An SPDE model for systemic risk with endogenous contagion

Abstract

We propose a dynamic mean-field model for ‘systemic risk’ in large financial systems, derived from a system of interacting diffusions on the positive half-line with an absorbing boundary at the origin. These diffusions represent the distances-to-default of financial institutions, and absorption at zero corresponds to default. As a way of modelling correlated exposures and herd behaviour, we consider a common source of noise and a form of mean-reversion in the drift. Moreover, we introduce an endogenous contagion mechanism whereby the default of one institution causes a drop in the distances-to-default of the other institutions. In this way, we aim to capture key ‘system-wide’ effects on risk. The resulting mean-field limit is characterised uniquely by a nonlinear SPDE on the half-line with a Dirichlet boundary condition. The density of this SPDE gives the conditional law of a non-standard ‘conditional’ McKean–Vlasov diffusion, for which we provide a novel upper Dirichlet heat kernel type estimate. Depending on the realisations of the common noise and the rate of mean-reversion, the SPDE can exhibit rapid accelerations in the loss of mass at the boundary. In other words, the contagion mechanism can give rise to periods of significant systemic default clustering.

Appendix

1.1 A.1 Some technical lemmas

Lemma A.1

Suppose$\nu$is a measure-valued process satisfying the conditions in Definition2.3and let$g_{s}(x)=g(s,x,\nu _{s},L_{s})$, where$g$is any of$b$, $\mathfrak{b}$, $\sigma$or$\sigma ^{2}$, with$L_{s}=1-\nu _{s}(0,\infty )$. Define the error term

$$ \mathcal{E}_{t,\varepsilon }^{g}(x):=\left \langle \nu _{t},g_{t}( \cdot )G_{\varepsilon }(x,\cdot )\right \rangle -g_{t}(x)(\mathcal{T} _{\varepsilon }\nu _{t})(x). $$

Then we have

$$ \mathbb{E}\bigg[\int _{0}^{T}\Vert \mathcal{E}_{t,\varepsilon }^{g} \Vert _{L^{2}(\mathbb{R})}^{2}\,dt\bigg]\longrightarrow 0 \qquad \textit{as } \varepsilon \downarrow 0. $$

Proof

This follows by a simple modification of Lemma 8.1 in [35]. The only thing to note is that the crude bound $|g_{t}(x)-g_{t}(y)| \leq 2\Vert g_{t}\Vert _{\infty }$ no longer holds, as $g_{t}$ need not be bounded. However, the arguments from [35] still apply if we instead rely on $|g_{t}(x)-g_{t}(y)|\leq \Vert \partial _{x}g_{t} \Vert _{\infty }|x-y|$. □

Lemma A.2

Suppose$\bar{\nu }$satisfies the whole-space analogues of conditions (iii), (iv) in Definition 2.3. Let$g_{s}=g(s,\cdot ,\nu _{s},L_{s})$, where$g$is any of$b$, $\mathfrak{b}$, $\sigma$or$\sigma ^{2}$, and define the error terms

$$\begin{aligned} \bar{\mathcal{E}}_{t,\varepsilon }^{g}(x) & :=\left \langle \bar{ \nu }_{t},g_{t}(\cdot )p_{\varepsilon }(x,\cdot )\right \rangle -g _{t}(x)\partial _{x}(\bar{\mathcal{T}}_{\varepsilon }\bar{\nu }_{t})(x)+ \partial _{x}g_{t}(x)\bar{\mathcal{H}}_{t,\varepsilon }^{g}(x), \\ \bar{\mathcal{H}}_{t,\varepsilon }^{g}(x) & :=\left \langle \bar{ \nu }_{t},(x-\cdot )\partial _{x}p_{\varepsilon }(x-\cdot )\right \rangle . \end{aligned}$$

Then for$k=1,2$, we have

$$ \mathbb{E}\bigg[\int _{0}^{T}\Vert \bar{\mathcal{E}}_{t,\varepsilon } ^{g}\Vert _{L^{2}(\mathbb{R})}^{2k}\,dt\bigg] \longrightarrow 0,\quad \mathbb{E}\bigg[\int _{0}^{T}\Vert x\bar{\mathcal{E}}_{t,\varepsilon } ^{g}\Vert _{L^{2}(\mathbb{R})}^{2}\,dt\bigg]\longrightarrow 0, \qquad \textit{as }\varepsilon \downarrow 0. $$

