A correction to Section 4 in [Reference Puhalskii1] is provided.
The expressions for
$I_{x_0}^Q(q)$
in the statement of Theorem 5 and for
$\overline I^Q_{q_0,x_0}(q)$
in [Reference Puhalskii1] are incorrect as the solutions of the associated variational problems do not account for the constraints
$\int_0^1\dot w^0(x)\,\mathrm{d}x=0$
and
$\int_0^1\dot k(x,t)\,\mathrm{d}x=0$
. The contents of Section 4 should be replaced with the following material.
4. Evaluating the deviation functions
This section is concerned with solving for
$ I^Q_{x_0}(q)$
and
$\overline I^Q_{q_0,x_0}(q)$
.
Theorem 5. Suppose that the c.d.f. F is an absolutely continuous function and
$I_{x_0}^Q(q)<\infty$
. Then q is absolutely continuous,
$(\dot q(t)-\int_0^t\dot q(s)\mathbf{1}{q(s)>0}F'(t-s)\,\mathrm{d}s,t\in\mathbb{R}_+)\in\mathbb L_2(\mathbb{R}_+)$
, the infimum in (2.5) is attained uniquely and
\begin{align*} I_{x_0}^Q(q)=\frac{1}{2}\,\int_0^\infty\hat{p}(t)\bigg(\dot q(t)-\int_0^t\dot q(s)\mathbf{1}{q(s)>0}F'(t-s)\,ds+(\beta-x_0^-) F_0'(t)\bigg)\,\mathrm{d}t,\end{align*}
where
$\hat{p}(t)$
represents the unique
$ \mathbb{L}_2(\mathbb{R}_+)$
solution p(t) of the Fredholm equation of the second kind,
\begin{multline} (\mu+\sigma^2)p(t)= \dot q(t)-\int_0^t\dot q(s)\mathbf{1}{q(s)>0}F'(t-s)\,\mathrm{d}s+(\beta-x_0^-) F_0'(t)\\+F'_0(t)\int_0^\infty F_0'(s)p(s)\,\mathrm{d}s+\sigma^2\int_0^\infty F'(\lvert s-t\rvert)p(s)\,\mathrm{d}s\\+(\mu-\sigma^2)\int_0^\infty\int_0^{s\wedge t}F'(s-\tilde s)F'(t-\tilde s)\,\mathrm{d}\tilde s\, p(s)\,\mathrm{d}s,\end{multline}
with
$\dot q$
,
$F_0'$
and F’ representing derivatives.
Proof. Using that
\begin{equation} \int_{\mathbb{R}_+^2} \mathbf{1}{x+s\le t}\,\dot k(F(x),\mu s)\,\mathrm{d}F(x)\,\mu\,\mathrm{d}s= \int_0^{ t}\int_0^{F(t-s)}\dot k(x,\mu s)\,\mathrm{d}x\,\mu\,\mathrm{d}s,\end{equation}
(2.6) can be expressed in the form
\begin{align*} q(t)=f(t)+\int_0^tq(t-s)^+\,\mathrm{d}F(s),\quad t\in\mathbb{R}_+,\end{align*}
with the functions f(t) and F(t) being absolutely continuous. The function q(t) is absolutely continuous by Lemma 8. In addition, (4.2) implies that, almost everywhere,
\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}\,\int_{\mathbb{R}_+^2} \mathbf{1}{x+s\le t}\,{\dot k}(F(x),\mu s)\,\mathrm{d}F(x)\,\mu\,\mathrm{d}s=\int_0^t{\dot k}(F(s),\mu (t-s))F'(s)\,\mu\,\mathrm{d}s.\end{equation}
The infimum in (2.5) is attained uniquely by coercitivity and strict convexity of the function being minimised, cf., Proposition II.1.2 in [5]. By differentiating equation (2.6), in light of equation (4.3), we find that, almost everywhere,
