Numerical algorithms for spline interpolation on space of probability density functions

Abstract

This paper addresses the problem of spline interpolations on $\mathcal {P}$, the space of probability density functions when only a few observations $p_i \in \mathcal {P}$ are available. Given a finite set of $n+1$ distinct time instants $t_i$ and corresponding data points $p_i \in \mathcal {P}$, we consider the general problem of estimating a spline as a special regularized function $\gamma$ on $\mathcal {P}$ with $\gamma (t_i)=p_i$. In particular, we focus on estimating missing data using smooth temporal splines to overcome the discrete nature of observations. In addition to generalizing splines on $\mathcal {P}$ with minimal squared-norm of the acceleration, we give numerical schemes for solving $C^1$ and $C^2$ splines from data points $p_i \in \mathcal {P}$. The two solutions are then shown to be computationally efficient, geometrically simpler, extensible, and can be transposed to other spaces and applications.

Appendices

Appendix 1. $C^{2}$ Bèzier spline on $\mathbb {R}^{m}$

Let us consider the Euclidean case $\mathbb {R}^{m}$. Given a list of $(n+1)$ interpolation points $p_{0},...,p_{n}$ and the control points ${\widehat{b}}_{i}^{+}$ and ${\widehat{b}}_{i}^{-}$ in the right and in the left of $p_{i}$, $i=0,...,n$. The $C^{1}$ differentiability condition at knots $p_{i}$ allows us to express control points ${\widehat{b}}_{i}^{+}$ in terms of ${\widehat{b}}_{i}^{-}$ as:

$$\begin{aligned} {\widehat{b}}_{1}^{+}&= \frac{5}{3}p_{1}-\frac{2}{3}{\widehat{b}}_{1}^{-}, \end{aligned}$$

(36)

$$\begin{aligned} {\widehat{b}}_{i}^{+}&= 2p_{i}-{\widehat{b}}_{i}^{-},i=2,...,n-2 \end{aligned}$$

(37)

$$\begin{aligned} {\widehat{b}}_{n-1}^{+}&= \frac{5}{2}p_{n-1}-\frac{3}{2}{\widehat{b}}_{n-1}^{-}, \end{aligned}$$

(38)

Hence, the task now is reduced to search only control points ${\widehat{b}}_{i}^{-}$, for $i=1,...,n-1$, that generate the $C^{1}$ Bézier spline $\beta$ in $\mathbb {R}^{m}$. Replacing the new optimization variables in problem (2) gives (24), which is merely the problem of minimization of the mean square acceleration of the Bézier curve $\beta$ in the Euclidean space $\mathbb {R}^{m}$. The optimal solution $Y=[{\widehat{b}}_{1}^{-},...,{\widehat{b}}_{n-1}^{-}]^{T} \in \mathbb {R}^{ (n-1) \times m}$ of that problem is the unique solution of a tridiagonal linear system

$$\begin{aligned} Y = A^{-1}CP = DP \; \text {with} \; \sum _{j=0}^{j=n+1}d_{ij}=1. \end{aligned}$$

(39)

where A is a tridiagonal sparse square matrix of size $(n-1)\times (n-1)$ with a dominant diagonal, C a matrix of size $(n-1)\times (n+1)$ and P the matrix of $p_i$’s of size $(n+1)\times m$ given by:

$$\begin{aligned} A_{(1,1:2)}&= [16 \; 6] \end{aligned}$$

(40)

$$\begin{aligned} A_{(2,1:3)}&=[6 \; 36 \; 9] \end{aligned}$$

(41)

$$\begin{aligned} A_{(i,i-1:i+1)}&= [9 \; 36 \;9], \; \end{aligned}$$

(42)

$$\begin{aligned} A_{(n-1,n-2:n-1)}&=[9 \; 36] \end{aligned}$$

(43)

$$\begin{aligned} C_{(1,1:2)}&= [16 \; 6] \end{aligned}$$

(44)

$$\begin{aligned} C_{(2,2:3)}&=[6 \; 36 \; 9] \end{aligned}$$

(45)

$$\begin{aligned} C_{(i,i:i+1)}&= [9 \; 36 \; 9], \; i=3,...,n-2 \end{aligned}$$

(46)

$$\begin{aligned} C_{(n-1,n-1:n+1)}&=[9 \; 36] \end{aligned}$$

(47)

