2. Preliminaries

Let $(\Omega,\mathscr{F},\mathbb{P})$ be a probability space, on which all random elements will be defined. We shall write, for simplicity, $L^2$ to denote $L^2(\Omega,\mathscr{F},\mathbb{P})$ , and ${\lVert{\cdot}\rVert}_2$ for its norm. For any $m \in \mathbb{R}$ and $\sigma \in \mathbb{R}_+$ , the sphere of $L^2$ of radius $\sigma$ centered in m will be denoted by $\mathscr{X}_{m,\sigma}$ , and just by $\mathscr{X}$ if $m=0$ and $\sigma=1$ . In other words, $\mathscr{X}_{m,\sigma}$ stands for the set of random variables X such that ${\mathbb{E}} X=m$ and $\operatorname{Var}(X) = {\mathbb{E}}(X-m)^2 = \sigma^2$ . It is clear that $\mathscr{X}_{m,\sigma} = m + \sigma \mathscr{X}$ . The intersection of $\mathscr{X}_{m,\sigma}$ with the cone of random variables bounded below by $\alpha \in \mathbb{R}$ will be denoted by $\mathscr{X}_{m,\sigma}^\alpha$ . It is easily verified that, for any $m \in \mathbb{R}$ and $\sigma \in \mathbb{R}_+$ ,

(2.1) \begin{equation} m + \sigma \mathscr{X}^\alpha = \mathscr{X}_{m,\sigma}^{m+\sigma\alpha}.\end{equation}

Recall that a random variable is said to be two-point distributed if it takes only two (distinct) values. The set of two-point distributed random variables belonging to $\mathscr{X}$ can be parametrized by the open interval $\mathopen]0,1\mathclose[$ : let X take the values $x,y \in \mathbb{R}$ , $x < y$ , and set $p \,:\!=\, \mathbb{P}(X=x)$ , $1-p = \mathbb{P}(X=y)$ . Then $X \in \mathscr{X}$ if and only if $px + (1-p)y = 0$ and $px^2 + (1-p)y^2 = 1$ , which implies

(2.2) \begin{equation} x = - \biggl( \frac{1-p}{p} \biggr)^{1/2}, \qquad y = \biggl( \frac{p}{1-p} \biggr)^{1/2}.\end{equation}

Note that $p=0$ and $p=1$ are not allowed, and hence $p \in \mathopen]0,1\mathclose[$ . This is also obvious a priori, as there is no degenerate random variable with mean zero and variance one. The following simple observations, the proofs of which are immediate consequences of (2.2) and elementary algebra, will be useful.

We shall also need the following elementary lattice identities.

Proof. The identities in (i) and (ii) are clear. The identity in (iii) can be verified ‘case by case’, but it can also be deduced from the identity $(a-b)^+ = (a-b) + (a-b)^-$ , where, using (i),

\begin{equation*} (a-b)^- = - \bigl( (a-b) \wedge 0 \bigr) = b - (a \wedge b), \end{equation*}

from which the claim follows immediately.

3. de la Peña–Ibragimov–Jordan bound

The following sharp estimates are proved in [Reference de la Peña, Ibragimov and Jordan5].

Theorem 3.1. (de la Peña, Ibragimov, and Jordan) Let $X \in \mathscr{X}_{m,\sigma}$ and $c \in \mathbb{R}$ . Setting $p_0 \,:\!=\, \mathbb{P}(X>c)$ , we have

(3.1) \begin{equation} (m-c)p_0 \leqslant {\mathbb{E}}(X-c)^+ \leqslant (m-c)p_0 + \sigma(p_0-p_0^2)^{1/2}. \end{equation}

The proof of (3.1) in [Reference de la Peña, Ibragimov and Jordan5] is very elegant, the main idea being the introduction of an independent copy of the random variable X. Here we give an alternative, entirely elementary, proof.

