Spatial prediction using bivariate exponential distribution

Abstract

In this paper, a certain bivariate exponential distribution is used for the spatial prediction. The unobserved random variable is predicted by the projection onto the space of all linear combinations of the powers, up to degree m, of the observed random variables plus the constant 1. We obtain a solution by assuming that all the bivariate distributions follow Gumbel’s type III or logistic form of bivariate exponential. The method is implemented on two data sets and the results are presented. The predictions are compared with the original values through Mean Structural Similarity (MSSIM) index of Wang et al. (IEEE Trans Image Process 13(4):600–612, 2004). Using the MSSIM index the proposed method is also compared with Ordinary Kriging and with Simple Kriging after normal score transform.

Appendices

Appendix 1

Suppose for some monotone function ${\phi}(\cdot),$ such that Y _α = ϕ(Z _α) has standard normal distribution. Also it is assumed that all bivariate distributions of Y _s and Y _(s+h) are given by bivariate normal distribution with correlation coefficient ρ _h . Now we will show that the projection of the unobserved r.v. Y ₀ on to the space ${{\mathcal{P}}}_n^m$ is same as the projection of Y ₀ on to the space of all linear combinations of {Y ₁, Y ₂,…,Y _n }. In other words, the predictor obtained by projecting Y ₀ on to the space ${{\mathcal{P}}}_n^m$ reduces to the linear form. This does not follow directly from the fact that, the conditional expectation function of the multivariate normal distribution is linear because here, we only assume that all the bivariate distributions are bivariate normal and do not make any assumption on the multivariate distribution of the process {Y _s }. To proceed, define the m-dimensional process $\{{{\mathbf{X}}}_{\varvec{\alpha}}\}$ such that ${{\mathbf{X}}}_{\varvec{\alpha}} = \{Y_{\alpha}, Y_{\alpha}^2, \ldots, Y_{\alpha}^m \}.$ Now we have data at n locations on m-dimensional process X _α. Note that X _α,i = Y ⁱ_α . The random variable X _α,1 is treated as a primary variable since the estimate of Y ¹_α is to be sought. The other random variables X _α,i for i > 1, are considered as the secondary variables. Autokrigeability condition (see, for example, Subramanyam and Pandalai 2004) states that we can drop all secondary variables from the cokriging equations if the following equation satisfies.

$$ {\frac{{\rm Cov}(X_{\alpha,1}, X_{\alpha + h, i})}{{\rm Cov}(X_{\alpha,1}, X_{\alpha + h, 1})}} = {\frac{{\rm Cov}(Y_{\alpha}, Y_{\alpha + h}^ i)}{{\rm Cov}(Y_{\alpha}, Y_{\alpha + h})}} = \hbox{constant}. $$

(A1.1)

From the bivariate normality of Y _α and Y _α+h , one can show that

$$ {{\mathbb{E}}}(Y_{\alpha} Y_{\alpha + h }^i) = \left\{\begin{array}{ll} 0, & (i+1) \; \hbox{is}\; \hbox{odd};\\ {\frac{2^{{\frac{i+1}{2}}}\Upgamma({\frac{i+2}{2}})} {\sqrt{\Uppi}}} \rho_h = c_i \rho_h & (i+1) \; \hbox{is} \;\hbox{even}, \end{array}\right. $$

where c _i depends only on i and hence independent of lag h. Since Y _α is a standard normal variate, ${\rm Cov}(Y_{\alpha}, Y_{\alpha + h }^i) = {{\mathbb{E}}}(Y_{\alpha} Y_{\alpha + h }^i) = c_i \rho_h.$ So the Eq. (A1.1) becomes

$$ {\frac{{\rm Cov}(Y_{\alpha}, Y_{\alpha + h}^i)}{{\rm Cov}(Y_{\alpha}, Y_{\alpha + h})}} ={\frac{c_i \rho_h}{\rho_h}} = c_i, \;\hbox{a constant}. $$

Hence all secondary variables i.e. X _α,i = Y ⁱ_α for i > 1, can be dropped from the estimation of Y ₀ and thus estimator reduces to the linear form.