\begin{multline*}\dot w^0(F_0(t))F_0'(t)+\sigma\,\dot w(t)-\int_0^tF'(t-s)\sigma\,\dot w(s)\,\mathrm{d}s+\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu\,\mathrm{d}s\\-\bigg( \dot q(t)-\int_0^t\dot q(s)\mathbf{1}{q(s)>0}F'(t-s)\,\mathrm{d}s+(\beta-x_0^-) F_0'(t)\bigg)=0.\end{multline*}
In addition, the requirements that
$w^0(0)=w^0(1)=0$
and
$k(0,t)=k(1,t)=0$
give rise to the constraints
\begin{equation} \int_0^1\dot w^0(x)\,\mathrm{d}x=0\end{equation}
and
\begin{equation} \int_0^1\dot k(x,t)\,\mathrm{d}x=0.\end{equation}
Introduce the map
\begin{multline*} \Phi:\, (\dot w^0,\dot w,\dot k)\to\bigg(\dot w^0(F_0(t))F_0'(t)+\sigma\,\dot w(t)-\int_0^tF'(t-s)\sigma\,\dot w(s)\,\mathrm{d}s\\+\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu\,\mathrm{d}s,t\in\mathbb{R}_+\bigg).\end{multline*}
Since
$F_0'(t)$
is bounded by (2.1),
$\Phi$
maps
$V=\mathbb L_2([0,1])\times\mathbb L_2(\mathbb{R}_+)\times\mathbb L_2( [0,1]\times\mathbb{R}_+)$
to
$\mathbb L_2(\mathbb{R}_+)$
. For instance, on using that
$\int_0^\infty F'(s)\,\mathrm{d}s=1$
,
\begin{align*} \int_0^\infty\bigg(\int_0^tF'(t-s)\dot w(s)\,\mathrm{d}s\bigg)^2\,\mathrm{d}t\le \int_0^\infty\int_0^tF'(t-s)\dot w(s)^2\,\mathrm{d}s\,\mathrm{d}t=\int_0^\infty\dot w(s)^2\,\mathrm{d}s<\infty\end{align*}
and
\begin{multline*} \int_0^\infty\bigg(\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu\mathrm{d}s\bigg)^2\,\mathrm{d}t\le \int_0^\infty\int_0^t\dot k(F(s),\mu (t-s))^2F'(s)\,\mu^2\mathrm{d}s\,\mathrm{d}t\\=\mu^2\int_0^\infty\int_0^1\dot k( x,t)^2\,\mathrm{d}x\,\mathrm{d}t<\infty.\end{multline*}
By Lagrange multipliers (see, e.g., Proposition III.5.2 in [5]) with
$Y=\mathbb L_2(\mathbb{R}_+)^2\times\mathbb{R}$
and the set of componentwise nonnegative functions as the cone
$\mathcal{C}$
,
\begin{multline} I_{x_0}^Q(q)={\sup}_{(p,\tilde p,r)\in \mathbb L_2(\mathbb{R}_+)^2\times\mathbb{R}}\;\inf_{\substack{(\dot w^0,\dot w,\dot k)\in\mathbb L_2([0,1])\\\times\mathbb L_2( \mathbb{R}_+)\times\mathbb L_2( [0,1]\times\mathbb{R}_+)}}\Bigg(\frac{1}{2}\int_0^1\dot w^0(x)^2\,\mathrm{d}x+\frac{1}{2}\int_0^\infty \dot w(t)^2\,\mathrm{d}t\\+\frac{1}{2}\int_0^\infty\int_0^1 \dot k(x,t)^2\,\mathrm{d}x\,\mathrm{d}t+\int_0^\infty p(t)\bigg(\dot q(t)+F'(t)x_0^++(\beta-x_0^-) F_0'(t)\\-\int_0^t\dot q(s)\mathbf{1}{q(s)>0}F'(t-s)\,\mathrm{d}s -\dot w^0(F_0(t))F_0'(t)-\sigma\,\dot w(t)+\int_0^tF'(t-s)\sigma\,\dot w(s)\,\mathrm{d}s\\-\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu\,\mathrm{d}s\bigg)\,\mathrm{d}t+r\int_0^1\dot w^0(x)\, \mathrm{d}x+\int_0^\infty\tilde p(t)\int_0^1\dot k(x,t)\,\mathrm{d}x\,\mathrm{d}t\Bigg).\end{multline}
