Space–Time Earthquake Prediction: The Error Diagrams

Abstract

The quality of earthquake prediction is usually characterized by a two-dimensional diagram n versus τ, where n is the rate of failures-to-predict and τ is a characteristic of space–time alarm. Unlike the time prediction case, the quantity τ is not defined uniquely. We start from the case in which τ is a vector with components related to the local alarm times and find a simple structure of the space–time diagram in terms of local time diagrams. This key result is used to analyze the usual 2-d error sets {n, τ _w } in which τ _w is a weighted mean of the τ components and w is the weight vector. We suggest a simple algorithm to find the (n, τ _w ) representation of all random guess strategies, the set D, and prove that there exists the unique case of w when D degenerates to the diagonal n + τ _w  = 1. We find also a confidence zone of D on the (n, τ _w ) plane when the local target rates are known roughly. These facts are important for correct interpretation of (n, τ _w ) diagrams when we discuss the prediction capability of the data or prediction methods.

Appendix

We are going to prove the statements 3.1–3.10.

Proof for 3.1, 3.2

Obviously, the projection γ _w preserves the property of convexity. Therefore, ${{\mathcal{E}}}_w$ and D _w are convex at the same time as are ${{\mathcal{E}}}$ and D. If D _w degenerates to the diagonal $\tilde{D}:n+\tau_w=1,$ then the simplex D is given by any of the two equations: $n+\sum^k_{i=1} w_i\tau_i = 1$ and $n+\sum^k_{i=1}\lambda_i \tau_i/\lambda =1.$Hence w _i = λ _i /λ.

Proof of 3.3.

The simplex D is the convex hull of $(n,{\varvec{\tau}})$ points of the form $Q(\varepsilon )=(1-\sum \lambda_i \varepsilon_i/\lambda , \varepsilon_1, \ldots , \varepsilon_k),$ where ɛ _i  = 0, 1. Accordingly, D _w is the convex hull of the Q _w (ɛ), see (12).

Proof of 3.4.

This statement follows intuitively from dimensionality considerations: The k-dimensional surface $n({\varvec{\tau}})$ with k > 1 is projected onto the (n, τ _w ) plane, hence its image cannot be single-dimensional in the generic case.

In order to prove (13), we note that a convex function on the simplex $S_n=\{\sum^k_{i=1}\tau_i w_i=u, 0\le \tau_i \le 1\}$ reaches its maximum at one of the edges, specifically, at a point of the form

$$ {\varvec{\tau}}=(\varepsilon_1, \ldots , \varepsilon_{i-1}, x, \varepsilon_{i+1}, \ldots , \varepsilon_k), \quad \varepsilon_j=0;1. $$

The use of (10) gives (13).

Suppose the upper and lower boundaries of the image of $n({\varvec{\tau}})$ are identical and the {n _i (τ)} are piecewise smooth functions. Consider all ${\varvec{\tau}} =(\tau_1, \ldots ,\tau_k)$ for which

$$ \sum^k_{i=1} \lambda_i n(\tau_i)/\lambda = n_0, \quad \sum^k_{i=1} w_i \tau_i = \tau_w, \quad n_0=n(\tau_w), $$

where τ _w is fixed.

Varying, e.g., τ ₁ and τ ₂, we have after differentiation:

$$ \lambda_1 n_1'(\tau_1) \tau_1' + \lambda_2 n_2'(\tau_2) = 0, \quad \tau_1' = -w_2/w_1. $$

(A1)

If τ ₁, τ ₂ are points of smoothness of n _i (τ), i = 1, 2, then repeated differentiation of (A1) will give

$$ \lambda_1 n_1''(\tau_1) (w_2/w_1)^2 + \lambda_2 n_2''(\tau_2) = 0. $$

However, n _i ′′(τ _i )≥ 0, i = 1, 2. Hence n _i ′′(τ _i ) = 0, i.e., n _i (τ) are locally linear at all points of smoothness. Since n _i (τ) are piecewise smooth, it follows that for any discontinuous point τ ₁ of n ₁(·) one can find a point τ ₂ where n ₂(·) will be smooth. Consequently, when n ₁ is discontinuous at τ, one should replace n ₁′(τ ₁) with n ₁′(τ ₁ + 0) and n ₁′(τ ₁ − 0) in equation (A1). But then we have from (A1) that n ₁′(τ) is continuous at τ ₁; hence all n _i (τ) are linear. Taking the boundary conditions n _i (0) = 1 and n _i (1) = 0 into account, we have n _i (τ) = 1 − τ. However, in that case one has ${{\mathcal{E}}}=D,$ and, in virtue of 3.2, w _i = λ _i /λ.

Proof of 3.5.

Let Q _w be the point where the convex set {ψ ≤ c} is tangent to the convex curve n(τ _w ). The function ψ reaches its minimum at the point Q _w on ${{\mathcal{E}}}_w,$ because the sets {ψ ≤c} increase with increasing c. Since $Q_w \in {{\mathcal{E}}}_w,$ the pre-image $Q=(n,{\varvec{\tau}}) \in {{\mathcal{E}}}.$ At this point φ (Q) = ψ (Q _w ) reaches its minimum on ${{\mathcal{E}}},$ hence Q belongs to the surface n(τ).

Proof of 3.6

follows from 2.3.

Proof of 3.7.

Let Q = (n ₀, τ ₀₁, ..., τ _0k ) belong to $n({\varvec{\tau}}).$ If $w_i = -{\frac{\partial n}{\partial\tau_i}} (Q)/c,$ then the equation

$$ n + c \sum^k_{i=1} w_i \tau_i = n_0 + c \sum^k_{i=1} w_i \tau_{i0} $$

(A2)

defines the tangent plane to $n({\varvec{\tau}}).$ Since $n({\varvec{\tau}})$ is convex and decreasing, it follows that w _i  ≥ 0 and ${{\mathcal{E}}}$ lie on the same side of the plane (A2). Consequently, a strategy having the characteristics Q = (n ₀, τ ₀₁,..., τ _0k ) optimizes the losses $\varphi =n + c \sum^k_{i=1} w_i \tau_i.$ Using 3.5, we complete the proof.