We may now write the $C^{2}$ differentiability condition. It is obvious that with this $C^{2}$ condition the position of the control points ${\widehat{b}}_{i}^{-}$ and ${\widehat{b}}_{i}^{+}$ that generate the curve $\beta$ will be modified. Therefore, it is more convenient to use another notation. Let us denote by $b^{-}_{i}$ and $b^{+}_{i}$ the new control points on the left and on the right hand side of the interpolation point $p_{i}$, for $i=1,...,n-1$. Computing the acceleration of $\beta$ on respective intervals and taking into account that $\beta$ is $C^{1}$, we shall replace $b_{1}^{+}$ by (36), $b_{i}^{+}$ by (37), and $b_{n-1}^{+}$ by (38). We deduce that:

$$\begin{aligned} b_{2}^{-}&= \frac{1}{3}p_{0}-\frac{1}{2}b_{1}^{-}+\frac{8}{3}p_{1}, \end{aligned}$$

(48)

$$\begin{aligned} b_{i+1}^{-}&= b_{i-1}^{+}+4p_{i}-4b_{i}^{-}, i=2,...,n-2 \end{aligned}$$

(49)

$$\begin{aligned} p_{n}&= 2p_{n-1}+ 2b_{n-1}^{+}-6b_{n-1}^{-}+3b_{n-2}^{+}, \end{aligned}$$

(50)

We see at once that points that will be modified by the additional $C^{2}$ condition are ${\widehat{b}}_{i}^{-}$ and hence ${\widehat{b}}_{i}^{+}$, for $i=2,...,n-1$. The point ${\widehat{b}}_{1}^{-}$ remains invariant and consequently it will be the case for ${\widehat{b}}_{1}^{+}$. According to the $C^{1}$ differentiability condition ensured at the first step, one can take $b_{1}^{-}={\widehat{b}}_{1}^{-}$, with ${\widehat{b}}_{1}^{-}$ is the first row of the matrix Y obtained as a solution of the optimization problem (24). However, the endpoint $p_{n}$ is affected as we can deduce from Eq. (50). Nevertheless, it follows that giving the control point $b_{1}^{-}$ allows us to find all the other control points including $b_{2}^{-}$ with Eq. (48) and hence $b_{2}^{+}$ with (37), then $b_{i+1}^{-}$ for $i=2,...,n-2$ with (49) and therefore $b_{i}^{+}$, for $i=3,...,n-2$ with (37) and $b_{n-1}^{+}$ with (38).

Appendix 2. Proof of Theorem 3

We now prove Theorem 3. The proof is based on the following two results given in Popiel and Noakes (2007).

Lemma 1

Let $\psi _{1} \in \mathcal {M}$.

• (i) $(d\varphi _{\psi _{1}})_{\psi _{2}}^{-1} = (d\varphi _{\psi _{1}})_{\varphi _{\psi _{1}}(\psi _{2})}, \; \text {for all} \; \psi _{2} \in \mathcal {M}$.

• (ii) $(d\varphi _{\psi _{1}})_{ \text {Exp}_{\psi _{1}}(H)} \circ (d \text {Exp}_{\psi _{1}})_{H} = - (d \text {Exp}_{\psi _{1}})_{-H} \; \text {for all} \; H \in T_{\psi _{1}} \mathcal {M}$.

Theorem 5

Let $t \longrightarrow \sigma _{j}(t,V_{0},...,V_{j})$ be the Bézier curve of order j on $\mathcal {M}$ with a number of control points $V_{j}$ for $i=0,...,j$. Then, $\sigma _{j}(t; V_{0},...,V_{j})$ satisfies:

• i) $\frac{D}{\textrm{d}t}|_{t=0} \; {\dot{\sigma }}_{j}(t; V_{0},...,V_{j})= j(j-1)\varOmega _{0}$, where

  $$\begin{aligned} {\varOmega _{0}:=} \left\{ \begin{array}{ll} {\dot{\alpha }}(0,V_{1},V_{2}), &{} if\,V_{0}=V_{1} \\ \ (d\text {Exp}_{V_{0}})^{-1}_{{\dot{\alpha }}(0,V_{0},V_{1})}\left( {\dot{\alpha }} (0,V_{1},V_{2})-{\dot{\alpha }}(1,V_{0},V_{1}) \right) , &{} if\,V_{0}\ne V_{1} \ \end{array}\right. \end{aligned}$$

