Erratum to: Finance Stoch (2001) 5:259–272 DOI 10.1007/PL00013534

1 Discussion

We sketch the counterexample in [4] in a slightly modified and self-contained way. In the sequel, we suppose that the reader is familiar with the paper [1] as well as with [3, Example 5.1 bis]. We briefly recall the notation of that example which is the basis of the counterexample in [4].

The stock price process \(S=(S_{0}, S_{1})\) is defined by \(S_{0}=1\) and by letting \(S_{1}\) assume the value \(x_{0}=2\) with probability \(p_{0}=1-\alpha\) and, for \(n\geq1\), the value \(x_{n}=\frac{1}{n}\) with probability \(p_{n}=\alpha2^{-n}\), for \(0 <\alpha< 1\) sufficiently small.

For logarithmic utility \(U(x)= \ln x\), we obtain, for given endowment \(x>0\), that it is optimal to invest the entire endowment into the stock \(S\) so that we end up at time \(t=1\) with the random wealth \(\hat {X}(x)=xS_{1}\). For the value function \(u(x)\), we thus obtain (see [3, Example 5.1 bis] for details) the expected utility of \(U(\hat {X}(x))\), i.e.,

$$ u(x)=\mathbb{E} \bigl[U \bigl(\hat{X}(x) \bigr) \bigr]=\ln x + \sum ^{\infty}_{n=0} p_{n} \ln x_{n}. $$
(1.1)

Using the notation from [1], the point of this example is that the dual optimizer \(\hat{Q}(y)\) is not an element of \(L^{1}\), but only of its bidual \((L^{\infty})^{*}\). In other words, \(\hat{Q}(y)\) defines a finitely additive probability measure on \((\varOmega, \mathcal{F})\) which fails to be countably (or \(\sigma\)-) additive. We write \(\hat{Q}(y)=\hat{Q}^{r}(y)+\hat {Q}^{s}(y)\) for the decomposition of \(\hat{Q}(y)\) into its regular and singular parts.

To verify that \(\|\hat{Q}^{r}(y)\| < \|\hat{Q}(y)\|\), fix \(x=y=1\). Writing \(\hat{X}\) for \(\hat{X}(1)\), \(\hat{Q}\) for \(\hat{Q}(1)\) and \(q_{n}\) for \(\hat{Q}[S_{1}=x_{n}]\), we have the relation [3, Theorem 2.2]

$$ U'(\hat{X})= \frac{d\hat{Q}^{r}}{d\mathbb{P}}, $$
(1.2)

so that

$$ (x_{n})^{-1}=\frac{q_{n}}{p_{n}}, \quad n \geq0, $$

which yields

$$ \sum^{\infty}_{n=0} q_{n} = \frac{1-\alpha}{2}+ \alpha\sum^{\infty}_{n=1}n p_{n}. $$

This term is smaller than 1 (recall that \(0 <\alpha<1\) is small), which readily shows that the regular part \(\hat{Q}^{r}\) has a smaller mass than \(\hat{Q}\).

So far, we have just recalled [3, Example 5.1 bis]. For the next step, we follow [2] and distinguish between the odd and even numbers \(n \in\mathbb{N}_{0}\) to define

$$A=\{S_{1} = x_{n} \text{ for odd } n \geq1\}, \qquad B=\{S_{1} = x_{n} \text{ for even } n \geq0\}. $$

Now comes the beautiful idea from [4]. Define the process \(\tilde{S}=(\tilde{S}_{0}, \tilde{S}_{1})\) by setting \(\tilde{S}_{0}=S_{0}=1\) and

S ˜ 1 = S 1 + 1 2 1 A .
(1.3)

We also consider the random variable \(e_{T}\) which we define as

e T =− 1 2 1 A .
(1.4)

For the utility maximization problem, subject to the additional random endowment \(e_{T}\), we define the value function \(\tilde{u}(x)\) as in [1]. This boils down to the formula

$$ \tilde{u}(x)= \sup_{\lambda\in\mathbb{R}} \mathbb{E} \bigl[\ln \bigl(x+\lambda(\tilde{S}_{1} - \tilde{S}_{0})+e_{T} \bigr) \bigr]. $$
(1.5)

For example, for \(x=1\), we again find that the optimizer \(\hat{\lambda }\) in (1.5) equals 1, i.e., it again is optimal to invest the entire initial endowment \(x=1\) into the stock \(\tilde{S}\); in this case, the terms 1 A in (1.3) and (1.4) cancel out perfectly. In fact, for all \(x\geq1\), we find that the optimal \(\hat{\lambda}\) in (1.5) equals \(\hat{\lambda} (x)=x\), just as in (1.1). Indeed, for \(x \geq1\), the crucial constraint is that we cannot invest more than the amount \(\hat{\lambda}=x\) into the stock due to the definition of \(\tilde{S}\) and \(e_{T}\) on the set \(B\), which corresponds to the even numbers \(n \in\mathbb{N}_{0}\).