Appendix 2

The expectation ${{\mathbb{E}}}(Y_{\alpha}^i Y_{\alpha +h}^j)$ given in the Eq. (7) is derived here. Let (X,Y) follows a bivariate exponential distribution of the Gumbel’s logistic form with the dependence parameter δ. Then the joint survival function is given by the Eq. (7) and the joint density function by

$$ g(x,y)={\rm exp}\left\{-\left(x^{\frac{1}{\delta}} + y^{\frac{1}{\delta}}\right)^{\delta}\right\} \left[x^{{\frac{1}{\delta}}-1} y^{{\frac{1}{\delta}}-1} \left(x^{{\frac{1}{\delta}}} + y^{{\frac{1} {\delta}}}\right)^{\delta - 2}\right]\left(\left(x^{{\frac{1}{\delta}}} + y^{{\frac{1}{\delta}}}\right)^\delta +\left({\frac{1}{\delta}}-1\right)\right). $$

Now consider the transformation,

$$ U = {\frac{X^{{\frac{1}{\delta}}}}{X^{{\frac{1}{\delta}}} + Y^{{\frac{1}{\delta}}}}} \quad S = \left(X^{{\frac{1}{\delta}}} + Y^{{\frac{1}{\delta}}}\right)^{\delta}. $$

Note that U ∈ [0,1] and S ∈ [0, ∞). The inverse transform is

$$ X = U^{\delta} S \quad Y = (1 - U)^\delta S.$$

The Jacobian of the transformation is

$$ |J| = \left|{\frac{\partial(X,Y)}{\partial(U,S)}} \right| = \left| \begin{array}{ll} \delta \; U^{\delta - 1} S & U^\delta\\ - \delta(1-U )^{\delta -1} S & (1-U )^\delta\\ \end{array}\right| =\delta \; S \; U^{\delta - 1} \; (1-U)^{\delta - 1}. $$

The joint density of the transformed variables (U, S) is given by

$$ \begin{aligned} h(u,s) & = g(x,y)|_{(u,s)} |J| \\ \,=\,& {\rm exp}\{-s\} \left(u^{1-\delta} s^{2\left({\frac{1}{\delta}} - 1\right)} (1-u)^{1-\delta} s^{1-{\frac{2}{\delta}}} \right)\left[s+\left({\frac{1}{\delta}}-1\right)\right]\delta \; s \; u^{\delta - 1} \; (1-u )^{\delta-1}\\ & = {\rm exp}\{-s\} \;\delta \;\left[s+\left({\frac{1}{\delta}}-1\right)\right] 1\\ & = h_1(s)\; \; h_2(u), \end{aligned} $$

(A2.1)

where $h_1(s) = {\rm exp}\{ -s \} \; \delta \; \left[s +({\frac{1} {\delta}} - 1) \right]$ is the marginal distribution of the variable S and h ₂(u) = 1 is the marginal distribution of standard Uniform variate U. From the Eq. (A2.1), we can say that the random variables U and S are independent. This implies

$$ \begin{aligned} {\mathbb E}\left( U^{i \delta} S^i (1 - U)^{j \delta} S^j \right) & = {\mathbb E} \left( X^i Y^j \right) \\ & = {\mathbb E}\left( U^{i \delta} (1 - U)^{j \delta} \right) {\mathbb E}\left( S^{(i+j)} \right).\\ \end{aligned} $$

(A2.2)

Now,

$$ \begin{aligned} {\mathbb E}\left( U^{i \delta} (1 - U)^{j \delta} \right) &= \int\limits_0^1 U^{i \delta} (1 - U)^{j \delta} du \\ & = {\frac{\Upgamma(i \delta+1) \Upgamma(j \delta + 1)} {\Upgamma(\delta (i+j) + 2)}}\\ \end{aligned} $$

(A2.3)

and the second term in the Eq. (A2.2) is

$$ \begin{aligned} {\mathbb E}\left( S^{(i+j)} \right) &= \int\limits_0^\infty s^{(i+j)} h_1(s) ds \\ &= \int\limits_0^\infty s^{(i+j)} {\rm exp}\{ -s \} \; \delta \; [ s + ({\frac{1} {\delta}} - 1) ] ds \\ & = \delta \left[ \int\limits_0^\infty s^{(i+j+1)} {\rm exp}\{ -s \} \; ds + \; ({\frac{1}{\delta}} - 1) \int\limits_0^\infty s^{(i+j)} {\rm exp}\{ -s \} ds \right] \\ & = \delta \left[\Upgamma(i+j+2)+\left({\frac{1}{\delta}}-1\right) \Upgamma(i+j+1)\right]. \end{aligned} $$

(A2.4)

From the Eqs. (A2.2), (A2.3), and (A2.4)

$$ {\mathbb E}( X^i Y^j) = {\frac{[({\frac{1}{\delta}}-1) \Upgamma(i+j+1) + \Upgamma(i+j+2)] \Upgamma(i \delta+1) \Upgamma(j \delta + 1)} {{\frac{1}{\delta}} \Upgamma( \delta (i+j)+ 2)}}. $$

Hence the expectation in the Eq. (7).