Proof. Let us start with the lower bound. We can assume, without loss of generality, that $m \geqslant c$ , otherwise there is nothing to prove. Since

\begin{equation*} {\mathbb{E}}(X-c)^+ = {\mathbb{E}}(X-c)\mathbf{1}_{\{X>c\}} = {\mathbb{E}} X\mathbf{1}_{\{X>c\}} - c\mathbb{P}(X>c), \end{equation*}

it suffices to show that ${\mathbb{E}} X \mathbf{1}_{\{X>c\}} \geqslant m \mathbb{P}(X>c)$ . To this purpose, note that

\begin{equation*} {\mathbb{E}} X \mathbf{1}_{\{X>c\}} = {\mathbb{E}} X - {\mathbb{E}} X \mathbf{1}_{\{X \leqslant c\}}, \end{equation*}

where, thanks to the assumption $m \geqslant c$ ,

\begin{equation*} {\mathbb{E}} X \mathbf{1}_{\{X \leqslant c\}} \leqslant {\mathbb{E}} c \mathbf{1}_{\{X \leqslant c\}} = c \mathbb{P}(X \leqslant c) \leqslant m \mathbb{P}(X \leqslant c), \end{equation*}

and hence $ {\mathbb{E}} X \mathbf{1}_{\{X>c\}} \geqslant m - m\mathbb{P}(X \leqslant c) = m \mathbb{P}(X>c) $ .

To prove the upper bound, note that

\begin{equation*} {\mathbb{E}}(X-c)\mathbf{1}_{\{X>c\}} - (m-c)p_0 = {\mathbb{E}}\bigl( X\mathbf{1}_{\{X>c\}} - m\mathbf{1}_{\{X>c\}} \bigr), \end{equation*}

hence, adding and subtracting ${\mathbb{E}} Xp_0 = mp_0$ on the right-hand side, the Cauchy–Schwarz inequality yields

\begin{equation*} {\mathbb{E}}(X-c)^+ - (m-c)p_0 = {\mathbb{E}}(X-m)(\mathbf{1}_{\{X>c\}}-p_0) \leqslant \sigma {\left\lVert{\mathbf{1}_{\{X>c\}}-p_0}\right\rVert}_2 = \sigma (p_0-p_0^2)^{1/2}, \end{equation*}

thus completing the proof.

Theorem 3.1 implies useful one-sided Chebyshev-like bounds.

Proof. The proof of Theorem 3.1 remains valid with $p_1 \,:\!=\, \mathbb{P}(X \geqslant c)$ in place of $p_0$ ; hence, as the right-hand side of (3.1) must be positive, $ \sqrt{p_1(1-p_1)} \geqslant cp_1 $ . If $c \geqslant 0$ , squaring both sides yields a linear inequality that is satisfied if and only if $p_1 \leqslant 1/(1+c^2)$ . This proves (i).

For (ii), if $c<0$ , $ \mathbb{P}(X \leqslant c) = \mathbb{P}({-}X \geqslant -c) = 1 - \mathbb{P}({-}X < -c) $ . Since $-X \in \mathscr{X}$ and $-c>0$ , (i) implies that $ \mathbb{P}({-}X < -c) \geqslant {c^2}/({1+c^2}) $ , and hence

\begin{equation*} \mathbb{P}(X \leqslant c) \leqslant 1 - \frac{c^2}{1+c^2} = \frac{1}{1+c^2}. \end{equation*}

4. Scarf–Lo bound

The following estimate is obtained in [Reference Scarf11].

Theorem 4.1. (Scarf) Let $c,m,\sigma$ be strictly positive real numbers. The infimum of the function $X \mapsto {\mathbb{E}}(X \wedge c)$ on the set $\mathscr{X}_{m,\sigma}^0$ is attained, i.e. it is a minimum, and is given by

\begin{equation*} \min_{X \in \mathscr{X}_{m,\sigma}^0} {\mathbb{E}}(X \wedge c) = \begin{cases} \displaystyle \frac{m^2}{m^2+\sigma^2}c & \text{if } \displaystyle c \leqslant \frac{m^2+\sigma^2}{2m},\\[12pt] \displaystyle \frac{c+m}{2} - \frac12 \bigl( (c-m)^2 + \sigma^2 \bigr)^{1/2} & \text{if } \displaystyle c \geqslant \frac{m^2+\sigma^2}{2m}. \end{cases} \end{equation*}

