Semi-nonparametric approximation and index options

Abstract

In an arbitrage-free securities market, all state-contingent claims and the stochastic discount factors can be approximated appropriately by index options with a semi-nonparametric method. These index options are constructed by efficient algorithms and uniform approximation error under these efficient algorithms are derived. This paper suggests a method to examine state-contingent claims and stochastic discount factors using index options in financial market regardless the market is complete or not.

Appendices

Appendix A: Proofs

Proof of Proposition 2.1

By its definition of $P_nf$, we have, for any real number $\alpha \in \mathbb {R}$,

$$\begin{aligned} ||r_n||^2= & {} ||f - P_{n}f||^2 \\\leqslant & {} ||f - P_{n-1}f - \alpha {<} r_{n-1}, g_n{>} g_{n}||^2\\\leqslant & {} ||r_{n-1}||^2 - 2 \alpha {<} r_{n-1}, g_n{>}^2 + \alpha ^2 {<}r_{n-1}, g_{n}{>}^2 {<}g_n, g_n{>} \\\leqslant & {} ||r_{n-1}||^2 - 2 \alpha {<} r_{n-1}, g_n{>}^2 + \alpha ^2 {<}r_{n-1}, g_{n}{>}^2 b^2. \end{aligned}$$

Then

$$\begin{aligned} ||r_n||^2 \leqslant ||r_{n-1}||^2 - {<} r_{n-1}, g_n{>}^2 (2 \alpha - \alpha ^2 b^2). \end{aligned}$$

(A-1)

Since $r_{n-1}$ is orthogonal to $Span\{g_1, \ldots , g_{n-1}\}$ and $f_{n-1} \in Span\{g_1, \ldots , g_{n-1}\}$, ${<}r_{n-1}, f_{n-1}{>} = 0$, then

$$\begin{aligned} ||r_{n-1}||^2= & {} {<}r_{n-1}, f - f_{n-1}{>} = {<} r_{n-1}, f{>} = {<} r_{n-1}, \sum _{g \in \mathcal{D}} b_g g {>}\\\leqslant & {} \sum |b_g| Sup |{<} r_{n-1}, g{>}| \leqslant ||f||_{\mathcal{L}_1} |{<} r_{n-1}, g_n{>}| \end{aligned}$$

where we choose any representation of $f \equiv \sum _{g \in \mathcal{D}} b_g g$ and the last inequality follows from the definition of $g_n$. Therefore,

$$\begin{aligned} |{<} r_{n-1}, g_n{>}| \geqslant \frac{||r_{n-1}||^2}{||f||_{\mathcal{L}_1}}. \end{aligned}$$

(A-2)

By considering $\alpha \in \mathbb {R}$ with $b(\alpha ):= 2 \alpha - \alpha ^2 b^2 {>} 0$, then we plug the inequality into inequality (A-1) in regard to $||r_{n-1}||^2$, we have

$$\begin{aligned} ||r_n||^2 \leqslant ||r_{n-1}||^2 \left( 1 - \frac{||r_{n-1}||^2}{||f||_{\mathcal{L}_1}^2} b(\alpha ) \right) . \end{aligned}$$

(A-3)

We prove, by induction on n, that

$$\begin{aligned} ||r_n||^2 \leqslant \frac{||f||_{\mathcal{L}_1}^2}{(n+1) b(\alpha )}. \end{aligned}$$

(A-4)

First, note that $b(\alpha )$ chooses the highest value when $\alpha = \frac{1}{b^2}$ for all real numbers $\alpha$, then $b(\alpha ) \leqslant \frac{1}{b^2}$ for all real numbers $\alpha$. Then, for $n = 0$,

$$\begin{aligned} ||r_0|| = ||f|| \leqslant b ||f||_{\mathcal{L}_1} \leqslant \frac{1}{\sqrt{b(\alpha )}}||f||_{\mathcal{L}_1} \end{aligned}$$

(A-5)

where the first inequality follows from

$$\begin{aligned} ||f||= \left| \sum _{g \in \mathcal{D}} b_g g\right| \left| \leqslant \sum _{g \in \mathcal{D}} \right| b_g| ||g|| \leqslant b \sum _{g \in \mathcal{D}} |b_g|. \end{aligned}$$

(A-6)

Therefore, the inequality (A-4) holds for $n=0$. Next, assuming (A-4) holds for $n-1$, and assume that $||r_{n-1}||^2 {>} \frac{||f||_{\mathcal{L}_1}^2}{(n+1) b(\alpha )}$. (Otherwise, $||r_n||^2 \leqslant ||r_{n-1}||^2 \leqslant \frac{||f||_{\mathcal{L}_1}^2}{(n+1) b(\alpha )}$. So (A-4) holds for n.) By invoke of (A-3),

$$\begin{aligned} ||r_n||^2 \leqslant \frac{||f||_{\mathcal{L}_1}^2}{n b(\alpha )} \left( 1 - \frac{||f||_{\mathcal{L}_1}^2}{(n+1) b(\alpha )} \frac{b(\alpha )}{ ||f||_{\mathcal{L}_1}^2 } \right) = \frac{||f||_{\mathcal{L}_1}^2}{(n+1) b(\alpha )}. \end{aligned}$$

