On the Growth Rate of a Linear Stochastic Recursion with Markovian Dependence

Abstract

We consider the linear stochastic recursion $x_{i+1} = a_{i}x_{i}+b_{i}$ where the multipliers $a_i$ are random and have Markovian dependence given by the exponential of a standard Brownian motion and $b_{i}$ are i.i.d. positive random noise independent of $a_{i}$. Using large deviations theory we study the growth rates (Lyapunov exponents) of the positive integer moments $\lambda _q = \lim _{n\rightarrow \infty } \frac{1}{n} \log \mathbb {E}[(x_n)^q]$ with $q\in \mathbb {Z}_+$. We show that the Lyapunov exponents $\lambda _q$ exist, under appropriate scaling of the model parameters, and have non-analytic behavior manifested as a phase transition. We study the properties of the phase transition and the critical exponents using both analytic and numerical methods.

Appendices

Appendix 1: Solution of the Euler–Lagrange Equation

We present in this Appendix the solution of the Euler–Lagrange Eq. (69) for $h(y)$. It is useful to introduce the notation

$$\begin{aligned} V(h) = 2\beta \log (1 + e^h). \end{aligned}$$

(164)

In terms of this function, the Eq. (69) is written as

$$\begin{aligned} h''(y) = - V'(h(y)), \end{aligned}$$

(165)

which is analogous to Newton’s law for a particle moving in the potential $V(h)$. Written in this form, it is easy to check that the following combination is a constant of motion of the Eq. (69)

$$\begin{aligned} E = \frac{1}{2} (h'(y))^2 + V(h(y)) = V(h(1)). \end{aligned}$$

(166)

The value of the constant of motion was determined from the boundary condition (70) at $y=1$ as $E = V(h(1))$.

The relation (166) will be useful to express $h'(y)$ in terms of $h(y)$. In particular, we can use it to write

$$\begin{aligned} \frac{dy}{dh} = \frac{1}{\sqrt{2(E-V(h(y)))}} = \frac{1}{2\sqrt{\beta }} \frac{1}{\sqrt{\log \frac{1+e^{h(1)}}{1+e^h}}}. \end{aligned}$$

(167)

Integrating this relation over $h$ from $h(0)$ to $h(1)$ we get an equation for $h(1)$

$$\begin{aligned} 1 = \frac{1}{2\sqrt{\beta }} \int _{h(0)=\log \rho }^{h(1)} \frac{dx}{\sqrt{\log \frac{1+e^{h(1)}}{1+e^x}}}. \end{aligned}$$

(168)

Thus we can find $h(1)$ by solving the equation

$$\begin{aligned} F(h(1);\rho ) \equiv \int _{\log \rho }^{h(1)} \frac{dx}{\sqrt{ \log \frac{1+e^{h(1)}}{1+e^x}}} = 2\sqrt{\beta }, \end{aligned}$$

(169)

Once $h(1)$ is found, the complete shape of the function $h(y)$ can be determined by solving the equation

$$\begin{aligned} \int _{\log \rho }^{h(y)} \frac{dx}{\sqrt{ \log \frac{1+e^{h(1)}}{1+e^x}}} = 2\sqrt{\beta } y. \end{aligned}$$

(170)

As the next step in the solution of the variational problem we would like to compute the functional $\varLambda [f]$ corresponding to a solution $f(x)$ of the Euler–Lagrange equation (62) with boundary condition (63). If the Euler–Lagrange equation has a unique solution $f(y)$, then $\lambda (\rho ,\beta ) = \varLambda [f]$. However, if it has several solutions, as is the case around the phase transition, the Lyapunov exponent is given by the supremum of $\varLambda [f]$ over these multiple solutions

$$\begin{aligned} \lambda (\rho ,\beta ) = \text{ sup }_{f} \varLambda [f]. \end{aligned}$$

(171)

The main result is summarized in Proposition 5. We present here the proof of this result.

Proof

We start by expressing the functional $\varLambda [f]$ in a simpler form as

$$\begin{aligned} \varLambda [f] = \frac{1}{2} \log \rho \int _0^1 dx f(x) - \frac{1}{2} \int _0^1 dx f(x) \log \frac{f(x)}{1-f(x)} - \int _0^1 dx \log (1-f(x)). \end{aligned}$$

(172)

This is obtained by eliminating the double integral over $K(z,y)$ using the Euler–Lagrange equation (59). Multiplying this equation with $f(y)$ and integrating over $y$, this allows us to solve for the double integral. Substituting into (58) gives the result above.

There are three integrals appearing in $\varLambda [f]$. We will show next that their sum can be expressed in terms of $h(1)$ alone. We consider them in turn.