Proof of 3.8.

By (10) and (16) one has

$$ \beta = n = \sum^k_{i=1} \lambda_i/\lambda \cdot n_i(\tau_i), \quad \alpha = \tau_w = \sum^k_{i=1} w_i \tau_i. $$

In the trivial case of I(t), one has n _i (τ) = 1 − τ and α + β = 1. Hence

$$ \beta = 1 - \sum^k_{i=1} \lambda_i/\lambda \cdot \tau_i, \quad \alpha = \sum^k_{i=1} w_i \tau_i = 1-\beta, $$

i.e., w _i  = λ _i /λ, i = 1, ..., k.

Suppose that {w _i } = {λ _i /λ}. The likelihood ratio of measures (17) and (18) at the point ω = (J(t), j) is

$$ L(\omega) = P(\hbox{d}\omega \vert H_1)/P(\hbox{d}\omega \vert H_0) = r_j(t)/\lambda_j. $$

Accepting the hypothesis H ₁ as soon as L(ω) > c and H ₀ otherwise, one has

$$ \begin{aligned} \alpha =& \int\limits_{L>c} P(\hbox{d}\omega \vert H_0) = \sum^k_{j=1} E {{{\mathbf{1}}}}_{(r_j/\lambda_j >c)} \cdot \lambda_j/\lambda = \sum^k_{j=1} \tau_j \lambda_j/\lambda = \tau_w , \\ \beta =& \int\limits_{L>c} L(w) P(\hbox{d}\omega \vert H_0) = \sum^k_{j=1} E r_j/\lambda_j \cdot {{{\mathbf{1}}}}_{(r_j/\lambda_j <c)} \cdot \lambda_j/\lambda = \sum^k_{j=1} n_j(\tau_j) \lambda_j/\lambda = n. \end{aligned} $$

Here we have used 2.1 and 2.2.

Proof of 3.9.

Let us consider the following testing problem: H ₁ versus H ₀ with the errors β = P ₁(L < c) and α = P ₀(L ≥ c) where L(ω) = dP ₁/dP ₀ is the likelihood ratio. Obviously

$$ E_0 f(L) : = \int\limits f(L(\omega)) \hbox{d}P_0(\omega) = \int\limits f(c) \hbox{d}F_L(c), $$

where F _L is the distribution of L with respect to the measure P ₀. But dβ = c dF(c) and dα =  −d F(c). Therefore

$$ E_0 f(L) = \int\limits_0^1 f\left(-{\frac{\hbox{d}\beta}{\hbox{d}\alpha }}\right) \hbox{d}\alpha. $$

Applying this relation to the case (16), (17), (18), one has

$$ \begin{aligned} \int\limits_0^1 f \left(-{\frac{\hbox{d}n_{\lambda}}{\hbox{d}\tau}}\right) \hbox{d}\tau =\,&E_0 f(L) = \sum^k_{i=1} E f \left({\frac{r_i(t)} {\lambda_i}}\right) p_i \\ =&\sum^k_{i=1} E f(L_i) p_i = \sum^k_{i=1} p_i \int\limits_0^1 f\left(-{\frac{\hbox{d}n_{i}}{\hbox{d}\tau}}\right) \hbox{d}\tau \end{aligned} $$

Here L _i is the likelihood ratio dP ₁/dP ₀ for G _i .

Proof of 3.10.

In virtue of 3.3 the set D _w is a convex hull of the points (12). Let $\tau_w(\varepsilon)=\sum^r_{i=1} w_i.$ We need to find the maximum of $y=\sum^r_{i=1}p_i$ under the conditions

$$ \sum^k_{i=1} p_i^2/\widehat{p}_i \le 1+q, $$

(A3)

$$ \sum^k_{i=1} p_i = 1, \quad p_i\ge 0. $$

(A4)

Because y is a linear function, it reaches its maximum at the boundary of the region (A3). Consequently, we will consider y as a function of the variables (p ₂ ..., p _k-1) given by the equation

$$ \sum^k_{i=1} p_i^2/\widehat{p}_i = 1+q $$

(A5)

where p ₁ = y − p ₂ − ... − p _r , p _k  = 1 − y − p _r+1 − ... − p _k-1. The point of maximum of y is formally defined by the following equations

$$ \partial y/\partial p_i =0, \quad 1<i<k. $$

This gives $p_i=c_1\widehat{p}_i, i=1,\ldots ,r;$ $p_j=c_2\widehat{p}_j,\, j=r+1,\ldots ,k.$ Taking (A5, A4) into account, we get two equations for c ₁ and c ₂. Finally we have

$$ \begin{aligned} y=&\widehat{y} + \sqrt{q \widehat{y}(1-\widehat{y})}, \quad \widehat{y} = \sum^r_{i=1} \widehat{p}_i,\\ p_i=&y\widehat{p}_i/\widehat{y},\quad i\le r, \\ p_j=&(1-y)\widehat{p}_j/(1-\widehat{y}), \quad j>r. \\ \end{aligned} $$

The conditions (A5, A4) hold, if 0 ≤ y ≤ 1, that is, when $\widehat{y}(1+q)\le 1.$ Otherwise we have to consider the vector

$$ (\widehat{p}_1/\widehat{y}, \ldots , \widehat{p}_r/\widehat{y}, 0, \ldots, 0). $$

This vector satisfies (A3, A4) and gives the maximum possible value for y, y = 1.

The proof is complete.