Hence, we have proved the inequality (A-4) which ensures that

$$\begin{aligned} ||r_n||^2 \leqslant \frac{||f||_{\mathcal{L}_1}^2}{(n+1)} \min _{\{ \alpha : b(\alpha ) {>} 0\}} \frac{1}{b(\alpha )} = \frac{b^2 ||f||_{\mathcal{L}_1}^2}{(n+1)}. \end{aligned}$$

(A-7)

The proof is completed. $\square$

Proof of Proposition 2.2

Given any $f \in \mathcal{X}, h \in \mathcal{L}_1$ and any $n \geqslant 1$, using ${<}r_{n-1}, f_{n-1}{>} = 0$,

$$\begin{aligned} ||r_{n-1}||^2= & {} {<} r_{n-1}, h+f - h{>} = {<} r_{n-1}, h{>} + {<}r_{n-1}, f-h{>} \\\leqslant & {} ||h||_{\mathcal{L}_1} |{<}r_{n-1}, g_n{>}| + ||r_{n-1}||\cdot || f-h|| \\\leqslant & {} ||h||_{\mathcal{L}_1} |{<}r_{n-1}, g_n{>}| + \frac{||r_{n-1}||^2 + ||f-h||^2}{2} \end{aligned}$$

where the first inequality follows from the definition of $g_n$ while the second and the third inequality follows from the Cauchy-Schwartz inequality. Then,

$$\begin{aligned} \frac{||r_{n-1}||^2 - ||f-h||^2}{2} \leqslant ||h||_{\mathcal{L}_1} |{<}r_{n-1}, g_n{>}| \end{aligned}$$

(A-8)

and

$$\begin{aligned} |{<}r_{n-1}, g_n{>}| \geqslant \frac{||r_{n-1}||^2 - ||f-h||^2}{2||h||_{\mathcal{L}_1}}. \end{aligned}$$

(A-9)

Let $a_{k} \equiv ||r_{k}||^2 - ||f-h||^2$ for each $k \geqslant 0$. If $||r_{n_0}|| \leqslant ||f-h||$ for a positive number $n_0$, then $||r_n|| \leqslant ||r_{n_0}|| \leqslant ||f-h||,$ for all $n \geqslant n_0$, and thus Proposition 2.2 holds naturally. Therefore, we assume that $a_n {>} 0$ for all n. Hence, (A-9) implies that

$$\begin{aligned} {<}r_{n-1}, g_n{>}^2 \geqslant \frac{a_{n-1}^2}{4 ||h||_{\mathcal{L}_1}^2}, \end{aligned}$$

(A-10)

which again yields (by using (A-1))

$$\begin{aligned} ||r_n||^2 \leqslant ||r_{n-1}||^2 - \frac{a_{n-1}^2}{4 ||h||_{\mathcal{L}_1}^2} b(\alpha ). \end{aligned}$$

(A-11)

Hence,

$$\begin{aligned} a_n \leqslant a_{n-1} \left( 1 - \frac{a_{n-1}}{4 ||h||_{\mathcal{L}_1}^2} b(\alpha ) \right) . \end{aligned}$$

(A-12)

Next, we prove that, by an induction argument, for any $n \geqslant 1$,

$$\begin{aligned} a_n \leqslant \frac{4 ||h||_{\mathcal{L}_1}^2}{n} \frac{1}{b(\alpha )} \end{aligned}$$

(A-13)

and then,

$$\begin{aligned} a_n \leqslant \frac{4 ||h||_{\mathcal{L}_1}^2}{n} \min _{\{\alpha : b(\alpha ) {>} 0\} }\left\{ \frac{1}{b(\alpha )} \right\} = \frac{4b^2 ||h||_{\mathcal{L}_1}^2}{n}, \end{aligned}$$

(A-14)

the proof is thus completed.

To the end, if $a_0 \leqslant \frac{4 ||h||_{\mathcal{L}_1}^2}{b(\alpha )}$, then $a_1 \leqslant a_0 \leqslant \frac{4 ||h||_{\mathcal{L}_1}^2}{b(\alpha )}$. If $a_0 {>} \frac{4 ||h||_{\mathcal{L}_1}^2}{b(\alpha )}$, then by using (A-12),

$$\begin{aligned} a_1 \leqslant a_0 \left( 1 - \frac{a_{0}}{4 ||h||_{\mathcal{L}_1}^2} b(\alpha ) \right) \leqslant 0 \end{aligned}$$

thus $a_1 \leqslant \frac{4 ||h||_{\mathcal{L}_1}^2}{b(\alpha )}$. Assuming the inequality (A-13) holds for $n-1$ and $a_{n-1} {>} \frac{4 ||h||_{\mathcal{L}_1}^2}{n b(\alpha )}$, then by (A-12), we obtain

$$\begin{aligned} a_{n} \leqslant \frac{4 ||h||_{\mathcal{L}_1}^2}{n-1} \frac{1}{b(\alpha )} \left( 1 - \frac{1}{n} \right) \leqslant \frac{4 ||h||_{\mathcal{L}_1}^2}{n} \frac{1}{b(\alpha )}. \end{aligned}$$

(A-15)

$\square$

Proof of Proposition 2.3

As $\mathcal{D}= - D$, any $f \in \mathcal{L}_1$ can be written as $f = \sum _{g \in \mathcal{D}} b_g g$ with $b_g \geqslant 0$ for all $g \in \mathcal{D}$. For any $\epsilon {>} 0$, by its definition of $\mathcal{L}_1$, there exists a finite subset $\hat{\mathcal{D}} \subseteq \mathcal{D}$ such that

• $f = \hat{f} + h, \hat{f} = \sum _{g \in \hat{\mathcal{D}}}b_g g$;

• $||h||_{\mathcal{L}_1} \leqslant \epsilon$;

• $||\hat{f}||_{\mathcal{L}_1} = \sum _{g \in \hat{\mathcal{D}}}b_g = ||f||_{\mathcal{L}_1} + \delta , |\delta | \leqslant \epsilon$.