The first integral is related to $f'(0)$ as

$$\begin{aligned} f'(0) = 2\beta f(0)(1-f(0)) \int _0^1 dz f(z) = \frac{2\beta \rho }{(1+\rho )^2} \int _0^1 dz f(z). \end{aligned}$$

(173)

This is obtained by taking $y=0$ in (61). This is uniquely determined by $h(1)$, as can be seen from Eq. (166)

$$\begin{aligned} \frac{1}{2} (h'(y))^2 + V(h(y)) = V(h(1)) \end{aligned}$$

(174)

and thus $h'(y) = \sqrt{2(V(h(1)) - V(h(y))} = 2\sqrt{\beta } \sqrt{\log \frac{1+e^{h(1)}}{1+e^{h(y)}}}$. Taking here $y=0$ we can express $h'(0)$ in terms of $h(1)$. As a result we have

$$\begin{aligned} I_0 = \int _0^1 dx f(x) = \frac{1}{2\beta } h'(0) = \frac{1}{\sqrt{\beta }} \sqrt{\log \frac{1+e^{h(1)}}{1+\rho }}. \end{aligned}$$

(175)

The second integral in $\varLambda [f]$ is

$$\begin{aligned} I_1&=\int _0^1 dx f(x) \log \frac{f(x)}{1-f(x)} = \int _0^1 dx h(x) \frac{e^{h(x)}}{1+e^{h(x)}}\nonumber \\&= -\frac{1}{2\beta } \int _0^1 dx h(x) h''(x) = \frac{1}{2\beta } \left( \int _0^1 dx (h'(x))^2 + h(0) h'(0) \right) , \end{aligned}$$

(176)

where we integrated by parts in the last step and used $h'(1) = 0$.

Finally, the third integral is

$$\begin{aligned} I_2&=\int _0^1 dx \log (1-f(x)) = - \int _0^1 dx \log (1 + e^{h(x)})\nonumber \\&= -\frac{1}{2\beta } \int _0^1 dx V(h(x)) = -\frac{1}{2\beta } \int _0^1 dy \left( E - \frac{1}{2} (h'(y))^2 \right) \nonumber \\&= - \log \big (1 + e^{h(1)}\big ) + \frac{1}{4\beta } \int _0^1 dy (h'(y))^{2}, \end{aligned}$$

(177)

where $V(h)$ is defined in (164). In the second line we used (166) to eliminate $V(h(y))$ in terms of $(h'(y))^2 = 4\beta \log \frac{1+e^{h(1)}}{1+e^{h(y)}}$.

The integrals $I_1$ and $I_2$ appear in $\varLambda [f]$ in the combination

$$\begin{aligned} \frac{1}{2} I_1 + I_2&= \frac{1}{2\beta } \int _0^1 dx \left\{ \frac{1}{2} (h'(x))^2 - V(h(x)) \right\} + \frac{1}{4\beta } h(0) h'(0) \nonumber \\&= \frac{1}{\sqrt{\beta }} \int _{\log \rho }^{h(1)} dx \sqrt{\log \frac{1+e^{h(1)}}{1+e^x}} - \log \big (1 + e^{h(1)}\big ) + \frac{1}{4\beta } \log \rho h'(0). \end{aligned}$$

(178)

Substituting this into (172) the last term cancels against the term proportional to the integral $I_0$, and we obtain the result (71). This concludes the proof of (71). $\square$

Appendix 2: Analytical Solution for $\beta d^2 \gg 1$

We present in this Appendix an analytical solution of the variational problem for $\varLambda (d)$ in the $\beta d^2 \rightarrow \infty$ limit. This is used to derive the properties of the phase transition in the same limit, which are summarized in Proposition 6.

We will prove here the following approximation for the functional $\varLambda (d)$

$$\begin{aligned}&\varLambda (d) = \beta d^2 - \frac{2}{3}\beta d^3 + d\log \left( \frac{\rho }{1+\rho }\right) + \log (1+\rho )+ o(a^{-1}), \end{aligned}$$

(179)

where $a = \beta d^2$.

The starting point is the observation that the integral appearing in (76) depends only on the combination $a=\beta d^2$. We denote it as

$$\begin{aligned} J(a;\rho ) = \int _0^1 \frac{y^2 dy}{1+\rho - e^{a(y^2-1)}} \end{aligned}$$

(180)

We would like to obtain an approximation for this integral for $a \gg 1$. This is given by the following result.

Lemma 1

The integral $J(a;\rho )$ has the following expansion for $a\gg 1$

$$\begin{aligned} J(a;\rho ) = \frac{1}{1+\rho } \left\{ \frac{1}{3} - \frac{1}{2a} \log \left( \frac{\rho }{1+\rho }\right) + o(a^{-2}) \right\} . \end{aligned}$$

(181)

Proof

This result is shown by proving matching lower and upper bounds for the integral $J(a;\rho )$. We start by deriving a lower bound for the integral. This bound follows from using the inequality $y^2 - 1 \ge 2(y-1)$, which holds for any $y\in (0,1)$, in the exponent in the denominator of the integral (180). Then the integral can be performed exactly with the result

$$\begin{aligned}&J(a;\rho ) \ge \frac{1}{3a^3(1+\rho )} \left\{ a^3 - \frac{3}{2} a^2 \log \left( \frac{\rho }{1+\rho }\right) \right. \nonumber \\&\left. \qquad - \frac{3}{2} a \text{ Li }_2 \left( \frac{1}{1+\rho }\right) + \frac{3}{4} \text{ Li }_3 \left( \frac{1}{1+\rho }\right) - \frac{3}{4} \text{ Li }_3 \left( \frac{e^{-2a}}{1+\rho }\right) \right\} \end{aligned}$$

(182)

Here $\text{ Li }_n(z)$ denotes the polylogarithm of order $n$ [1].