Proof

This is immediate from the obvious modifications of [35, Lemma 8.2]. □

Lemma A.3

Fix any positive integer$k \geq 1$. Let$(\nu _{t}^{{N}})$be as defined in (2.1) and let$(\tilde{\nu }_{t})$be any measure-valued process satisfying Definition2.3. Then for every$a>0$, we have

$$ \mathbb{E}\bigg[\int _{0}^{T}\langle \tilde{\nu }_{t},x^{k}\mathbf{1} _{[\lambda ,\infty )}(x)\rangle\,dt\bigg]=o(\lambda ^{k}e^{-a\lambda }) \qquad \textit{as }\lambda \rightarrow \infty , $$

(A.1)

and likewise, it holds uniformly in$N\geq 1$and$t\in [0,T]$that

$$ \mathbb{E}[\langle \nu _{t}^{{N}},x^{k}\mathbf{1}_{[\lambda ,\infty )}(x) \rangle ] =o(\lambda ^{k}e^{-a\lambda }) \qquad \textit{as }\lambda \rightarrow \infty . $$

(A.2)

Proof

Fix an arbitrary $a>0$ and let $\varepsilon >0$ be given. By Corollary 3.4, there exists $\lambda _{0}\geq 0$ such that

$$ \mathbb{E}[\nu _{t}^{{N}}(\lambda ,\infty )]\leq \varepsilon e^{-a \lambda }, \qquad \forall \lambda \geq \lambda _{0}, $$

(A.3)

uniformly in $N\geq 1$ and $t\in [0,T\text{]}$. Given any $\lambda \geq \lambda _{0}$, we let $\{ s_{0},s_{1},\ldots \}$ denote the partition of $[\lambda , \infty )$ with $s_{0} = \lambda$ and $s_{i} - s_{i-1} = 1/a$. By (A.3) and monotone convergence, we get

$$\begin{aligned} \mathbb{E}[\langle \nu _{t}^{{N}},x\mathbf{1}_{[\lambda ,\infty )} \rangle ] & \leq \sum _{i=1}^{\infty }s_{i}\mathbb{E}[\nu _{t}^{{N}}(s _{i-1},\infty )]\leq \varepsilon \sum _{i=1}^{\infty }s_{i}e^{-as_{i-1}}= \varepsilon e\sum _{i=1}^{\infty }s_{i}e^{-as_{i}}. \end{aligned}$$

Now $x\mapsto xe^{-ax}$ is decreasing for $x\geq 1/a$; so taking $\lambda _{0}^{\prime }:=\max \{ 1/a,\lambda _{0}\}$ and noting that $s_{0}=\lambda$, it holds for all $\lambda >\lambda _{0}^{\prime }$ that

$$\begin{aligned} \mathbb{E}[\langle \nu _{t}^{{N}},x\mathbf{1}_{[\lambda ,\infty )} \rangle ] & \leq \varepsilon e\int _{\lambda }^{\infty }xe^{-ax}\,dx= \varepsilon e\bigg(\frac{\lambda }{a}+\frac{1}{a^{2}}\bigg)e^{-a \lambda }. \end{aligned}$$

This proves (A.2) for $k=1$, and the work for $k\geq 2$ is analogous. The claim in (A.1) follows similarly by relying on the exponential tail property from Definition 2.3. □

Lemma A.4

Let$\nu$be a limit point from Theorem 2.4and$\tilde{\nu }$any measure-valued process satisfying Definition2.3. Then almost surely (as$\lambda \rightarrow \infty$),

$$ \lim _{\varepsilon \downarrow 0}\frac{\nu _{t}(0,\varepsilon )}{\varepsilon }=0, \qquad \nu _{t}(\lambda ,\infty )=O(e^{-\lambda }), \qquad \int _{0}^{T}\tilde{\nu }_{t}(\lambda ,\infty )\,dt=O(e^{-\lambda }). $$