By the definition of $f_n$, for any $\beta \in \mathbb {R}$ and $g \in \mathcal{D}$, we have

$$\begin{aligned} ||r_n||^2\leqslant & {} ||f - \alpha _{n} f_{n-1} - \beta g||^2 \\= & {} ||f - \alpha _n (f - r_{n-1}) - \beta g||^2 = ||\alpha _n r_{n-1} + (1 - \alpha _n) f - \beta g||^2 \\= & {} \alpha _n^2 ||r_{n-1}||^2 + 2 \alpha _n {<} r_{n-1}, (1- \alpha _n)f - \beta g{>} + ||(1 - \alpha _n)f - \beta g||^2 \\= & {} \alpha _n^2 ||r_{n-1}||^2 + 2 \alpha _n (1-\alpha _n) {<} r_{n-1}, f{>} - 2 \alpha _n \beta {<}r_{n-1}, g{>} \\&+\, (1- \alpha _n)^2 ||f||^2 - 2(1- \alpha _n) \beta {<}f, g{>} + \beta ^2 ||g||^2. \end{aligned}$$

The inequality holds for all $g \in \mathcal{D}$, so it holds on the average with weights $\frac{b_g}{\sum _{g \in \hat{\mathcal{D}}} b_g}$, using $||g|| \leqslant b$ for all $g \in \mathcal{D}$, we have

$$\begin{aligned} ||r_n||^2\leqslant & {} \alpha _{n}^2 ||r_{n-1}||^2 + 2 \alpha _n (1 - \alpha _n) {<}r_{n-1}, f{>} - 2 \alpha _n \beta {<} r_{n-1}, \frac{ \sum _{g \in \hat{\mathcal{D}}} b_g g}{\sum _{g \in \hat{\mathcal{D}}} b_g }{>} \\&+\, (1- \alpha _n)^2 ||f||^2 - 2(1-\alpha _n) \beta {<} f, \frac{ \sum _{g \in \hat{\mathcal{D}}} b_g g}{\sum _{g \in \hat{\mathcal{D}}} b_g } {>} + \beta ^2 b^2\\\leqslant & {} \alpha _{n}^2 ||r_{n-1}||^2 + 2 \alpha _n (1 - \alpha _n) {<}r_{n-1}, f{>} - 2 \alpha _n \frac{\beta }{\sum _{g \in \hat{\mathcal{D}}} b_g} {<}r_{n-1}, f - h{>} \\&+\, (1- \alpha _n)^2 ||f||^2 - 2 (1-\alpha _n) \frac{\beta }{\sum _{g \in \hat{\mathcal{D}}} b_g}{<}f, f - h{>} + \beta ^2 b^2. \end{aligned}$$

Choosing $\alpha _n = 1 -\frac{1}{n}, \beta = \frac{1}{n} \sum _{g \in \hat{\mathcal{D}}}b_g$, we obtain

$$\begin{aligned} ||r_n||^2\leqslant & {} \alpha _{n}^2 ||r_{n-1}||^2 + 2 \alpha _n \frac{1}{n}{<} r_{n-1}, h{>} \\&-\, (1-\alpha _n)^2 ||f||^2 + \beta ^2 b^2 + 2 (1-\alpha _n)^2 {<} f, h{>}. \end{aligned}$$

Since

$$\begin{aligned} {<}r_{n-1}, h{>}= & {} {<} r_{n-1}, \sum _{g \in \mathcal{D}- \hat{\mathcal{D}}} b_g g{>} = \sum _{g \in \mathcal{D}- \hat{\mathcal{D}}} b_g {<} r_{n-1}, g{>} \\= & {} \sum _{g \in \mathcal{D}- \hat{\mathcal{D}}} \frac{b_g}{2} \left( ||r_{n-1}||^2 + b^2 \right) \leqslant \frac{\epsilon }{2}\left( ||r_{n-1}||^2 + b^2\right) \end{aligned}$$

and similarly,

$$\begin{aligned} {<}f, h{>} \leqslant \frac{\epsilon }{2}\left( ||f||^2 + b^2\right) , \end{aligned}$$

we obtain

$$\begin{aligned} ||r_n||^2\leqslant & {} \alpha _{n}^2 ||r_{n-1}||^2 -\frac{1}{n^2}||f||^2 + \frac{(||f||_{\mathcal{L}_1}^2 + \epsilon )}{n^2} b^2 + O(\epsilon ) \end{aligned}$$

(A-16)

where $O(\epsilon )$ represents some quantity which converges to zero when $\epsilon \rightarrow 0$. Letting $\epsilon \downarrow 0$,

$$\begin{aligned} ||r_n||^2 \leqslant \left( 1 - \frac{1}{n} \right) ^2 ||r_{n-1}||^2 + \frac{||f||_{\mathcal{L}_1}^2 b^2 - ||f||^2}{n^2}. \end{aligned}$$