• ii) $\frac{D}{\textrm{d}t}|_{t=1} \; {\dot{\sigma }}_{j}(t; V_{0},...,V_{j})= j(j-1)\varOmega _{j}$, where

  $$\begin{aligned} {\varOmega _{j}:=} \left\{ \begin{array}{ll} -{\dot{\alpha }}(0,V_{j-2},V_{j-1}), &{} if\,V_{j-1}=V_{j} \ \\ (d\text {Exp}_{V_{j}})^{-1}_{-{\dot{\alpha }}(1,V_{j-1},V_{j})}\left( {\dot{\alpha }}(0,V_{j-1},V_{j}) -{\dot{\alpha }}(1,V_{j-2},V_{j-1}) \right) , &{} if\,V_{j-1}\ne V_{j} \ \end{array}\right. \end{aligned}$$

We will exploit a modified form of the Theorem 5 to obtain the proof of the Theorem 3.

Proof of Theorem 3

Part i) follows from theorem 2. We now prove ii). The Bézier curve $\sigma$ is $C^{2}$ on $\mathcal {M}$ if and only if it satisfies the $C^{2}$ differentiability condition at joint points $\tilde{\psi _{i}}$, for $i=1,...,n-1$. At the point $\tilde{\psi _{1}}$, this means:

$$\begin{aligned} \frac{D}{\textrm{d}t}|_{t=1} \; {\dot{\sigma }}_{2}(t;\tilde{\psi _{0}},\chi _{1}^{-},\tilde{\psi _{1}}) = \frac{D}{\textrm{d}t}|_{t=0} \; {\dot{\sigma }}_{3}(t;\tilde{\psi _{1}},\chi _{1}^{+}, \chi _{2}^{-},\tilde{\psi _{2}}). \end{aligned}$$

(51)

Applying Theorem. 5 yields: $\sigma$ is $C^{2}$ on $\tilde{\psi _{1}}$ if and only if $\varOmega _{2}-3\varOmega _{0}=0$ with:

$$\begin{aligned} \varOmega _{2}-3\varOmega _{0}&= (d \text {Exp}_{\tilde{\psi _{1}}})^{-1}_{-{\dot{\alpha }}(1,\chi _{1}^{-},\tilde{\psi _{1}})} \left( {\dot{\alpha }}(0,\chi _{1}^{-},\tilde{\psi _{1}})-{\dot{\alpha }}(1,\tilde{\psi _{0}},\chi _{1}^{-}) \right) \nonumber \\&- 3 (d \text {Exp}_{\tilde{\psi _{1}}})^{-1}_{{\dot{\alpha }}(0,\tilde{\psi _{1}},\chi _{1}^{+})} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-})-{\dot{\alpha }}(1,\tilde{\psi _{1}},\chi _{1}^{+}) \right) . \end{aligned}$$

(52)

Since $\beta _{1}$ is a $C^{1}$ Bézier curve on $T_{\psi _{1}}\mathcal {M}$, we get that ${\dot{\alpha }}(1,\chi _{1}^{-},\tilde{\psi _{1}}) = {\dot{\alpha }}(0,\tilde{\psi _{1}},\chi _{1}^{+})$. By Lemma 1, we have

$$\begin{aligned}&(d \text {Exp}_{\tilde{\psi _{1}}})^{-1}_{{\dot{\alpha }}(0,\tilde{\psi _{1}},\chi _{1}^{+})} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-})-{\dot{\alpha }}(1,\tilde{\psi _{1}},\chi _{1}^{+}) \right) \\&= - (d \text {Exp}_{\tilde{\psi _{1}}})^{-1}_{-{\dot{\alpha }}(0,\tilde{\psi _{1}},\chi _{1}^{+})} \left( (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-})-{\dot{\alpha }}(1,\tilde{\psi _{1}},\chi _{1}^{+}) \right) \right) . \end{aligned}$$