Minimising in (4.6) yields, with
$(\dot{\hat{ w}}^{0}(t),\dot{\hat w}(t), \dot{\hat{k}}(x,t))$
being optimal,
\begin{align*} \dot{\hat w}^0(x)-p(F_0^{-1}(x))+r=0,\\\dot{\hat w}(t)-\sigma p(t)+\sigma\int_0^\infty p(t+s)F'(s)\,\mathrm{d}s=0,\\\dot{\hat k}(x,t)-p\bigg(\frac{t}{\mu}+F^{-1}(x)\bigg)+\tilde p(t)=0.\end{align*}
For the latter, note that
\begin{multline*} \int_0^\infty p(t)\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu\,\mathrm{d}s\,\mathrm{d}t=\int_0^\infty\int_s^\infty p(t)\dot k(F(s),\mu (t-s))F'(s)\,\mu\,\mathrm{d}t\, \mathrm{d}s\\=\int_0^\infty\int_0^\infty p(t+s)\dot k(F(s),\mu t)F'(s)\,\mu\,\mathrm{d}t\,\mathrm{d}s=\int_0^\infty\int_0^1 p\bigg(\frac{t}{\mu}+F^{-1}(x)\bigg)\dot k( x,t)\,\mathrm{d}x\,\mathrm{d}t.\end{multline*}
Hence,
\begin{multline*} \!\!I_{x_0}^Q(q)={\sup}_{(p,\tilde p,r)\in \mathbb L_2(\mathbb{R}_+)^2\times\mathbb{R}}\bigg(\!\int_0^\infty\!\! p(t)\bigg(\dot q(t)-\!\!\int_0^t\!\!\dot q(s)\mathbf{1}{q(s)>0}F'(t-s)\,\mathrm{d}s+(\beta-x_0^-) F_0'(t)\!\bigg)\mathrm{d}t\\-\frac{1}{2}\bigg(\int_0^1(p(F_0^{-1}(x))-r)^2\,\mathrm{d}x+\sigma^2\int_0^\infty\bigg( p(t)-\int_0^\infty p(t+s)F'(s)\,\mathrm{d}s\bigg)^2\,\mathrm{d}t\\+\int_0^\infty\int_0^1 \bigg(p\bigg(\frac{t}{\mu}+F^{-1}(x)\bigg)-\tilde p(t))^2\,\mathrm{d}x\,\mathrm{d}t\bigg)\bigg).\end{multline*}
Given p, the optimal r is
$\hat r=\int_0^1 p(F_0^{-1}(x))\,\mathrm{d}x$
and the optimal
$\tilde p(t)$
is
$\hat{\tilde p}(t)=\int_0^1p(t/\mu+F^{-1}(x))\,\mathrm{d}x$
. (As a by-product,
$\hat w^0$
and
$\hat k$
satisfy the constraints in (4.4) and (4.5).) Therefore,
\begin{multline} I_{x_0}^Q(q)=\sup_{p\in \mathbb L_2(\mathbb{R}_+)}\bigg(\int_0^\infty p(t)\bigg(\dot q(t)-\int_0^t\dot q(s)\mathbf{1}{q(s)>0}F'(t-s)\,\mathrm{d}s+(\beta-x_0^-) F_0'(t)\bigg)\,\mathrm{d}t\\-\frac{1}{2}\bigg(\int_0^1(p(F_0^{-1}(x))-\int_0^1 p(F_0^{-1}(\tilde x))\,\mathrm{d}\tilde x)^2\,\mathrm{d}x+\sigma^2\int_0^\infty\bigg( p(t)-\int_0^\infty p(t+s)F'(s)\,\mathrm{d}s\bigg)^2\,\mathrm{d}t\\+\int_0^\infty\int_0^1 \bigg(p\bigg(\frac{t}{\mu}+F^{-1}(x)\bigg)-\int_0^1p\bigg(\frac{t}{\mu}+F^{-1}(\tilde x)\bigg)\,\mathrm{d}\tilde x\bigg)^2\,\mathrm{d}x\,\mathrm{d}t\bigg)\bigg).\end{multline}
As the function in the
$\sup$
is strictly concave in p, a maximiser in (4.7) is specified uniquely; see, e.g., Proposition II.1.2 in [5]. The existence of the maximiser follows from Proposition III.5.2 in [5].