On the other hand, for \(x < 1\), the picture changes; now the binding constraint is given by the odd numbers \(n \in\mathbb{N}_{0}\), i.e., by the behaviour of \(\tilde{S}\) and \(e_{T}\) on the set \(A\). For \(\frac{1}{2} < x \leq1\), we find that the optimizer \(\hat{\lambda}\) in (1.5) equals \(\hat{\lambda}(x)=1-2(1-x)=2x-1\). The remaining amount \(x-(2x-1)=1-x\) of the initial wealth \(x\) is kept in the bond. Note that for \(x\leq\frac{1}{2}\), there is no admissible solution \(\lambda\) in (1.5), i.e., \(\tilde{u}(x)=-\infty\) in (1.5). We thus obtain for the optimal terminal wealth \(\hat{X}(x)=x+\hat {\lambda} (\tilde{S}_{1}-\tilde{S}_{0}) + e_{T}\), for \(\frac{1}{2} < x \leq1\),

$$ \hat{X}(x)= \textstyle\begin{cases} (2x-1) x_{n} + (1-x) & \mbox{for }n\mbox{ even,}\\ (2x-1) (x_{n} + \frac{1}{2}) + (\frac{1}{2}-x) & \mbox{for }n\mbox{ odd} \end{cases} $$
(1.6)

and, for \(1\leq x <\infty\),

$$ \hat{X}(x)= \textstyle\begin{cases} x x_{n} & \mbox{for }n\mbox{ even,}\\ x(x_{n}+\frac{1}{2})-\frac{1}{2} & \mbox{for }n\mbox{ odd}. \end{cases} $$
(1.7)

Note that we always have that \(\hat{X}(x)\) is an a.s. strictly positive random variable whose essential infimum is zero. The latter property is obtained by considering the sets \(\{S_{n}=x_{n}\}\) with \(n\) tending to infinity, where we have to consider the odd \(n\) in the case \(\frac {1}{2} < x\leq1\), and the even \(n\) in the case \(x \geq1\).

Clearly, the definitions (1.6) and (1.7) coincide for \(x=1\) in which case we obtain \(\hat{X}(1)=S_{1}\). In particular, the value function

$$\tilde{u}(x)=\mathbb{E}[\ln\hat{X}(x)] $$

is continuous at \(x=1\), as must be the case. We shall see that \(\tilde{u}\) has a kink at \(x=1\). Indeed, we may calculate the derivative of \(\tilde{u}(x)\), for \(x \in (\frac{1}{2},1)\) as well as for \(x \in(1,\infty)\), by using the formula

$$ \frac{d}{dx} \tilde{u}(x)=\mathbb{E} \biggl[\frac{d}{dx} \ln\hat {X}(x) \biggr], \quad x \in (1 / 2,1) \cup(1, \infty). $$

Hence the difference \(\Delta\tilde{u}'(1)=\lim_{x\searrow1} (\frac {d}{dx} \tilde{u} (x)) -\lim_{x\nearrow1} (\frac{d}{dx} \tilde{u}(x))\) of the right and left derivatives of \(\tilde{u}\) at \(x=1\) can be explicitly computed as

$$ \Delta\tilde{u}'(1)= \sum^{\infty}_{n=0} p_{n} \biggl(\frac{1}{x_{n}} - \frac {2}{x_{n}} \biggr)=- \biggl( \frac{p_{0}}{2} + \sum^{\infty}_{n=1} np_{n} \biggr), $$

which clearly shows that the function \(\tilde{u}(x)\) fails to be differentiable at \(x=1\).

Summing up, following [4], we have constructed an example where the value function \(\tilde{u}(\cdot)\) fails to be differentiable.

We still want to have a closer look at the dual problem associated to the above example. In particular, we want to spot precisely where the erroneous argument in [1] has occurred. Define

$$ y_{1}=\lim_{x\searrow1} \biggl(\frac{d}{dx} \tilde{u}(x) \biggr) \quad\mbox{and} \quad y_{2}=\lim _{x\nearrow1} \biggl(\frac{d}{dx} \tilde{u}(x) \biggr). $$

As the dual value function \(\tilde{v}\) (see [1] for the definition)

$$ \tilde{v}(y)=\min_{Q \in\mathcal{D}} \biggl\{ \mathbb{E} \biggl[V \biggl(y \frac{dQ^{r}}{d\mathbb{P}} \biggr) \biggr]+y \langle Q, e_{T} \rangle \biggr\} $$
(1.8)

is conjugate to \(\tilde{u}\) (see [1]), we know from the fact that \(\tilde{u}(x)\) has a kink at \(x=1\) that \(\tilde{v}(y)\) is an affine function with slope −1 on the interval \([y_{1}, y_{2}]\), in view of the basic relation

$$\tilde{v}(y)=\sup_{x} \bigl(\tilde{u}(x) -xy \bigr). $$

What are the dual optimizers \(\hat{Q}_{y}\) for \(y \in[y_{1}, y_{2}]\), given by [1, Theorem 3.1 and Lemma 4.1]? We know from [1] that the regular parts \(\hat{Q}^{r}_{y}\) are unique and given by the formula