For $\rho \rightarrow 0$ and $a\gg 1$ the polylogarithms approach constant values

$$\begin{aligned}&\lim _{\rho \rightarrow 0} \text{ Li }_2 \left( \frac{1}{1+\rho }\right) = \frac{\pi ^2}{6} \end{aligned}$$

(183)

$$\begin{aligned}&\lim _{\rho \rightarrow 0} \text{ Li }_3 \left( \frac{1}{1+\rho }\right) = \zeta (3) \simeq 1.202 \end{aligned}$$

(184)

$$\begin{aligned}&\lim _{x \rightarrow 0} \text{ Li }_3 (x) = x + o(x^2). \end{aligned}$$

(185)

We get thus the lower bound

$$\begin{aligned} J(a;\rho ) \ge \frac{1}{3(1+\rho )} -\frac{1}{2a(1+\rho )} \log \left( \frac{\rho }{1+\rho }\right) + o(a^{-2}). \end{aligned}$$

(186)

Next we prove the upper bound for the integral $J(a;\rho )$

$$\begin{aligned} J(a;\rho ) \le \frac{1}{1+\rho } \left\{ \frac{1}{3} - \frac{1}{2a} \log \left( \frac{\rho }{1+\rho - e^{-a}}\right) \right\} . \end{aligned}$$

(187)

This is obtained by considering the difference

$$\begin{aligned} \frac{1}{1 + \rho - e^{a(y^2-1)}} - \frac{1}{1+\rho } = \frac{1}{(1+\rho )[(1+\rho )e^{a(1-y^2)}-1]} \end{aligned}$$

(188)

Multiplying with $y^2$ and integrating over $y:(0,1)$ we have

$$\begin{aligned}&J(a;\rho ) - \frac{1}{3(1+\rho )} \le \frac{1}{1+\rho } \int _0^1 \frac{ydy}{(1+\rho ) e^{a(1-y^2)}-1} \nonumber \\&= -\frac{1}{2a(1+\rho )} \log \left( \frac{\rho }{1+\rho -e^{-a}}\right) < -\frac{1}{2a(1+\rho )} \log \left( \frac{\rho }{1+\rho }\right) . \end{aligned}$$

(189)

This proves the upper bound (187). Comparing with the lower bound (186) we obtain the expansion (181). $\square$

The final result (179) follows directly from using the expansion (181) of the integral $J(a;\rho )$ into the expression for $\varLambda (d)$.

We use next the approximation (179) to study the solution of the variational problem for $\varLambda (d)$ for $\beta \rightarrow \infty$. We would like to find the supremum over $d$ of the cubic polynomial in (179). This supremum is reached at $d = 0$ or $d_* = \frac{1}{2}\Big (1 +\sqrt{1+\frac{2}{\beta }\log \frac{\rho }{1+\rho }}\Big ) > 0$. according to the following condition

$$\begin{aligned} \text{ sup }_d \varLambda (d) = \left\{ \begin{array}{cc} \varLambda (0) = \log (1+\rho ) &{} \text{ if } \beta \le - \frac{8}{3}\log \frac{\rho }{1+\rho } \\ \varLambda (d_*) >\log (1+\rho ) &{} \text{ if } \beta > - \frac{8}{3}\log \frac{\rho }{1+\rho } \\ \end{array} \right. \end{aligned}$$

(190)

The supremum switches branches at the point

$$\begin{aligned} \beta _\mathrm{cr}(\rho ) = - \frac{8}{3}\log \Big (\frac{\rho }{1+\rho }\Big ) \end{aligned}$$

(191)

which thus defines the phase transition curve for $\beta \rightarrow \infty$. This proves the result (79). As $\beta \rightarrow \infty$ or equivalently $\rho \rightarrow 0$, the slope of this curve in $(-\log \rho ,\beta )$ coordinates approaches the value $8/3$. This is smaller than the slope of the phase transition curve in the mean-field approximation, which is equal to 3.

The values of $d_1, d_2$ along the phase transition curve approach

$$\begin{aligned} d_1 = 0,\quad d_2 = \frac{3}{4} = 0.75 \end{aligned}$$

(192)

as $\beta \rightarrow \infty$. This proves the result (78). For $\beta \gg \beta _\mathrm{cr}(\rho )$, we have $\lim _{\beta \rightarrow \infty } d = 1_-$ and the Lyapunov exponent becomes

$$\begin{aligned} \lim _{\beta \rightarrow \infty } \lambda (\rho ,\beta ) = \lim _{\beta \rightarrow \infty } \varLambda (d) = \frac{1}{3} \beta + \log \rho . \end{aligned}$$

(193)

This result is in agreement with the large $\beta$ asymptotic behavior of the Lyapunov exponent proven in Proposition 1, see Eq. (24).