Note that the constraint $X \geqslant 0$ is dictated by the structure of the practical inventory problem considered by Scarf. It is not needed, however, to avoid the infimum being minus infinity. In fact,

\begin{equation*} |{{\mathbb{E}}(X \wedge c)}| \leqslant {\mathbb{E}}|{X \wedge c}| \leqslant {\mathbb{E}}|{X}| + |{c}| \leqslant \bigl({\mathbb{E}} X^2 \bigr)^{1/2} + |{c}|,\end{equation*}

where $ \bigl({\mathbb{E}} X^2 \bigr)^{1/2} = \lVert{X}\rVert_2 = \lVert{X-m+m}\rVert_2 \leqslant \sigma + |{m}|$ , which implies

\begin{equation*} \inf_{X \in \mathscr{X}_{m,\sigma}} {\mathbb{E}}(X \wedge c) \geqslant -(\sigma + \lvert{m}\rvert + \lvert{c}\rvert).\end{equation*}

As observed in [Reference Lo7], Lemma 2.2(iii) immediately yields the following result.

Corollary 4.1. (Lo) Let $c,m,\sigma$ be strictly positive real numbers. The supremum of the function $X \mapsto {\mathbb{E}}(X - c)^+$ on the set $\mathscr{X}_{m,\sigma}^0$ is attained, i.e. it is a maximum, and is given by

\begin{equation*} \max_{X \in \mathscr{X}_{m,\sigma}^0} {\mathbb{E}}(X - c)^+ = \begin{cases} \displaystyle m - \frac{m^2}{m^2+\sigma^2}c & \text{if } \displaystyle c \leqslant \frac{m^2+\sigma^2}{2m},\\[10pt] \displaystyle \frac{m-c}{2} + \frac12 \bigl( (m-c)^2 + \sigma^2 \bigr)^{1/2} & \text{if } \displaystyle c \geqslant \frac{m^2+\sigma^2}{2m}. \end{cases} \end{equation*}

The remainder of this section is dedicated to showing that Theorem 4.1, and hence also its corollary, are consequences of Theorem 3.1. We shall argue by a sequence of elementary lemmas and propositions. The first is a reduction step that, in spite of its simplicity, considerably reduces the burden of symbolic calculations. Throughout the section we assume that $c,m,\sigma \in \mathbb{R}$ , with $\sigma>0$ . Further constraints (that do not imply any loss of generality) will be introduced as needed.

Lemma 4.1. Let $ \widetilde{c} \,:\!=\, ({c-m})/{\sigma} $ . Then

\begin{equation*} \inf_{X \in \mathscr{X}_{m,\sigma}^0} {\mathbb{E}}(X \wedge c) = m + \sigma \inf_{X \in \mathscr{X}^{-m/\sigma}} {\mathbb{E}}(X \wedge \widetilde{c}). \end{equation*}

Proof. Since $\mathscr{X}^0_{m,\sigma} = m + \sigma \mathscr{X}^{-m/\sigma}$ by (2.1), we have

\begin{equation*} \inf_{X \in \mathscr{X}_{m,\sigma}^0} {\mathbb{E}}(X \wedge c) = \inf_{Y \in \mathscr{X}^{-m/\sigma}} {\mathbb{E}}( (m+\sigma Y) \wedge c), \end{equation*}

where, by Lemma 2.2,

\begin{equation*} {\mathbb{E}}( (m+\sigma Y) \wedge c) = m + \sigma {\mathbb{E}}\biggl( Y \wedge \frac{c-m}{\sigma} \biggr), \end{equation*}

which immediately yields the conclusion.