The function in the
$\sup$
can be simplified if one notes that
\begin{multline*} \int_0^1(p(F_0^{-1}(x))-\int_0^1 p(F_0^{-1}(\tilde x))\,\mathrm{d}\tilde x)^2\,\mathrm{d}x=\int_0^\infty p(s)^2F_0'(s)\,\mathrm{d}s-\bigg(\int_0^\infty p(s)F'_0(s)\,\mathrm{d} s\bigg)^2,\end{multline*}
\begin{multline*} \int_0^1 \bigg(p\bigg(\frac{t}{\mu}+F^{-1}(x)\bigg)-\int_0^1p\bigg(\frac{t}{\mu}+F^{-1}(\tilde x)\bigg)\,\mathrm{d}\tilde x\bigg)^2\,\mathrm{d}x\\=\int_0^\infty p\bigg(\frac{t}{\mu}+s\bigg)^2F'(s)\,\mathrm{d}s-\bigg(\int_0^\infty p\bigg(\frac{t}{\mu}+s\bigg)F'(s)\,\mathrm{d} s\bigg)^2,\end{multline*}
and that
\begin{align*} \int_0^\infty p(s)^2F_0'(s)\,\mathrm{d}s+\mu\int_0^\infty\int_0^\infty p(t+s)^2\,F'(s)ds\,\mathrm{d}t=\mu\int_0^\infty p(s)^2\,\mathrm{d}s.\end{align*}
As a result,
\begin{multline} I_{x_0}^Q(q)=\sup_{p\in\mathbb L_2(\mathbb{R}_+)}\bigg(\int_0^\infty p(t)\bigg(\dot q(t)-\int_0^t\dot q(s)\mathbf{1}{q(s)>0}F'(t-s)\,\mathrm{d}s+(\beta-x_0^-) F_0'(t)\bigg)\,\mathrm{d}t\\-\frac{1}{2}\bigg(\mu\int_0^\infty p(s)^2\,\mathrm{d}s-\bigg(\int_0^\infty p(s)F'_0(s)\,\mathrm{d} s\bigg)^2+\sigma^2\int_0^\infty\bigg(p(t)-\int_0^\infty p(t+s)F'(s)\,\mathrm{d}s\bigg)^2\,\mathrm{d}t\\-\mu\int_0^\infty\bigg(\int_0^\infty p(t+s)F'(s)\,\mathrm{d} s\bigg)^2\,\mathrm{d}t\bigg)\bigg).\end{multline}
Varying p in (4.8) implies (4.1). As the maximiser in (4.8) is unique, so is an
$\mathbb L_2(\mathbb{R}_+)$
solution of the Fredholm equation (4.1).
It is noteworthy that the integral operator on
$\mathbb L_2(\mathbb{R}_+)$
that appears on the right-hand side of (4.1) is not generally either Hilbert–Schmidt or compact, so the existence and uniqueness of
$\hat p(t)$
is not a direct consequence of the general theory. Solving Fredholm equations numerically, such as (4.1), is discussed at quite some length in the literature. For instance, the collocation method with a basis of ‘hat’ functions could be tried: for
$i\in\mathbb{N}$
and
$n\in\mathbb{N}$
, let
$t_i=i/n$
and
$\ell_i(t)=(1-\lvert t-t_i\rvert)\mathbf{1}{t_{i-1}\le t\le t_i}$
, with
$t_0=0$
. Then an approximate solution is
\begin{align*} p_n(t)=\sum_{i=1}^{n^2}p_n(t_i)\ell_i(t)\end{align*}
where the
$p_n(t_i), i=1,\ldots,n^2$
, satisfy the linear system
\begin{multline*} (\mu+\sigma^2)p_n(t_i)-\sigma^2\sum_{j=1}^{n^2}p_n(t_j)\int_0^{n^2} \tilde K(t_i,s)\ell_j(s)\,\mathrm{d}s\\= \dot q(t_i)-\int_0^{t_i}\dot q(s)\mathbf{1}{q(s)>0}F'(t_i-s)\,\mathrm{d}s+(\beta-x_0^-) F_0'(t_i),\end{multline*}
with
$\tilde K$
denoting the kernel of the integral operator in (4.1). For more background, see [3] and references therein.