(A-17)

By an easy induction argument with respect to n, we have

$$\begin{aligned} ||r_n||^2 \leqslant \frac{||f||_{\mathcal{L}_1}^2 b^2 - ||f||^2}{n}. \end{aligned}$$

(A-18)

$\square$

Proof of Proposition 2.4

As $h \in \mathcal{L}_1$, for any $\epsilon {>} 0$, there exists a decomposition $h = \hat{h} + h^*$ where

• $\hat{h}= \sum _{g \in \hat{\mathcal{D}}} b_g g, b_g {>} 0$ and $\hat{\mathcal{D}}$ is a finite subset of $\mathcal{D}$;

• $\sum _{g \in \hat{\mathcal{D}}} b_g = ||h||_{\mathcal{L}_1} + \delta , |\delta | \leqslant \epsilon$;

• $||h^*||_{\mathcal{L}_1} \leqslant \epsilon$.

Given any $\alpha \in (0,1)$ and $\beta \in \mathbb {R}$, by using the same argument in Proposition 2.3, we have

$$\begin{aligned} ||r_n||^2\leqslant & {} \alpha ^2 ||r_{n-1}||^2 + 2 \alpha (1 - \alpha ) {<}r_{n-1}, f{>} - 2 \alpha \frac{\beta }{\sum _{g \in \hat{\mathcal{D}}} b_g} {<}r_{n-1}, f - h{>} \\&+ (1- \alpha )^2 ||f||^2 - 2 (1-\alpha ) \frac{\beta }{\sum _{g \in \hat{\mathcal{D}}} b_g}{<}f, f - h{>} + \beta ^2 b^2. \end{aligned}$$

By using the above equality for each $g \in \hat{D}$ on the average with weights $\frac{b_g}{\sum _{g \in \hat{\mathcal{D}}} b_g}$, choosing $\beta := (1-\alpha )\sum _{g \in \hat{\mathcal{D}}}b_g$ and using the same argument as in Proposition 2.3 while letting $\epsilon \downarrow 0$, we have

$$\begin{aligned} ||r_n||^2\leqslant & {} \alpha ^2 ||r_{n-1}||^2 + 2 \alpha (1 - \alpha ) {<}r_{n-1}, f-h{>} + (1- \alpha )^2||f-h||^2 \\&-\, (1 - \alpha )^2 ||h||^2 + (1- \alpha )^2 ||h||_{\mathcal{L}_1}^2 b^2 \\\leqslant & {} \alpha ^2 ||r_{n-1}||^2 + \alpha (1 - \alpha ) \left( ||r_{n-1}||^2 + ||f-h||^2 \right) \\&+\, (1- \alpha )^2||f-h||^2 + (1-\alpha ^2)M\\\leqslant & {} \alpha ||r_{n-1}||^2 + (1 - \alpha )||f-h||^2 + (1-\alpha )^2 M \end{aligned}$$

where $M:= b^2 ||h||_{\mathcal{L}_1}^2 - ||h||^2$. Hence, for any $\alpha \in (0,1)$,

$$\begin{aligned} ||r_n||^2 - ||f-h||^2 \leqslant \alpha (||r_{n-1}||^2 - ||f-h||^2) + (1- \alpha )^2M. \end{aligned}$$

(A-19)

Choosing $\alpha = 0$ for $n=1$, then (A-19) implies that $||r_1||^2 - ||f-h||^2 \leqslant M$. Assuming for $n \geqslant 2$,

$$\begin{aligned} ||r_{n-1}||^2 - ||f-h||^2 \leqslant \frac{4M}{n-1} \end{aligned}$$

then by using $\alpha = 1 - \frac{2}{n}$ and (A-19), we derive

$$\begin{aligned} ||r_n||^2 - ||f-h||^2 \leqslant \left( 1 - \frac{2}{n} \right) ^2 \frac{4M}{n-1} + \frac{4}{n^2} M \leqslant \frac{4M}{n}. \end{aligned}$$

(A-20)

Therefore, the proof is completed by an induction argument. $\square$.

Proof of Proposition 3.1

See Tian (2014), Theorem 2.1, for $p=2$. This approximation theorem is inspired by the uniform approximation theorem in artificial neural network. See Cybenko (1989) and Hornik (1991). $\square$

Proof of Proposition 3.2

Since $\mathcal{F}= \mathcal{F}_x$, by Proposition 3.1, any state-contingent claim in $L^2(\Omega , \mathcal{F},P)$ can be approximated well by an element of $Span\mathcal{H}$. If the index options $\sigma (m \cdot x + \theta )$ are marketable assets with a corresponding (market) price functional $\pi : \mathcal{H}\rightarrow \mathbb {R}$, then the securities market $\left( \Omega , \mathcal{F}, \mathcal{H}, \pi \right)$ is a complete market in the sense of Jarrow et al. (1999). Equivalently, the market is complete in the sense of Harrison and Kreps (1979), that is, there exists a unique stochastic discount factor $\phi \in L^2(\Omega , \mathcal{F})$ such that the price of any contingent claim, x, at time zero, is $\mathbb {E}[\phi x]$. Therefore, there exists a unique dual $\phi$, in $L^2(\Omega , \mathcal{F}, P)$ such that $\mathbb {E}[\phi g] = \pi (g)$ for any $g \in \mathcal{H}$. Moreover, if the securities market is arbitrage free in the sense of Harrison and Kreps (1979), the price of any state-contingent claim $f \in L^2(\Omega , \mathcal{F}, P)$ is given by $\mathbb {E}[\phi f]$.