Proof

For the first claim, recall from Proposition 4.3 that there exist $\delta \in (0,1]$ and $\beta >0$ such that $\mathbb{E}[\nu _{t}(0,\varepsilon )]=t^{-\frac{ \delta }{2}}O(\varepsilon ^{1+\beta })$ as $\varepsilon \rightarrow 0$. Using Markov’s inequality, we thus deduce that for any $\theta >0$ and $n$ sufficiently large,

$$ \mathbb{P}[n^{2/\beta }\nu _{t}(0,n^{-2/\beta })>\theta ]\leq \theta ^{-1}n ^{2/\beta }\mathbb{E}[\nu _{t}(0,n^{-2/\beta }]\leq Ct^{-\frac{ \delta }{2}}\theta ^{-1}n^{-2}. $$

Hence the Borel–Cantelli lemma gives $\limsup _{n}n^{2/\beta }\nu _{t}(0,n ^{-2/\beta })=0$ almost surely. Now, given $\varepsilon >0$, we have $(n+1)^{-2/\beta }<\varepsilon \leq n ^{-2/\beta }$ for some $n\geq 1$; so we deduce that

$$ \limsup _{\varepsilon \downarrow 0}\frac{\nu _{t}(0,\varepsilon )}{ \varepsilon }\leq \limsup _{n\to \infty }\frac{\nu _{t}(0,n^{-2/\beta })}{(n+1)^{-2/ \beta }}\leq \limsup _{n\to \infty }\frac{\nu _{t}(0,n^{-2/\beta })}{n ^{-2/\beta }}\bigg(\frac{n+1}{n}\bigg)^{\frac{2}{\beta }}. $$

Since the latter is zero with probability 1, this proves the first claim. The two remaining results follow by analogous considerations for the tail probabilities, using the exponential tail properties from Proposition 4.3 and Definition 2.3. □

1.2 A.2 Proofs of Propositions 3.5 and 3.6

As an immediate consequence of Corollary 6.2, we have the following result.

Lemma A.5

Let$\varLambda _{t}^{i,{N}}=|X_{t}^{i}|+{ \sum _{j=1}^{{N}}}a_{j}^{{N}}|X _{t}^{j}|$as in (6.1). Then

$$ \mathbb{P}\bigg[\sup _{t\leq T}\varLambda _{t}^{i,{N}}\geq \lambda \bigg]=o(1) \qquad \textit{as }\lambda \rightarrow \infty , $$

uniformly in$N\geq 1$and$i=1,\ldots ,N$.

Based on this lemma, we can adapt the arguments from [35, Sect. 4] to prove Propositions 3.5 and 3.6.

Proof of Proposition 3.5

Arguing as in the proof of [35, Proposition 4.6], for all $a>0$ and $\theta :=\frac{1}{2}(1-r)$, we have

$$ \mathbb{P}[L_{t+h}^{{N}}-L_{t}^{{N}}< \delta ,L_{t}^{{N}}< r]\leq \mathbb{P}[L_{t+h}^{{N}}-L_{t}^{{N}}< \delta ,\;\nu _{t}^{{N}}(0,a)> \theta ]+o(e^{-a}). $$

Let $E:=\{ L_{t+h}^{{N}}-L_{t}^{{N}}<\delta ,\;\nu _{t}^{{N}}(0,a)> \theta \}$ and define a random index set ℐ by

$$ \mathcal{I}:=\{1\leq i\leq N:X_{t}^{i}< a,\;t< \tau _{i},\;a_{i}^{{N}}> \theta /2N\}. $$

Note that the particles with index in $\{ 1\leq i\leq N:a_{i}^{{N}} \leq { \frac{1}{2}}\theta N^{-1}\}$ can contribute at most ${ \frac{1}{2}}\theta$ towards $\nu _{t}^{{N}}(0,a)$. Recalling also that $a_{i}^{{N}}\leq m/N$ for some $m>0$, see (2.2), it follows that on the event $E$, we must have $|\mathcal{I}|>N\theta /2m$. Hence

$$ \mathbb{P}[E]\leq \sum _{\mathcal{I}_{0}:\left |\mathcal{I}_{0}\right |> \frac{\theta N}{2m}}\mathbb{P}[E\,|\,\mathcal{I}=\mathcal{I}_{0}] \mathbb{P}[\mathcal{I}=\mathcal{I}_{0}]. $$