The lemma implies that it suffices to study the problem of minimizing the function $X \mapsto {\mathbb{E}}(X \wedge c)$ over the set $\mathscr{X}^{-m}$ . We shall first study the minimization problem without the lower-boundedness constraint, i.e. on $\mathscr{X}$ rather than on $\mathscr{X}^{-m}$ . We shall need some more notation: the subset of $\mathscr{X}_{m,\sigma}$ such that $\mathbb{P}(X \leqslant c) = p$ is denoted by $\mathscr{X}_{m,\sigma}(p;c)$ . Note that, in view of Corollary 3.1, these sets are nonempty only for certain combinations of the parameters p and c. Let $L_c \colon \mathopen]0,1\mathclose[ \mapsto \mathbb{R}$ be the function defined by $ L_c\colon p \longmapsto - (p-p^2)^{1/2} + c(1-p)$ .

Proof. Lemma 2.2 (iii) implies $ {\mathbb{E}} (X-c)^+ = -{\mathbb{E}}(X \wedge c) $ for any X with mean zero; hence, by Theorem 3.1,

\begin{equation*} \inf_{X \in \mathscr{X}(c;p)} {\mathbb{E}}(X \wedge c) = - \sup_{X \in \mathscr{X}(c;p)} {\mathbb{E}} (X-c)^+ \geqslant - (p-p^2)^{1/2} + c(1-p). \end{equation*}

We are now going to show that the infimum in Lemma 4.2 is achieved, and that the minimizer is a two-point distributed random variable. We shall only consider, without loss of generality, those values of p such that $\mathscr{X}(p;c)$ is nonempty, that is, by Corollary 3.1, setting

\begin{equation*}\Pi_c \,:\!=\,\begin{cases} \displaystyle \biggl] 0 , \frac{1}{1+c^2} \biggr] &\quad \text{if } c<0,\\[17pt] \displaystyle \biggl[ \frac{c^2}{1+c^2} , 1 \biggr[ &\quad \text{if } c \geqslant 0,\end{cases}\end{equation*}

to $p \in \Pi_c$ .

Lemma 4.3. Let $p \in \Pi_c$ , and $X_0 \in \mathscr{X}$ be the two-point distributed random variable with parameter p. Then $ {\mathbb{E}}(X_0 \wedge c) = L_c(p) $ , and hence

\begin{equation*} \inf_{X \in \mathscr{X}(c;p)} {\mathbb{E}}(X \wedge c) = \min_{X \in \mathscr{X}(c;p)} {\mathbb{E}}(X \wedge c) = {\mathbb{E}}(X_0 \wedge c) = L_c(p). \end{equation*}

The following result essentially shows that Theorem 3.1 implies Theorem 4.1 in the unconstrained case (i.e. without assuming that the minimizer should be bounded from below).

Proof. The decomposition $ \mathscr{X} = \bigcup_{p \in \Pi_c} \mathscr{X}(p;c) $ implies (see, e.g., [Reference Bourbaki3, p. III.11])

\begin{equation*} \inf_{X \in \mathscr{X}} {\mathbb{E}}(X \wedge c) = \inf_{p \in \Pi_c} \inf_{X \in \mathscr{X}(p;c)} {\mathbb{E}}(X \wedge c), \end{equation*}

so that Lemma 4.3 implies the claim.

This clearly indicates that the next step should be to find the minimum of the function $L_c$ .

Proof. The argument is elementary, so it will be sketched only (the details can be found in [Reference Marinelli8]). The function $L_c$ is smooth and its derivative is the function

\begin{equation*} L^{\prime}_c\colon p \mapsto - \frac12 \bigl( p(1-p) \bigr)^{-1/2} (1-2p) - c; \end{equation*}

hence, standard calculus yields the claims (i)–(iii). It only remains to show that $p_\ast \in \Pi_c$ ; if $c \geqslant 0$ , this is equivalent to

\begin{equation*} p_\ast = \frac12 + \frac12 \frac{c}{(1+c^2)^{1/2}} \geqslant \frac{c^2}{1+c^2}. \end{equation*}