It follows that

$$\begin{aligned} \begin{array}{l} \varOmega _{2}-3\varOmega _{0}\\ =(d\text {Exp}_{\tilde{\psi _{1}}})^{-1}_{-{\dot{\alpha }}(0,\tilde{\psi _{1}},\chi _{1}^{+})} \left( {\dot{\alpha }}(0,\chi _{1}^{-}, \tilde{\psi _{1}})-{\dot{\alpha }}(1,\tilde{\psi _{0}},\chi _{1}^{-}) \right) \\ +3(d\text {Exp}_{\tilde{\psi _{1}}})^{-1}_{-{\dot{\alpha }}(0,\tilde{\psi _{1}},\chi _{1}^{+})} \left( (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-})-{\dot{\alpha }}(1,\tilde{\psi _{1}},\chi _{1}^{+}) \right) \right) \\ =(d\text {Exp}_{\tilde{\psi _{1}}})^{-1}_{-{\dot{\alpha }}(0,\tilde{\psi _{1}},\chi _{1}^{+})}\Bigg [ 3(\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-}) \right) - 3 (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}} \left( {\dot{\alpha }}(1,\tilde{\psi _{1}},\chi _{1}^{+}) \right) \\ + {\dot{\alpha }}(0,\chi _{1}^{-},\tilde{\psi _{1}})-{\dot{\alpha }}(1,\tilde{\psi _{0}},\chi _{1}^{-}) \Bigg ]. \end{array} \end{aligned}$$

Hence, $\varOmega _{2}-3\varOmega _{0}=0$ if and only if

$$\begin{aligned} \begin{array}{l} = 3(\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-}) \right) - 3 (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}} \left( {\dot{\alpha }}(1,\tilde{\psi _{1}},\chi _{1}^{+}) \right) \\ + {\dot{\alpha }}(0,\chi _{1}^{-},\tilde{\psi _{1}})-{\dot{\alpha }}(1,\tilde{\psi _{0}},\chi _{1}^{-}) = 0 \end{array} \end{aligned}$$

Nevertheless $\varphi _{\tilde{\psi _{1}}} \left( \alpha (t,\tilde{\psi _{1}},\chi _{1}^{+}) \right) = \alpha (1-t,\chi _{1}^{-},\tilde{\psi _{1}}), \; \forall t\in [0,1]$. Differentiate this identity with respect to t, we obtain

$$\begin{aligned} (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}} \left( {\dot{\alpha }}(1,\tilde{\psi _{1}},\chi _{1}^{+}) \right) =-{\dot{\alpha }}(0,\chi _{1}^{-},\tilde{\psi _{1}}). \end{aligned}$$

Accordingly, Eqn. ( ) becomes

$$\begin{aligned} 3(\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-}) \right)= & {} {\dot{\alpha }}(1,\tilde{\psi _{0}},\chi _{1}^{-}) -4 {\dot{\alpha }}(0,\chi _{1}^{-}, \tilde{\psi _{1}}). \end{aligned}$$

(53)

Now, Lemma. 1 shows that

$$\begin{aligned} (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{+}}\left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-}) \right)= & {} (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\varphi _{\tilde{\psi _{1}}}(\chi _{1}^{-})} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-})\right) \\= & {} (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{-}}^{-1} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-}) \right) . \end{aligned}$$

It follows that $(\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{-}}^{-1} \left( {\dot{\alpha }}(0,\chi _{1}^{+},\chi _{2}^{-})\right) = \frac{1}{3} \left( {\dot{\alpha }}(1,\tilde{\psi _{0}},\chi _{1}^{-}) -4 {\dot{\alpha }}(0,\chi _{1}^{-}, \tilde{\psi _{1}}) \right)$.

Consequently, with the exponential map at the point $\chi _{1}^{+}$, we get

$$\begin{aligned} \chi _{2}^{-}= & {} \text {Exp}_{\chi _{1}^{+}}\left( \frac{1}{3} \left( (\textrm{d}\varphi _{\tilde{\psi _{1}}})_{\chi _{1}^{-}} \left( {\dot{\alpha }}(1,\tilde{\psi _{0}},\chi _{1}^{-}) \right) - 4{\dot{\alpha }}(0,\chi _{1}^{-},\tilde{\psi _{1}})\right) \right) . \end{aligned}$$

(54)

The proof of Part iii) follows in much the same way as Part ii). $\square$