(A.4)

Furthermore, since $L_{t+h}^{{N}}-L_{t}^{{N}}<\delta$ on $E$ while $a_{i}^{{N}}>\theta /2N$ for $i\in \mathcal{I}$, we deduce that

$$ \mathbb{P}[E\,|\,\mathcal{I}=\mathcal{I}_{0}]\leq \mathbb{P}\big[ \bigl|\bigl\{ i\in \mathcal{I}_{0}:\inf _{u\leq h}X_{t+u}^{i}\leq 0 \bigr\} \bigr|< 2\delta N/\theta \big| \mathcal{I}=\mathcal{I}_{0} \big]. $$

(A.5)

In order to estimate (A.5), we let $Z_{t}^{i}=\varUpsilon _{t}(X_{t}^{i})$ as in Lemma 6.3 and recall that $(Z_{t}^{i})$ then satisfies

$$ dZ_{t}^{i}=\hat{b}_{t}^{i}\,dt+dB_{t}^{i}, \qquad |\hat{b}_{t}^{i}|\leq c(1+\varLambda _{t}^{i,{N}}), \qquad |Z_{t}^{i}|\leq c'(1+|X_{t}^{i}|). $$

(A.6)

Using that $Z_{t}^{i}\leq 0$ if and only if $X_{t}^{i}\leq 0$, the estimate in (A.5) implies

$$ \mathbb{P}[E\,|\,\mathcal{I}=\mathcal{I}_{0}] \leq \mathbb{P}\big[ \bigl|\bigl\{ i\in \mathcal{I}:\inf _{u\leq h}Z_{t+u}^{i}\leq 0\bigr\} \bigr|< 2\delta N/\theta \big|\mathcal{I}=\mathcal{I}_{0}\big]. $$

Moreover, it is immediate from (A.6) that

$$ Z_{t+u}^{i}\leq Z_{t}^{i}+c_{1}h(1+\tilde{\varLambda }_{h}^{i,{N}})+ \tilde{B}_{u}^{i}, \qquad \tilde{B}_{u}^{i}:=B_{t+u}^{i}-B_{t}^{i}, \qquad \tilde{\varLambda }_{h}^{i,{N}}:=\sup _{u\leq h}\varLambda _{t+u}^{i,{N}}. $$

In particular, if $\inf _{u\leq h}\tilde{B}_{u}^{i}\leq -Z_{t}^{i}-ch(1+ \tilde{\varLambda }_{h}^{i,{N}})$, then $\inf _{u\leq h}Z_{t+u}^{i}\leq 0$ and hence

$$ \mathbb{P}[E\,|\,\mathcal{I}=\mathcal{I}_{0}]\leq \mathbb{P}\bigg[ \bigl|\bigl\{ i\in \mathcal{I}:\inf _{u\leq h}\tilde{B}_{u}^{i}\leq -Z _{t}^{i}-ch(1+\tilde{\varLambda }_{h}^{i,{N}})\bigr\} \bigr|< { \frac{2 \delta }{\theta }} N \bigg|\mathcal{I}=\mathcal{I}_{0}\bigg]. $$

Using the bound $|Z_{t}^{i}| \leq c' |X_{t}^{i}|$ from (A.6), the definition of ℐ implies that there is a constant $\tilde{c}>0$ such that

$$ \mathbb{P}[E\,|\,\mathcal{I}=\mathcal{I}_{0}]\leq \mathbb{P}\bigg[ \bigl|\bigl\{ i\in \mathcal{I}:\inf _{u\leq h}\tilde{B}_{u}^{i}\leq - \tilde{c}a-\tilde{c}(1+\tilde{\varLambda }_{h}^{i,{N}})\bigr\} \bigr|< { \frac{2\delta }{\theta }} N \bigg| \mathcal{I}=\mathcal{I}_{0} \bigg]. $$