Setting $x \,:\!=\, c/(1+c^2)^{1/2} \in [0,1\mathclose[$ , this reduces to $1+x \geqslant 2x^2$ , which is satisfied if $x \in [0,1]$ . If $c<0$ , $p_\ast$ belongs to $\Pi_c$ if and only if

\begin{equation*} p_\ast = \frac12 + \frac12 \frac{c}{(1+c^2)^{1/2}} = \frac12 - \frac12 \frac{\lvert{c}\rvert}{(1+c^2)^{1/2}} \leqslant \frac{1}{1+c^2}, \end{equation*}

that is, setting $\langle c \rangle \,:\!=\, (1+c^2)^{1/2}$ for convenience, if and only if $\langle c \rangle \lvert{c}\rvert \geqslant \langle c \rangle^2 - 2$ . This inequality is easily seen to be satisfied for every $c \in [-1,0]$ . If $c \leqslant -1$ , the inequality is equivalent to $ \langle c \rangle^2 c^2 \geqslant (c^2-1)^2 $ , which simplifies to $3c^2 \geqslant 1$ , i.e. it is verified for every $\lvert{c}\rvert \geqslant 1/\sqrt{3} \vee 1$ , that is for every $c \leqslant -1$ . Finally, the expression for $L_c(p_\ast)$ follows by elementary algebra.

We have thus solved the problem of of minimizing the function $X \mapsto {\mathbb{E}}(X \wedge c)$ on $\mathscr{X}$ .

Proposition 4.2. Let $p_\ast$ be defined as in Lemma 4.4, and

\begin{equation*} x_\ast \,:\!=\, - \biggl( \frac{1-p_*}{p_*} \biggr)^{1/2}, \qquad y_\ast \,:\!=\, \biggl( \frac{p_*}{1-p_*} \biggr)^{1/2}. \end{equation*}

The two-point distributed random variable $X_0$ with $\mathbb{P}(X_0 = x_\ast) = p_\ast$ , $\mathbb{P}(X_0 = y_\ast) = 1-p_\ast$ is a minimizer of $\inf_{X \in \mathscr{X}} {\mathbb{E}}(X \wedge c)$ , i.e.

\begin{equation*} \inf_{X \in \mathscr{X}} {\mathbb{E}}(X \wedge c) = L_c(p_\ast) = {\mathbb{E}}(X_0 \wedge c). \end{equation*}

If the parameters of the problem are such that $X_0$ , as defined in Proposition 4.2, is bounded below by $-m$ , the original minimization problem is clearly solved. We shall assume, until further notice, that $m\geqslant 0$ and $c>-m$ . This comes at no loss of generality, as $\mathscr{X}^{-m}$ is empty if $m<0$ , and the problem degenerates if $c \leqslant -m$ , in the sense that ${\mathbb{E}}(X \wedge c) = c$ for every $X \geqslant -m$ .

Corollary 4.2. Let $X_0$ be defined as in Proposition 4.2. Then

\begin{equation*} \inf_{X \in \mathscr{X}^{-m}} {\mathbb{E}}(X \wedge c) = {\mathbb{E}}(X_0 \wedge c) \end{equation*}

if and only if $ c \geqslant ({1-m^2})/{2m} $ .

Proof. By definition of $X_0$ , the lower bound $X_0 \geqslant -m $ holds if and only if $p_\ast \geqslant 1/(1+m^2)$ , i.e. if and only if

\begin{equation*} \frac{1}{1+m^2} \leqslant \frac12 + \frac12 \frac{c}{(1+c^2)^{1/2}}, \end{equation*}

or, equivalently,

\begin{equation*} 2 \frac{1}{1+m^2} - 1 \leqslant \frac{c}{(1+c^2)^{1/2}}, \end{equation*}

where the left-hand side takes values in the interval $\mathopen]-1,1\mathclose]$ . Elementary algebra shows that the inequality

\begin{equation*} \beta \leqslant \frac{c}{(1+c^2)^{1/2}}, \qquad \beta \in \mathopen]-1,1\mathclose], \end{equation*}

is verified if and only if

\begin{equation*} c \geqslant \frac{\beta}{(1-\beta^2)^{1/2}}. \end{equation*}

Replacing $\beta$ by $2/(1+m^2)-1$ finally implies that $X_0 \geqslant -m$ if and only if

\begin{equation*} c \geqslant \frac{1-m^2}{2m}, \end{equation*}

from which the claim follows immediately.

