}{c}K-1\\ |A|\end{array}}\right) }U_{A}^{\text {cumul}}}\right] \\&=\frac{1}{K} \sum _{A\subset \{1,\dots ,K\}}\left[ {|A|\frac{1}{\left( {\begin{array}{c}K-1\\ |A|-1\end{array}}\right) } -(K-|A|)\frac{1}{\left( {\begin{array}{c}K-1\\ |A|\end{array}}\right) }}\right] U_{A}^{\text {cumul}} \\&=U^{\text {cumul}}_{\{1,\dots ,K\}}, \end{aligned}$$

where the last equality follows from the fact that

$$\begin{aligned}&|A|\frac{1}{\left( {\begin{array}{c}K-1\\ |A|-1\end{array}}\right) } -(K-|A|)\frac{1}{\left( {\begin{array}{c}K-1\\ |A|\end{array}}\right) } \\&\quad =|A|\frac{(K-|A|)!(|A|-1)!}{(K-1)!} -(K-|A|)\frac{(K-|A|-1)!|A|!}{(K-1)!}=0 \end{aligned}$$

for any A such that $|A|\le K-1$.

The nonnegativity of the stage-wise uncertainties clearly follows from the nonnegativity of the model-wise uncertainties defined in (4.7), which is proven in Sect. 4.2. $\square$

1.2.3 Proof of Theorem 2

For the proof of of Theorem 2, we need the following technical lemma.

Lemma 1

For any $r\in \mathbb {N}\cup \{0\}$ and $s\in \mathbb {N}\cup \{0\}$, we have

$$\begin{aligned} \sum _{k=0}^{r}\frac{(s+k)!}{k!} = \frac{1}{s+1}\frac{(r+s+1)!}{r!}. \end{aligned}$$

Proof

From the identity $\left( {\begin{array}{c}n\\ k\end{array}}\right) =\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) +\left( {\begin{array}{c}n-1\\ k\end{array}}\right) ,$ we have

$$\begin{aligned} \sum _{k=0}^r\left( {\begin{array}{c}s+k\\ k\end{array}}\right)= & {} \sum _{k=0}^r\left( {\begin{array}{c}s+k\\ s\end{array}}\right) \\= & {} \sum _{k=0}^r\left( {\left( {\begin{array}{c}s+k+1\\ s+1\end{array}}\right) - \left( {\begin{array}{c}s+k\\ s+1\end{array}}\right) }\right) \\= & {} \left( {\begin{array}{c}s+r+1\\ s+1\end{array}}\right) . \end{aligned}$$

Therefore

$$\begin{aligned} \sum _{k=0}^{r}\frac{(s+k)!}{k!}= & {} s!\sum _{k=0}^r\left( {\begin{array}{c}s+k\\ k\end{array}}\right) \\= & {} s!\left( {\begin{array}{c}r+s+1\\ s+1\end{array}}\right) \\= & {} \frac{1}{s+1}\frac{(r+s+1)!}{r!}, \end{aligned}$$

which completes the proof. $\square$

We are ready to prove Theorem 2. Note that

$$\begin{aligned}&U_k^{cumul,stage }\\&\quad =\frac{1}{K}\sum _{r=1}^K \frac{1}{\left( {\begin{array}{c}K-1\\ r-1\end{array}}\right) }\sum _{A\subset \{1,\ldots ,K\}: A\ni k, |A|=r} U_{A, k}^{\text {cumul}}\\&\quad =\sum _{A\subset \{1,\ldots ,K\}: A\ni k}\frac{1}{K}\frac{1}{\left( {\begin{array}{c}K-1\\ |A|-1\end{array}}\right) }\sum _{B\subset \{1,\ldots ,K\}: B\cap (A\setminus \{k\}) =\emptyset , B\ni k} VAR(\beta ^B(\mathbf {X})). \\&\quad =\sum _{B\subset \{1,\ldots ,K\}: B\ni k}VAR(\beta ^B(\mathbf {X}))\left( {\sum _{A\subset \{1,\ldots ,K\}:B\cap (A\setminus \{k\}) =\emptyset , A\ni k}\frac{1}{K}\frac{1}{\left( {\begin{array}{c}K-1\\ |A|-1\end{array}}\right) }}\right) . \end{aligned}$$

For each B and each $j=0,\dots , K-|B|$, we need to obtain the number of subsets A with size $j+1$ such that $B\cap (A\setminus \{k\}) =\emptyset$ and $A\ni k$, which is given by

$$\begin{aligned}&|\{A:|A|=j+1, B\cap (A\setminus \{k\})=\emptyset , A\ni k\}| \\&\quad = |\{A:|A|=j+1,A\setminus \{k\}\subset B^c, A\ni k\}| = \left( {\begin{array}{c}K-|B|\\ j\end{array}}\right) . \end{aligned}$$

This implies that

$$\begin{aligned}&\sum _{A\subset \{1,\ldots ,K\}:B\cap (A\setminus \{k\}) =\emptyset , A\ni k}\frac{1}{K}\frac{1}{\left( {\begin{array}{c}K-1\\ |A|-1\end{array}}\right) } = \frac{1}{K}\sum _{j=0}^{ K-|B|}\sum _{A:|A|=j+1, B\cap (A\setminus \{k\})=\emptyset , A\ni k}\frac{1}{\left( {\begin{array}{c}K-1\\ j\end{array}}\right) }\\&\quad = \frac{1}{K}\sum _{j=0}^{ K-|B|}\frac{\left( {\begin{array}{c}K-|B|\\ j\end{array}}\right) }{\left( {\begin{array}{c}K-1\\ j\end{array}}\right) } \\&\quad = \frac{1}{K}\frac{(K-|B|)!}{(K-1)!}\sum _{j=0}^{ K-|B|} \frac{(K-1-j)!}{(K-|B|-j)!}\\&\quad = \frac{(K-|B|)!}{K!}\sum _{k=0}^{ K-|B|}\frac{(|B|-1+k)!}{k!}\\&\quad =\frac{(K-|B|)!}{K!}\frac{1}{|B|}\frac{K!}{(K-|B|)!}=\frac{1}{|B|}, \end{aligned}$$

where the fifth equality is obtained by applying Lemma 1 with $r=K-|B|$ and $s=|B|-1$. This completes the proof. $\square$