Recalling the fact that

$$ \tilde{B}_{u}^{i}=\int _{t}^{t+u}\rho _{s}\,dW_{s}^{0}+\int _{t}^{t+u}\sqrt{1- \rho _{s}^{2}}\,dW_{s}^{i}=:I_{u}+J_{u}^{i}, $$

we split the above probability on $\{\sup _{u\leq h}|I_{u}|<\tilde{c}a \}\cap \{\sup _{u\leq h}\varLambda _{t+u}^{i,{N}}<a\}$ and its complement. In this way, we get

$$\begin{aligned} \mathbb{P}[E \,|\,\mathcal{I}=\mathcal{I}_{0}] &\leq \mathbb{P}\bigg[ \bigl|\bigl\{ i\in \mathcal{I}:\inf _{u\leq h}J_{u}^{i}\leq -3 \tilde{c}a-\tilde{c}\bigr\} \bigr|< { \frac{2\delta }{\theta }} N \bigg|\mathcal{I}=\mathcal{I}_{0}\bigg] \\ & \phantom{=:}+\mathbb{P}\bigg[\,\sup _{u\leq h}\left |I_{u}\right | \geq \tilde{c}a\bigg]+\mathbb{P}\bigg[\,\sup _{u\leq h}\varLambda _{t+u} ^{i,{N}}\geq a\bigg]. \end{aligned}$$

From Lemma A.5, we know that the last term is $o(1)$ as $a\rightarrow \infty$ uniformly in $N\geq 1$. Similarly, $(I_{u})$ is a martingale, and hence the second term is also $o(1)$ as $a\rightarrow \infty$ by Doob’s maximal inequality. Concerning the first term, we can introduce a time-change to make each $J^{i}$ an independent Brownian motion. Recalling (A.4), the proof can then be completed via a law of large numbers argument as in [35, Proposition 4.6] by choosing the free parameter $a$ as a suitable function of $\delta$. □

Proof of Proposition 3.6

Arguing as in [35, Proposition 4.7], it suffices to show that

$$ \lim _{\delta \rightarrow 0}\lim _{N\rightarrow \infty }\mathbb{P}[L _{t+\delta }^{{N}}-L_{t}^{{N}}\geq \eta ,\;\nu _{t}^{{N}}(0,\varepsilon )< \eta /2]=0. $$

Let $E:=\{ L_{t+\delta }^{{N}}-L_{t}^{{N}}\geq \eta ,\;\nu _{t}^{{N}}(0, \varepsilon )<\eta /2\}$ and define the random index set

$$ \mathcal{I}:=\{ 1\leq i\leq N:X_{t}^{i}\geq \varepsilon \text{ or } t \geq \tau _{i}\} . $$

By (2.2), there exists $m$ such that $a_{i}^{{N}}\leq m/N$; so noting that $\{ \nu _{t}^{{N}}(0,\varepsilon )<\eta /2\}$ is contained in $\{ |\mathcal{I}|\geq \frac{N}{m}(1-\frac{ \eta }{2})\}$, we have

$$ \mathbb{P}[E]\leq \sum _{\mathcal{I}_{0}:|\mathcal{I}_{0}|\geq \frac{N}{m}(1-\frac{ \eta }{2})}\mathbb{P}[E\,|\,\mathcal{I}=\mathcal{I}_{0}]\mathbb{P}[ \mathcal{I}=\mathcal{I}_{0}]. $$

The conditional probabilities can be estimated by

$$\begin{aligned} \mathbb{P}[E \,|\,\mathcal{I}=\mathcal{I}_{0}] &\leq \mathbb{P}\bigg[ \bigl|\bigl\{ i\in \mathcal{I}_{0}:\inf _{s\in [t,t+\delta ]}X_{s} ^{i}\leq 0,\quad X_{t}^{i}\geq \varepsilon \bigr\} \bigr|\geq \frac{N \eta }{2m}\bigg|\mathcal{I}=\mathcal{I}_{0}\bigg] \\ & \leq \mathbb{P}\bigg[\bigl|\bigl\{ i\in \mathcal{I}_{0}: \inf _{s\in [t,t+\delta ]}(X_{s}^{i}-X_{t}^{i})\leq -\varepsilon \bigr\} \bigr|\geq \frac{N\eta }{2m}\bigg|\mathcal{I}=\mathcal{I}_{0} \bigg]. \end{aligned}$$

(A.7)