In view of the corollary, we only need to consider the problem under the condition

\begin{equation*} c < \frac{1-m^2}{2m}.\end{equation*}

Note that this implies $c < 1/m$ .

Let us start by observing that, for any $X \geqslant -m$ ,

(4.1) \begin{align} {\mathbb{E}}(X \wedge c) &= {\mathbb{E}}\bigl( X \mathbf{1}_{\{X \leqslant c\}} + c\mathbf{1}_{\{X>c\}} \bigr)\nonumber\\ &\geqslant -m \mathbb{P}(X \leqslant c) + c\mathbb{P}(X>c)\\ &= {\mathbb{E}}(Y \wedge c), \nonumber\end{align}

where Y is a random variable taking values in $\{-m,y\}$ , $y \geqslant c$ , with

\begin{equation*}\mathbb{P}(Y = -m) = \mathbb{P}(X \leqslant c), \qquad \mathbb{P}(Y=y) = \mathbb{P}(X>c).\end{equation*}

In order for the random variable Y to belong to $\mathscr{X}$ , it is sufficient and necessary, in view of (2.2) and Lemma 2.1, that

\begin{equation*} \mathbb{P}(Y=-m) = \mathbb{P}(X \leqslant c) = \frac{1}{1+m^2},\end{equation*}

and either $c \leqslant 0$ or $c \geqslant 0$ and

\begin{equation*} \mathbb{P}(X \leqslant c) \geqslant \frac{c^2}{1+c^2}.\end{equation*}

In other words, Y belongs to $\mathscr{X}$ and takes values in $\{-m,y\}$ with $y \geqslant c$ if and only if $c \leqslant 0$ or $c>0$ and

\begin{equation*} \frac{1}{1+m^2} \geqslant \frac{c^2}{1+c^2}.\end{equation*}

As this inequality is satisfied if and only if $cm \leqslant 1$ , which holds by assumption, Y satisfies the abovementioned conditions if and only if $\mathbb{P}(X \leqslant c) = 1/(1+m^2)$ . Let us then define the random variable $Y_0 \in \mathscr{X}^{-m}$ as the (unique) random variable in $\mathscr{X}$ identified by the parameter $p_m = {1}/({1+m^2})$ . We are going to show that $Y_0$ is in fact the minimizer of the problem.

Proposition 4.3. If $c < (1-m^2)/(2m)$ , then

\begin{equation*} \inf_{X \in \mathscr{X}^{-m}} {\mathbb{E}}(X \wedge c) = {\mathbb{E}}(Y_0 \wedge c) = c - (m+c)\frac{1}{1+m^2}. \end{equation*}

Proof. Let us rewrite (4.1) as $ {\mathbb{E}}(X \wedge c) \geqslant -c + (m+c) \mathbb{P}(X \leqslant c) $ , which holds for every $X \in \mathscr{X}^{-m}$ . Since $m+c>0$ by assumption, the function $p \mapsto -c + (m+c)p$ is decreasing. Therefore, for every $X \in \mathscr{X}^{-m}$ such that $\mathbb{P}(X \leqslant c) \leqslant p_m$ , we have

\begin{equation*} {\mathbb{E}}(X \wedge c) \geqslant c - (m+c)p_m = {\mathbb{E}}(Y_0 \wedge c). \end{equation*}

Let $X \in \mathscr{X}^{-m}$ be such that $p\,:\!=\,\mathbb{P}(X \leqslant c)>p_m$ . Then Lemma 4.2 yields

\begin{equation*} {\mathbb{E}}(X \wedge c) \geqslant -\bigl( p(1-p) \bigr)^{1/2} +c(1-p) = L_c(p). \end{equation*}

Since $p_* < p_m$ by assumption and the function $L_c$ is increasing on $\mathopen]p_*,1\mathclose[$ by Lemma 4.4, it follows that $ {\mathbb{E}}(X \wedge c) \geqslant L_c(p) \geqslant L_c(p_m) = {\mathbb{E}}(X_0 \wedge c) $ , which concludes the proof.