We first prove this proposition for $f \in \mathcal{L}_1$. By Proposition 2.3, there exists a portfolio of index options, $g_n$, with at most n elements in $\mathcal{D}$ such that

$$\begin{aligned} ||f - g_n||^2 \leqslant \frac{C}{n} \end{aligned}$$

(A-21)

where $C = ||f||_{\mathcal{L}_1}^2 b^2 - ||f||^2$ is a nonnegative number. Thus,

$$\begin{aligned} \left| \mathbb {E}[\phi f] - \pi (g_n)\right|= & {} \left| \mathbb {E}\left[ \phi (f - g_n)\right] \right| \leqslant \mathbb {E}[\phi ^2]^{1/2} ||f-g_n|| \leqslant \frac{\mathbb {E}[ C\phi ^2]^{1/2}}{\sqrt{n}}.\nonumber \\ \end{aligned}$$

(A-22)

Next, with a general contingent claim $f \in L^2(\Omega , \mathcal{F}, P)$, Proposition 3.1 ensures that f can be approximated arbitrary well by elements of $Span \mathcal{H}$, thus, elements of $\mathcal{L}_1$. Then there exists an element $h \in \mathcal{L}_1$ such that $||f - h|| \leqslant \frac{\epsilon }{\mathbb {E}\left[ \phi ^2\right] ^{1/2}}$. Then, we use the first part of this proposition for h to obtain

$$\begin{aligned} \left| \mathbb {E}\left[ \phi f\right] - \pi (g_n)\right|= & {} \left| \mathbb {E}\left[ \phi (f - g_n)\right] \right| \leqslant \mathbb {E}[\phi ^2]^{1/2} ||f-g_n||\\\leqslant & {} \mathbb {E}[\phi ^2]^{1/2} \left( ||f-h|| + ||g_n - h|| \right) \\\leqslant & {} \mathbb {E}[\phi ^2]^{1/2} \left( ||f-h|| + \frac{C}{\sqrt{n}} \right) \\\leqslant & {} \mathbb {E}[\phi ^2]^{1/2} ||f-h|| + \mathbb {E}[\phi ^2]^{1/2} \frac{C}{\sqrt{n}} \\\leqslant & {} \epsilon + O\left( \frac{1}{\sqrt{n}}\right) \end{aligned}$$

where C is a nonnegative number in (A-21) in which the element $f \in \mathcal{L}_1$ is replaced by $h \in \mathcal{L}_1$. Finally, given the greedy algorithms in Sect. 2, the construction of the portfolio of index options with at most n index options, $g_n$, is effective. $\square$

Proof of Proposition 3.3

This result belongs to Hornik (1990), Corollary 3.6. By its definition, any function in $S^{1}_{2}\left( \mathbb {R}^{k}, \mu \right)$ has continuous first-order derivative. But many contingent claims such as call-type derivatives have no first-order derivative everywhere. Therefore, the function space $S^{1}_{2}\left( \mathbb {R}^{k}, \mu \right)$ might be too strong to be applied to the contingent claim analysis. For this purpose, we consider the Sobolev space $W^{1}_{2}(\mathbb {R}^{k})$ in which each function has distribution derivatives $\partial ^{\alpha }f, \forall |\alpha | \leqslant 1$. The Sobolev space $W^{1}_{2}(\mathbb {R}^{k})$ is large enough to include most contingent claims. Moreover, it is a Hilbert space associated with a norm $|| \cdot ||_{1,2,S}$. By applying Hornik (1990), Theorem 2.1 and Corollary 3.8, $Span\mathcal{H}(\sigma )$ generates the Sobolev space $W^{1}_{2}(\mathbb {R}^{k})$ under the norm $|| \cdot ||_{1,2,S}$. $\square$

Proof of Proposition 4.1

We first prove that for any $x, y \in L^2$, the following inequality holds:

$$\begin{aligned} ||x^{+} - y^{+}||^2 \leqslant 2 ||x-y||^2. \end{aligned}$$

(A-23)

To show it, let $A = \left\{ x^{+} \geqslant y^{+} \right\}$ denotes the event on which $x^{+}$ is larger than $y^{+}$. Since $x^{+} - y^{+} \leqslant (x-y)^{+}$, we obtain

$$\begin{aligned} ||x^{+} - y^{+}||^2= & {} \int _{\Omega } \left( x^{+} - y^{+}\right) ^2 dP\\= & {} \int _{A} \left( x^{+} - y^{+}\right) ^2 dP + \int _{A^{c}} \left( x^{+} - y^{+}\right) ^2 dP\\= & {} \int _{A} \left( x^{+} - y^{+}\right) ^2 dP + \int _{A^{c}} \left( y^{+} - x^{+}\right) ^2 dP\\\leqslant & {} \int _{A} \left( (x-y)^{+}\right) ^2 dP + \int _{A^{c}} \left( (y - x)^{+}\right) ^2 dP\\\leqslant & {} \int _{\Omega } \left( (x-y)^{+}\right) ^2 dP + \int _{\Omega } \left( (y - x)^{+}\right) ^2 dP\\\leqslant & {} \int _{\Omega } \left( x-y\right) ^2 dP + \int _{\Omega } \left( y- x\right) ^2 dP\\\leqslant & {} 2||x-y||^2. \end{aligned}$$