Using the scale transform $\varUpsilon$ from Lemma 6.3, we let $U_{s}^{i}:=\varUpsilon _{t+s}(X_{t+s}^{i})-\varUpsilon _{t+s}(X_{t}^{i})$ and note that as in Lemma 6.3,

$$\begin{aligned} dU_{s}^{i}= u_{s}^{i}\,ds+\rho _{t+s}\,dW_{t+s}^{0}+(1-\rho _{t+s}^{2})^{ \frac{1}{2}}\,dW_{t+s}^{i}=:u_{s}^{i}\,ds+dI_{s}+dJ_{s}^{i}, \end{aligned}$$

(A.8)

where the drift satisfies $|u_{s}^{i}|\leq c_{1}(1+\varLambda _{t+s}^{i, {N}})$. By the construction of $\varUpsilon$ and the boundedness from below of $1/\sigma$, say $1/\sigma \geq c_{2}>0$, it follows from (A.7) that

$$ \mathbb{P}[E \,|\,\mathcal{I}=\mathcal{I}_{0}]\leq \mathbb{P}\bigg[ \bigl|\bigl\{ i\in \mathcal{I}_{0}:\inf _{s\leq \delta }U_{s}^{i} \leq -c_{2}\varepsilon \bigr\} \bigr|\geq \frac{N\eta }{2m}\bigg| \mathcal{I}=\mathcal{I}_{0}\bigg]. $$

Given $\delta >0$, fix a constant $a=a(\delta )>0$ to be specified later. Using the decomposition (A.8) of $U_{s}^{i}$ and the growth estimate for its drift $u_{s}^{i}$, we see that on the event

$$ \bigg\{ \sup _{s\leq \delta }\varLambda _{t+s}^{i,{N}}< a/\delta \bigg\} \cap \bigg\{ \sup _{s\leq \delta }|I_{s}|< a\bigg\} , $$

(A.9)

if $i\in \mathcal{I}_{0}$ is such that $\inf _{s\leq \delta }U_{s}^{i} \leq -c_{2}\varepsilon$, then

$$\begin{aligned} \inf _{s\leq \delta }J_{s}^{i} & \leq -c_{2}\varepsilon +\delta c_{1} \bigg(1+\sup _{s\leq \delta }\varLambda _{t+s}^{i,{N}}\bigg)+ \sup _{s\leq \delta }\left |I_{s}\right | \\ & \leq -c_{2}\varepsilon +c_{1}\delta +(1+c_{1})a\:\leq -c_{3}( \varepsilon -\delta -a). \end{aligned}$$

Consequently, splitting up the desired probability on the event (A.9) and its complement, we get

$$\begin{aligned} \mathbb{P}[E \,|\,\mathcal{I}=\mathcal{I}_{0}] & \leq \mathbb{P} \bigg[\bigl|\bigl\{ i\in \mathcal{I}_{0}:\inf _{s\leq \delta }J_{s} ^{i}\leq -c_{3}(\varepsilon -\delta -a)\bigr\} \bigr|\geq \frac{N \eta }{2m}\bigg|\mathcal{I}=\mathcal{I}_{0}\bigg] \\ & \phantom{=:}+\mathbb{P}\bigg[\,\sup _{s\leq \delta }\varLambda _{t+s} ^{i,{N}}\geq a/\delta \bigg]+\mathbb{P}\bigg[\,\sup _{s\leq \delta } \left |I_{s}\right |\geq a\bigg]. \end{aligned}$$

By Doob’s maximal inequality, the last term is bounded by $\delta a ^{-2}$; so choosing $a=a(\delta )$ such that $\delta a^{-2} \rightarrow 0$ as $\delta \rightarrow 0$, we can ensure that it vanishes as $\delta \rightarrow 0$. Moreover, requiring also that $\delta /a \rightarrow 0$ as $\delta \rightarrow 0$, it follows from Lemma A.5 that the second term vanishes (uniformly in $N$) as $\delta \rightarrow 0$. These requirements are satisfied by $a=a(\delta ):=\delta ^{1/2}\log \log (1/\delta )$, and with this choice for $a$, the proof can now be finished by the same arguments as in [35, Proposition 4.7]. □