We have therefore proved the main result, which reads as follows.

Theorem 4.2. Let $c,m,\sigma \in \mathbb{R}$ with $m \geqslant 0$ , $\sigma>0$ , and $c>-m$ . Then

\begin{equation*} \inf_{X \in \mathscr{X}^{-m}} {\mathbb{E}}(X \wedge c) = \begin{cases} \displaystyle {\mathbb{E}}(Y_0 \wedge c) = L_c(p_m) = c - (m+c)\frac{1}{1+m^2} &\quad \displaystyle \text{if } c \leqslant \frac{1-m^2}{2m}, \\[10pt] \displaystyle {\mathbb{E}}(X_0 \wedge c) = L_c(p_*) = \frac12 c - \frac12 (1+c^2)^{1/2} &\quad \displaystyle \text{if } c \geqslant \frac{1-m^2}{2m}. \end{cases} \end{equation*}

Using the notation X(p) to denote a two-point distributed random variable in $\mathscr{X}$ with parameter p, we could write, more concisely,

\begin{equation*} \inf_{X \in \mathscr{X}^{-m}} {\mathbb{E}}(X \wedge c) = {\mathbb{E}}(X(p_\ast \vee p_m) \wedge c) = L_c(p_\ast \vee p_m).\end{equation*}

The bound by Scarf, and hence the one by Lo, i.e. Theorem 4.1 and its corollary, follow immediately by the previous theorem and Lemma 4.1.

5. Applications to option prices

In order to also consider bounds for put options, we record an easy consequence of Theorem 3.1.

Corollary 5.1. Under the hypotheses of Theorem 3.1, let $p \,:\!=\, \mathbb{P}(X \leqslant c)$ . Then

\begin{equation*} (c-m)p \leqslant {\mathbb{E}}(c-X)^+ \leqslant (c-m)p + \sigma (p-p^2)^{1/2}. \end{equation*}

Let us also note that a simple but useful sharpening of the lower bound in Theorem 3.1 can be given: Jensen’s inequality implies that

\begin{equation*}{\mathbb{E}}(X-c)^+ \geqslant ({\mathbb{E}} X - c)^+ = (m - c)^+\end{equation*}

as well as ${\mathbb{E}}(c-X)^+ \geqslant (c-m)^+$ .

Assume that the probability space $(\Omega,\mathscr{F},\mathbb{P})$ is equipped with a filtration $(\mathscr{F}_t)_{t\in[0,T]}$ satisfying the so-called usual conditions. Let $\widehat{S}$ and $\beta$ be the price processes of two traded assets, the latter of which is strictly positive and is used as numéraire, so that $S\,:\!=\,\beta^{-1}\widehat{S}$ is the discounted price process of the former asset. We assume that no asset pays dividends. A classical result [Reference Delbaen and Schachermayer6] asserts that a suitable version of no-arbitrage holds (precisely, no free lunch with vanishing risk) if and only if there exists a probability measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ such that S is a $\sigma$ -martingale with respect to $\mathbb{Q}$ . For simplicity of notation, let us assume that $\mathbb{P}$ is already an equivalent $\sigma$ -martingale measure that is used for pricing. We are then interested in upper and lower bounds for

\begin{equation*} \pi_c \,:\!=\, {\mathbb{E}} \beta_T^{-1} \bigl( \widehat{S}_T - K \bigr)^+ = {\mathbb{E}} \bigl( S_T - \beta_T^{-1}K \bigr)^+, \quad \pi_p \,:\!=\, {\mathbb{E}} \beta_T^{-1} \bigl( K - \widehat{S}_T \bigr)^+ = {\mathbb{E}} \bigl( \beta_T^{-1}K - S_T \bigr)^+.\end{equation*}

We first establish an easy consequence of the $\sigma$ -martingale property of S.