(1) For any $\phi \in (L^{2})^{+}$, by Proposition 3.2, there exists $h_n \in Span\mathcal{H}$ such that $||\phi - h_n|| \leqslant \frac{1}{n}$. Then, since $\phi = \phi ^{+}$, we have

$$\begin{aligned} ||\phi - h_n^{+}||^2 = ||\phi ^{+} - h_{n}^{+}||^2 \leqslant 2 ||\phi - h_{n}||^2 \leqslant \frac{2}{n^2}. \end{aligned}$$

(A-24)

Therefore, we have proved that $(L^{2})^{+} = \overline{(Span \mathcal{H})^{+}}$.

(2). By the effective construction of the index option approximation follows from the effective construction of $\phi \in L^2$ in terms of the index options, by Proposition 2.2 and Proposition 2.4.

(3). It follows from the first part and the Lebesgue dominated convergence theorem. $\square$

Proof of Proposition 4.2

The proof depends on a frame theory in a finite dimensional space, and we refer to Christensen (2003), Chapter 1 for details. Let $V = L^{2}(\Omega , P)$, V is a finite dimensional space with dimension n. Since V is spanned by $f_l, l = l_1, \ldots , l_m$, then the family $\left\{ f_l: l = l_1, \ldots , l_m \right\} \subseteq V$ is a frame (Christensen (2003), Corollary 1.1.3). We define the frame operator

$$\begin{aligned} S: V \rightarrow V, S(f) = \sum \mathbb {E}[f f_l] f_l. \end{aligned}$$

(A-25)

Equivalently, $S (f) = A \cdot f$. By Christensen (2003), Theorem 1.1.5, S is invertible and self-adjoint, and for any $f \in V$,

$$\begin{aligned} f = \sum _{l} \mathbb {E}[f f_l] S^{-1}(f_l) \end{aligned}$$

(A-26)

In particular, for the unique stochastic discount factor $f = \phi$, we obtain

$$\begin{aligned} \phi = \sum _{l} \mathbb {E}[\phi f_l] S^{-1}(f_l) = \sum _{l} a_l S^{-1}(f_l) \end{aligned}$$

(A-27)

since $a_l = \mathbb {E}[\phi f_l]$ by the equation of the stochastic discount factor. Notice that $A^{-1} \cdot f = S^{-1}(f)$. The proof of this proposition is completed. $\square$

Proof of Proposition 4.3

By Proposition 3.1, for any $\epsilon {>} 0$, there exists an index option portfolio, $\phi _0$, in $Span \mathcal{H}$ such that

$$\begin{aligned} || \phi - \phi _0 || \leqslant 2 \epsilon . \end{aligned}$$

(A-28)

$\phi _0$ is merely an element of $Span \mathcal{H}$ satisfying (A-28), which might not be entirely positive let alone a stochastic discount factor; but it can still be used to price a general state-contingent claim via index options up to a uniform error. To see this, choose an index option portfolio $g_n$ in $Span_{n} \mathcal{H}$ such that (by Proposition 3.2)

$$\begin{aligned} ||f - g_n|| \leqslant \epsilon + \frac{C}{\sqrt{n}}. \end{aligned}$$

Henceforth, we have,

$$\begin{aligned} \left| \mathbb {E}\left[ \phi f \right] - \mathbb {E}\left[ \phi _0 g_n \right] \right|\leqslant & {} \left| \mathbb {E}\left[ (\phi - \phi _0 ) f \right] \right| + \left| \mathbb {E}\left[ \phi _0 (f - g_n) \right] \right| \\\leqslant & {} ||\phi - \phi _0 || \cdot ||f|| + ||\phi _0|| \cdot ||f - g_n|| \\\leqslant & {} 2 \epsilon \cdot ||f|| + (||\phi || + 2\epsilon ) \cdot ||f - g_n|| \\\leqslant & {} 2 \epsilon \cdot ||f|| + (||\phi || + 2\epsilon ) \cdot \left( \epsilon + \frac{C}{\sqrt{n}} \right) \\\leqslant & {} 2 \epsilon \left( ||f|| + ||\phi || \right) + \frac{C'}{\sqrt{n}}, \end{aligned}$$

where we choose $\epsilon \leqslant ||\phi ||$ and both C and $C'$ are two positive numbers independent of n. $\square$

Proof of Proposition 4.4

It is the same as Proposition 3.2 since $\mathbb {E}[\phi ^2]$ is bounded by a constant A, for all $\phi \in \Phi$. $\square$