$$ U' \bigl(\hat{X}(x) \bigr)=y\frac{d\hat{Q}^{r}_{y}}{d\mathbb{P}}, $$

as in (1.2) above. The number \(x\) is associated to \(y\) via the relation \(-\tilde{v}'(y)=x\), which yields \(x=1\) for \(y\in[y_{1}, y_{2}]\). This implies the amazing fact that the regular parts \(y \hat {Q}^{r}_{y}\) of the dual optimizers \(y\hat{Q}_{y}\) are identical, for all \(y \in[y_{1}, y_{2}]\). Note that the total mass of the elements \(y \hat{Q}_{y} \in(L^{\infty})^{*}\) equals \(\|y \hat{Q}_{y}\|=y\). If we pass as usual to the normalized finitely additive probability measures \(\hat{Q}_{y}\), their regular parts \(\hat{Q}_{y}^{r}\) scale by the factor \(y^{-1}\).

As regards the singular part \(\hat{Q}^{s}_{y}\) of \(\hat{Q}_{y}\), it is clear that \(\hat{Q}^{s}_{y}\) is supported by each of the sets

$$C_{N}=\bigcup^{\infty}_{n=N} \{S_{n}=x_{n}\}. $$

Indeed, for each \(\epsilon>0\), the singular measure \(\hat{Q}^{s}_{y}\) is supported by the set \(\{\hat{X}(x) < \epsilon\}\), where \(-\tilde {v}'(y)=x\). This follows from the analysis in [3, Example 5.1 bis]. But now the additional aspect of the odd and even \(n\) arises: How much of this singular mass sits on \(C_{N} \cap A\), and how much on \(C_{N} \cap B\)?

It follows from (1.6), (1.7) and the subsequent discussion that for \(\frac{1}{2} < x <1\) and \(\tilde{u}'(x)=y\), the singular measure \(\hat{Q}_{y}^{s}\) is supported by \(A\), while for \(1< x<\infty\) and \(\tilde{u}'(x)=y\), the singular measure \(\hat{Q}_{y}^{s}\) is supported by \(B\). One may also pass to the limits \(x\nearrow1\) and \(x \searrow1\) to show that \(\hat{Q}^{s}_{y_{1}}\) is supported by \(A\), while \(\hat {Q}^{s}_{y_{2}}\) is supported by \(B\). It turns out that for general \(y \in [y_{1}, y_{2}]\) of the form \(y=\mu y_{1} + (1-\mu)y_{2}\), we have the affine relations

$$ y\hat{Q}^{s}_{y}[A] = \mu y_{1} \hat{Q}^{s}_{y_{1}} [A] $$
(1.9)

and

$$ y\hat{Q}^{s}_{y}[B] = (1-\mu) y_{2} \hat{Q}^{s}_{y_{2}} [B]. $$
(1.10)

Indeed, because \(\mathbb{E}_{Q}[\tilde{S}_{1} - \tilde{S}_{0}]=0\) for any equivalent martingale measure \(Q\) for \(\tilde{S}\), we also obtain \(\langle\hat{Q}_{y}, \tilde{S}_{1} - \tilde{S}_{0} \rangle =0\) by weak-star continuity. As \(y\hat{Q}^{r}_{y}\) does not depend on \(y\), we obtain that \(\langle y\hat{Q}_{y}^{s}, \tilde{S}_{1} - \tilde{S}_{0}\rangle\) does not depend on \(y\) either, for \(y \in[y_{1}, y_{2}]\). On the set \(C_{N} \cap A\) (resp. \(C_{N} \cap B\)), the random variable \(\tilde{S}_{1}-\tilde {S}_{0}\) equals \(-\frac{1}{2}\) (resp. −1), up to an error of at most \(\frac{1}{N}\) which disappears in the limit \(N \mapsto\infty\). This implies that \(-\frac{1}{2}y \hat{Q}^{s}_{y}[A]-y \hat{Q}_{y}^{s}[B]\) is constant when \(y\) varies in \([y_{1}, y_{2}]\) and readily yields the affine relations (1.9) and (1.10).

Finally, let us have a closer look where the mistake in Remark 4.2 of [1] occurred. In that argument, we fixed numbers \(0 < y_{1} <y_{2}\) (which may or may not coincide with the \(y_{1}, y_{2}\) considered above) and considered the value function \(\tilde{v}(y)\) as in (1.8). For \(y_{1} \neq y_{2}\), we have that \(y_{1} \hat{Q}_{y_{1}}\) is different from \(y_{2}\hat{Q}_{y_{2}}\), as shown in [1]. Up to this point, the reasoning was correct. We then tacitly (and incorrectly) assumed that this implies that their regular parts \(y_{1} \hat {Q}^{r}_{y_{1}}\) and \(y_{2} \hat{Q}^{r}_{y_{2}}\) must be different too! This would imply the strict inequality claimed in Remark 4.2 of [1]. But as we just have seen, it may happen that these two measures coincide. In addition, the singular parts \(y\hat{Q}_{y}^{s}\) satisfy the affine relations (1.9) and (1.10), which also prevent the inequality in Remark 4.2 of [1] becoming strict.