Proof. Recall first that a $\sigma$ -martingale bounded from below is a local martingale thanks to the Ansel–Stricker lemma [Reference Ansel and Stricker1, p. 309], thus also a supermartingale by an application of Fatou’s lemma. This implies that ${\mathbb{E}} S_T \leqslant S_0$ , and hence, by Jensen’s inequality,

\begin{equation*} \pi_p = {\mathbb{E}} \bigl( \beta_T^{-1}K - S_T \bigr)^+ \geqslant \bigl( K {\mathbb{E}}\beta_T^{-1} - {\mathbb{E}} S_T \bigr)^+ \geqslant \bigl( K {\mathbb{E}}\beta_T^{-1} - S_0 \bigr)^+, \end{equation*}

which proves (i). The proof of (ii) is entirely analogous.

It is clear that the estimates of Theorem 3.1 and Corollary 4.1 cannot be directly applied, as $\beta_T$ is a random variable. In some cases, however, this is indeed possible. For instance, apart from the trivial case where the price process $\beta$ of the numéraire is non-random, if the random variables $\beta_T^{-1}$ and $\widehat{S}_T^+$ belong to $L^2$ , and $\beta_T^{-1}$ and $(S_T-K)^+$ are uncorrelated, so that

\begin{equation*} {\mathbb{E}} \beta_T^{-1} \bigl( \widehat{S}_T - K \bigr)^+ = {\mathbb{E}}\beta_T^{-1} {\mathbb{E}} \bigl( \widehat{S}_T - K \bigr)^+,\end{equation*}

estimates on $\pi_c$ can be obtained by Corollary 4.1 in terms of the mean of $\beta_T^{-1}$ and the mean and variance of $\widehat{S}_T$ . In order to apply Theorem 3.1, the value of the distribution function of $\widehat{S}_T$ at K is also needed. In the following we make the stronger assumption that the random variables $\beta_T$ and $\widehat{S}_T$ are independent. Furthermore, we assume that S is a supermartingale. By independence,

\begin{equation*} S_0 \geqslant {\mathbb{E}} S_T = {\mathbb{E}}\beta_T^{-1} \widehat{S}_T = {\mathbb{E}}\beta_T^{-1} {\mathbb{E}}\widehat{S}_T,\end{equation*}

and hence ${\mathbb{E}}\widehat{S}_T \leqslant S_0 / {\mathbb{E}}\beta_T^{-1}$ . Setting $k \,:\!=\, K {\mathbb{E}}\beta_T^{-1}$ and $\widehat{\sigma} \,:\!=\, \lVert{\widehat{S}_T - {\mathbb{E}}\widehat{S}_T}\rVert_2$ , we thus have the bounds

\begin{align*} (k-S_0)^+ \leqslant \pi_p &\leqslant (k - {\mathbb{E}} S_T)p + {\mathbb{E}}\beta_T^{-1} \widehat{\sigma} (p-p^2)^{1/2},\\ ({\mathbb{E}} S_T - k)^+ \leqslant \pi_c &\leqslant (S_0 - k)p_0 + {\mathbb{E}}\beta_T^{-1} \widehat{\sigma} (p_0-p_0^2)^{1/2}.\end{align*}

If S is a martingale, then ${\mathbb{E}} S_T=S_0$ , so the bounds for $\pi_p$ and $\pi_c$ assume a more symmetric form.

Adapting Lo’s estimate to option prices can be done along the same lines. It does not seem possible, though, to exploit the supermartingale property of S to obtain bounds involving $S_0$ rather than ${\mathbb{E}} S_T$ . On the other hand, if S is a martingale, and if the constraint $\widehat{S}_T \geqslant 0$ is not enforced, it follows from the proofs in Section 4 and elementary computations that

\begin{equation*} \pi_c \leqslant \frac{S_0-k}{2} + \frac12 \bigl( (S_0-k)^2 + \bigl({\mathbb{E}}\beta_T^{-1}\bigr)^2\widehat{\sigma}^2 \bigr)^{1/2},\end{equation*}

from which a corresponding upper bound for put options can be obtained by put–call parity.