Proof of Proposition 5.1

By Tian (2014), Theorem 2.1, there exists a $f_1 \in Span \mathcal{H}\subseteq \mathcal{L}_1$ such that $||f - f_1|| \leqslant \frac{\epsilon }{||\phi ||}$. Since $f_1 \in \mathcal{L}_1$, there exists a positive number L such that $f_1 = L f_2$ with $f_2 \in \overline{co \mathcal{H}}$, the closure of the convex hull of $\mathcal{H}$. Since $\sigma (t)$ is bounded by a positive number b, we see easily that $||f_2 - g|| \leqslant K \equiv 2 b$ for any $g \in \mathcal{H}$. By using Donahue et al. (1997), Corollary 3.6, there exists a sequence $h_n \in Span_{n} \mathcal{H}$ such that

$$\begin{aligned} ||f_2 - h_n|| \leqslant \frac{2^{1/p}(K + \epsilon )}{n^{1- 1/p}} \left[ 1 + \frac{(p-1) log_{n} n}{n} \right] ^{1/p} \end{aligned}$$

for $1 {<} p \leqslant 2$ and

$$\begin{aligned} ||f_2 - h_n|| \leqslant \frac{(2p-2)^{1/2}(K+\epsilon )}{\sqrt{n}} \left[ 1 + \frac{log_{2} n}{n} \right] ^{1/2} \end{aligned}$$

for $2 {<} p {<} \infty$. Then $g_n = L h_n \in Span_{n} \mathcal{H}$ is what is desired. The proof for the incomplete market in a good-deal setting is similar and omitted. $\square$

Proof of Proposition 5.2

Given any $\phi \in (L^{p})^{+}$, by virtue of Proposition 5.1, there exists a sequence $r_n \in Span\mathcal{H}$ such that $||\phi - r_n|| \rightarrow 0$. Therefore, the set $\left\{ ||r_{n}||: n = 1, \ldots \right\}$ as well as the set $\left\{ ||r_{n}^{+} ||: n = 1, 2, \right\}$ is bounded. Then, there exists a convergent subsequence of $r_n$, we write it as $h_n$, and $h_n \rightarrow \phi$ almost surely. Moreover, $h_n^{+} \rightarrow \phi$ almost surely since $\phi = \phi ^{+}$. We next apply the Banach-Saks theorem for the Banach space $L^p$, there exists a subsequence of $h_n^{+}$, written as $h_{n_k}^{+}$ such that

$$\begin{aligned} \frac{h_{n_1}^{+} + \cdots + h_{n_k}^{+}}{k} \end{aligned}$$

converges to some $z \in L^{p}$ under the norm topology when $k \rightarrow \infty$. Write $s_k = h_{n_k}^{+} \in (Span \mathcal{H})^{+}$. Since $h_{n}^{+}$ converges to $\phi$ almost surely,

$$\begin{aligned} \frac{h_{n_1}^{+} + \cdots + h_{n_k}^{+}}{k} \rightarrow \phi \end{aligned}$$

almost surely, then $\phi = z$ with probability one. Therefore,

$$\begin{aligned} \frac{s_1 + \cdots + s_k}{k} \end{aligned}$$

converges to $\phi$ under the norm topology. We have thus proved that $(L^{p})^{+} = \overline{co((Span \mathcal{H})^{+}) }$. $\square$

Appendix B: Numerical examples

We present the details of the construction of the universal approximation in Examples.

Example 3.2.

We implement the orthogonal greedy algorithm for the contingent claim $f(x,y) = e^{x}y$, where (x, y) has a standard bivariate Gaussian distribution, and the activation function is $\sigma (t) = 1, t \geqslant 0; \sigma (t) = 0, t {<} 0$.

The index is written as $w_1 x + w_2 y + \theta$. Because $\sigma (at) = \sigma (t), \forall a {>} 0$, we can assume that $|w_i| \leqslant 1, |\theta | \leqslant 1$ (See Tian 2014, Theorem 2.7). Moreover, we can assume that $w_2 = 0$, or $w_2 = 1$. When $w_2 = 0$, the index is written as $x - \theta$; when $w_2 = 1$, the index is written as $y - \beta x - \theta$ for $|\beta | \leqslant 1, |\theta | \leqslant 1$.

Step 1.

Let $f_0 = 0$. We choose $g \in \mathcal{H}$ to maximize $\mathbb {E}\left[ fg\right]$. When $g = \sigma (x - \theta )$, $\mathbb {E}[fg] = 0$. When $g = \sigma (y - \beta x - \theta )$, it is straightforward to derive that

$$\begin{aligned} \mathbb {E}\left[ e^{x}y \sigma (y - \beta x - \theta )\right] = \frac{1}{\sqrt{1+\beta ^2}}exp\left( -\frac{1}{2} \frac{\theta ^2 + 2 \beta \theta - 1}{1+\beta ^2} \right) . \end{aligned}$$

(B-1)

Then, it reduces to calculate

$$\begin{aligned} \max _{|\beta | \leqslant 1, |\theta | \leqslant 1} \frac{1}{\sqrt{1+\beta ^2}}exp\left( -\frac{1}{2} \frac{\theta ^2 + 2 \beta \theta - 1}{1+\beta ^2} \right) , \end{aligned}$$

and the optimum values are $\beta = 0, \theta = 0$. In other words, $g_{1}(x,y) = \sigma (y)$. Therefore, the projection $f_1 = Pr{<}g_1{>}$ is

$$\begin{aligned} f_1 = \frac{\mathbb {E}[f g_1]}{\mathbb {E}[g_1g_1]}g_1 = \sqrt{\frac{2 e}{\pi }}\sigma (y). \end{aligned}$$

(B-2)

Step 2.