Evaluating
$\overline I^Q_{q_0,x_0}$
is done similarly:
\begin{multline*} \overline I^Q_{q_0,x_0}(q)=\sup_{(p,\tilde p,r)\in \mathbb{L}_2(\mathbb{R}_+)^2\times \mathbb{R}}\; \inf_{ \substack{(\dot w^0,\dot w,\dot k)\in \mathbb{L}_2([0,1])\\\times\mathbb L_2( \mathbb{R}_+)\times \mathbb L_2([0,1]\times\mathbb{R}_+)}}\bigg(\frac{1}{2}\int_0^1\dot w^0(x)^2\,\mathrm{d}x+\frac{1}{2}\int_0^\infty \dot w(t)^2\,\mathrm{d}t\\+\frac{1}{2}\int_0^\infty\int_0^1 \dot k(x,t)^2\,\mathrm{d}x\,\mathrm{d}t+\int_0^\infty p(t)\bigg(\dot q(t) +x_0 F_0'(t) -\sqrt{q_0}\dot w^0(F_0(t))F_0'(t) -\sigma\,\dot w(t)\\ +\int_0^tF'(t-s)\sigma\,\dot w(s)\,\mathrm{d}s -\int_0^t\dot k(F(s),\lambda (t-s))F'(s)\lambda\, \mathrm{d}s\bigg)\,\mathrm{d}t+r\int_0^1\dot w^0(x)\,\mathrm{d}x\\+\int_0^\infty\tilde p(t)\int_0^1\dot k(x,t)\,\mathrm{d}x\,\mathrm{d}t\bigg).\end{multline*}
Calculations as previously imply that
\begin{multline*} \overline I^Q_{q_0,x_0}(q)=\sup_{p\in \mathbb{L}_2(\mathbb{R}_+)}\bigg(\int_0^\infty p(t)(\dot q(t) +x_0 F_0'(t))\,\mathrm{d}t-\frac{1}{2}\bigg(q_0\int_0^\infty p(s)^2F_0'(s)\,\mathrm{d}s\\-q_0\bigg(\int_0^\infty p(s)F'_0(s)\,\mathrm{d} s\bigg)^2+\lambda\int_0^\infty p(t+s)^2F'(s)\,\mathrm{d}s-\lambda\bigg(\int_0^\infty p(t+s)F'(s)\,\mathrm{d} s\bigg)^2\\+\sigma^2\int_0^\infty\bigg( p(t)-\int_0^\infty p(t+s)F'(s)\,\mathrm{d}s\bigg)^2\,\mathrm{d}t\bigg)\bigg)=\frac{1}{2}\int_0^\infty\overline p(t)(\dot q(t) +x_0 F_0'(t))\,\mathrm{d}t,\end{multline*}
with
$\overline p(t)$
being the
$\mathbb L_2(\mathbb{R}_+)$
solution p(t) to the Fredholm equation of the second kind
\begin{align*} (\lambda+\sigma^2)p(t)=\dot q(t)+x_0F^{\prime}_0(t)+q_0F^{\prime}_0(t)\int_0^{\infty}F^{\prime}_0(s)p(s)\,\mathrm{d}s+\sigma^{2}\int_0^\infty F^{\prime}(\lvert s-t\rvert)p(s)\,\mathrm{d}s\\+(\lambda-\sigma^2)\int_0^\infty \int_0^{s\wedge t}F'(s-\tilde s)F^{\prime}(t-\tilde s)\,\mathrm{d}\tilde s\, p(s)\,\mathrm{d}s.\end{align*}
Competing interests
There were no competing interests to declare which arose during the preparation or publication process of this article.
