Production and financial linkages in inter-firm networks: structural variety, risk-sharing and resilience

Abstract

The paper analyzes how (production and financial) inter-firm networks can affect firms’ default probabilities and observed default rates. A simple theoretical model of shock transfer is built to investigate some stylized facts on how firm-idiosyncratic shocks are allocated in the network, and how this allocation changes firm default probabilities. The model shows that the network works as a perfect “risk-pooling” mechanism, when it is both strongly connected and symmetric. But the “risk-sharing” does not necessarily reduce default rates, unless the shock firms face is lower on average than their financial capacity. Conceived as cases of symmetric inter-firm networks, industrial districts might have a comparative disadvantage in front of heavy crises.

Appendix

Proof of Proposition 1

When T is strongly connected, it is a standard result of the theory of Markov chains that aperiodicity is a necessary and sufficient condition for T to be convergent (e.g. Kemeny and Snell 1960). Moreover, when this happens, T is also primitive, i.e. T ^t has only strictly positive entries for some t ≥ 1 (e.g. Perkins 1961), and there is a unique (up to scale) left eigenvector s of T, corresponding to the unit eigenvalue, such that for any v:

$$ \lim\limits_{t \rightarrow \infty} \mathbf{T}^t \mathbf{v} = \mathbf{s}' \mathbf{v}. $$

Since T is convergent, $\mathbf{S} \equiv \lim_{t \rightarrow \infty} \mathbf{T}^t$ exists and hence:

$$ \mathbf{S} \mathbf{T} = \lim\limits_{t \rightarrow \infty} \mathbf{T}^t\, \mathbf{T} = \lim\limits_{t \rightarrow \infty} \mathbf{T}^t = \mathbf{S} $$

where each row of S is equal to s′.

It follows that:

$$ \hat{\mathbf{x}}' = \mathbf{x}'_0 \left ( \lim\limits_{t \rightarrow \infty} \mathbf{T}^t \right ) = \mathbf{x}'_0 \mathbf{S} = \mathbf{x}'_0 \left ( \begin{matrix} {ccc}\mathbf{s}' \cr \vdots \cr \mathbf{s}' \end{matrix} \right ) = \mathbf{s}' \left(\sum\limits_i x_{i0}\right). $$

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Proof of Proposition 2

A symmetric network implies T = T′ and therefore:

$$ \mathbf{S}' = \left(\lim\limits_{t \rightarrow \infty} \mathbf{T}^t\right)' = \lim\limits_{t \rightarrow \infty} \mathbf{T}^t = \mathbf{S} $$

i.e. S must be symmetric too (s _ij  = s _ji ). As in S by definition s _ji  = s _ii , the symmetry implies s _ii  = s _ij .

Moreover, since the sum by column of each row is one, it follows that:

$$ \sum\limits_{j=1}^n s_{ij} = n\ s_{ii} = 1 $$

for each i and all the elements of S are equal to 1/n. Hence:

$$ \hat{x}_i = \mathbf{x}'_0 \left ( \begin{matrix}{ccc} \frac{1}{n} \cr \vdots \cr \frac{1}{n} \end{matrix} \right ) = \frac{\sum_i x_{i0}}{n} = \bar x $$

for each i ∈ N.□

Proof of Proposition 3

Given that, in a connected-symmetric network $\hat x_i = \bar x_0$ and this variable is asymptotically normally distributed with variance σ ²/n and mean μ, when n gets larger it converges in probability toward μ. Therefore, we have:

$$ \lim\limits_{n \rightarrow \infty} \Pr ( \hat x_i > \theta_i ) = \left \{ \begin{array}{ll} 1 & \mbox{if } \theta_i > \mu \\ 0 & \mbox{if } \theta_i < \mu . \end{array} \right . $$

By contrast, since σ ² > 0 , there is always a ε > 0 such that $\epsilon < \Pr (x_{i0} > \theta_i) < 1 - \epsilon$ and this probability is so strictly bound between 0 and 1.□

Proof of Proposition 4

Assuming that θ _i are identically and independently distributed, and so are x _i0, the number of firm defaults follows a binomial distribution with expected value $n \Pr(\theta_i - x_{i0} < 0)$. The expected value of the default rate is, therefore, simply $\Pr(\theta_i - x_{i0} < 0)$.

For firms in a symmetric-connected network, the expected value of the binomial is instead: $n \Pr(\theta_i - \bar x < 0)$, with an expected default rate equals to $\Pr(\theta_i - \bar x < 0)$.

Given that $\bar x \stackrel{p}{\rightarrow} \mu$ we have that:

$$ \lim\limits_{n \rightarrow \infty} \Pr(\theta_i - \bar x < 0) = \Pr(\theta_i < \mu) $$

Hence, the expected default rate of firms when the number of firms gets large is higher (lower) in isolation than in a symmetric-connected network if $\Pr(\theta_i - x_{i0} < 0) > \Pr(\theta_i < \mu)$ ($\Pr(\theta_i - x_{i0} < 0) < \Pr(\theta_i < \mu)$).□

Proof of Proposition 5

For a common threshold (θ _i  = θ), we have :

$$ \lim\limits_{n \rightarrow \infty} \Pr(\bar x < \theta) = \left \{ \begin{array}{ll} 1 & \mbox{if } \theta > \mu \\ 0 & \mbox{if } \theta < \mu \end{array} \right . $$

while $\Pr(x_{i0} < \theta)$ remains strictly bound between 0 and 1.□

Proof of Proposition 6

The probability of default in a strongly connected asymmetric network for firm i is:

$$ \Pr ( \hat x_i > \theta_i ) = \Pr \left( s_i \sum\limits_i x_{i0} > \theta_i \right) = \Pr \left( s_i n \frac{\sum_i x_{i0}}{n} > \theta_i \right) = \Pr \left(\bar x > \frac{\theta_i}{s_i n}\right) $$

$\bar x$ converges in probability toward μ, therefore we have that $\Pr ( \hat x_i > \theta_i )$ tends to 1 if $\bar x > \frac{\theta_i}{s_i n}$ and 0 if instead $\bar x < \frac{\theta_i}{s_i n}$.

By contrast, since σ ² > 0 , there is always a ε > 0 such that $\epsilon < \Pr (x_{i0} > \theta_i) < 1 - \epsilon$ and the probability in this case is strictly bound between 0 and 1.□