Let $r_1 = f - f_1$. Since $\mathbb {E}[f \sigma (x - \theta )] = 0, \mathbb {E}[r_1 \sigma (x -\theta )] {<} 0$, we reduce to compute

$$\begin{aligned} Argmax_{|\beta | \leqslant 1, |\theta | \leqslant 1} \mathbb {E}\left[ r_1 \sigma (y - \beta x - \theta )\right] \end{aligned}$$

(B-3)

By straightforward calculation,

$$\begin{aligned} \mathbb {E}\left[ \sigma (y) \sigma (y - \beta x - \theta )\right] = \left\{ \begin{array}{l} 1 - N(max(\theta ,0)), \ \ \ \ \ \text { if } \beta = 0; \\ 1 - \frac{1}{2}N(-\theta /\beta ) - \int _{x \geqslant -\theta /\beta } n(x) N(\beta x + \theta )dx, \text { if } \beta {>} 0; \\ 1 - \frac{1}{2}N(\theta /\beta ) - \int _{x \leqslant -\theta /\beta } n(x) N(\beta x + \theta ) dx, \text { if } \beta {<} 0. \end{array} \right. \end{aligned}$$

By computation, the optimum values are $\beta ^* = 0.03, \theta ^* = -0.05$. Therefore, the subspace to approximate f is spanned by $g_1 = \sigma (y), g_2(y) = \sigma (y - 0.03x + 0.05)$. Therefore, $f_2 = Pr(f)$, the projection of f to this subspace is

$$\begin{aligned} f_2 = -24.6826 \cdot \sigma (y) + 26.0059 \cdot \sigma (y - 0.03x + 0.05). \end{aligned}$$

Let $r_2 = f - f_2$. Similarly, we find

$$\begin{aligned} Argmax_{|\beta | \leqslant 1, |\theta | \leqslant 1} \mathbb {E}\left[ r_2 \sigma (y - \beta x - \theta )\right] , \end{aligned}$$

and continue to find $f_3(x)$ and so on.

Example 3.5.

We implement the orthogonal greedy algorithm for the contingent claim $f(x) = e^{2x} 1_{\{x \leqslant 0\} } + 1_{\{ x {>} 0 \}}$, where x has a standard normal distribution, and the activation function is $\sigma (t) = 1, t \geqslant 0; \sigma (t) = 0, t {<} 0$. The index is written as $\beta x + \theta$. Similar to Example 3.2, it suffices to consider all index options with the payoff $\sigma (x - \theta ), |\theta | \leqslant 1$.

Step 1.

Let $f_0 = 0$, $r_0 = f -f_0$. We choose $g = \sigma (x - \theta ) \in \mathcal{H}$ to maximize $\mathbb {E}\left[ r_0 g\right]$. By straightforward calculation,

$$\begin{aligned} {<}r_0, g{>} \equiv \mathbb {E}\left[ r_0(x) \sigma (x - \theta )\right] = \left\{ \begin{array}{l} 1 - N(\theta ), \ \ \ \ \ \text { if } \theta \geqslant 0; \\ \frac{1}{2} + e^{2}\left( N(-2) - N(\theta - 2)\right) , \text { if } \theta {<} 0. \end{array} \right. \end{aligned}$$

We find that $\theta _1^* = Argmax_{|\theta | \leqslant 1} \mathbb {E}\left[ f(x) \sigma (x - \theta )\right] = -1$. Then $g_1(x) = \sigma (x - \theta _1^*) = \sigma (x + 1)$. By calculation, the first-order approximation is

$$\begin{aligned} f_1 = \frac{{<}f,g_1{>}}{{<}g_1, g_1{>}}g_1 = 0.7822 g_1(x). \end{aligned}$$

(B-4)

Step 2.

Let $r_1(x) = f(x) - f_1(x)$. By similar calculation, we obtain

$$\begin{aligned} Argmax_{|\theta | \leqslant 1} {<} r_1(x), \sigma (x - \theta ){>} = -0.12. \end{aligned}$$

Therefore, $f_2 \in Span\left\{ g_1(x), g_2(x) \right\}$ where $g_2(x) \equiv \sigma (x + 0.12)$. Precisely, $f_2(x) = c_1 g_1(x) + c_2 g_2(x)$, and the coefficients $c_1, c_2$ are determined by the following linear system:

$$\begin{aligned} {<}f(x) - f_2(x), g_1(x){>} = 0, {<} f(x) - f_2(x), g_2(x){>} = 0. \end{aligned}$$

We find $c_1 = 0.3940$ and $c_2 = 0.5964$. Then, the second-order approximation is $f_2(x) = 0.3940 \sigma (x+1) + 0.5964 \sigma (x+0.12)$.

Step 3.

Let $r_2(x) = f(x) - f_2(x)$. We obtain

$$\begin{aligned} Argmax_{|\theta | \leqslant 1} {<} r_2(x), \sigma (x - \theta ){>} = -0.14. \end{aligned}$$

Then, $f_3(x)$ is the projection of the function f(x) to the space $Span \left\{ g_1(x), g_2(x), g_3(x) \right\}$ where $g_3(x) = \sigma (x+0.14)$. A straightforward calculation derives that the third-order approximation is $f_3(x) = 0.2488\sigma (x+1) + 0.4213\sigma (x+0.12) + 0.3203 \sigma (x + 0.14)$. $\square$