Let $m$ be a probability measure on $Y \subseteq \mathbb{R}^p$, so that $m(Y)=1$.

Consider a function $f: \mathbb{R}^n \times Y \rightarrow \mathbb{R}^n$, continuous on the first arguments, measurable in the second. Assume that $f(0,y) = 0$ for all $y \in Y$, and that there exists a neighborhood $\mathcal{X}$ of $0$ such that $\mathbb{E} \left[ \max_{x \in \mathcal{X}} f(x,\cdot) \right]$ is finite.

Consider the discrete-time stochastic process $$ x_{k+1} = f(x_k,y_k), \ k = 0, 1, ..., $$ with $x(0) = x_0 \in \mathbb{R}^n$, and with $y_k \in Y$ random variable associated to the probability measure $m$. The random variables $y_0, y_1, ...$ are i.i.d.. So, for instance, the expected value of each $y_k$ is $\mathbb{E}(y_k) := \int_Y y m(dy)$.

We study the "attractivity" of the origin.

Assume that for all $\epsilon>0$ and $\lambda \in (0,1)$ there exists integer $K>0$ such that

$$ \mathbb{E} \left[ \mathbb{1}_{ \epsilon \mathbb{B} }( x_K ) \right] \geq 1-\lambda. $$

In other words, for $k$ big enough the process $x_k$ reaches a neighborhood of $0$ with probability $1$.

Under these assumptions, I am wondering if also for the non-stochastic process $$x_{k+1} = \mathbb{E} \left[ f(x_k,y_k) \right], \ k=0,1,..., $$ the origin is attractive, namely that for all $\epsilon>0$ there exists integer $K>0$ such that $x_K \in \epsilon \mathbb{B}$: $$ \mathbb{1}_{ \epsilon \mathbb{B} }( \mathbb{E} \left[ x_K \right] ) = 1.$$

Comment. The above question relies on relations between the asymptotic stability in probability and asymptotic stability of the averaged process. The setting is similar to this one.


Here is a counterexample. Assume that $f(x,y)=x2^y$ for every $x$ in $\mathbb R$ and $y$ in $\{-1,+1\}$, and call $p=P[y_0=+1]$. Then the stochastic process $x_{k+1}=f(x_k,y_k)$ is solved by $x_{k+1}=x_02^{y_0+\cdots+y_k}$. Assume that $E[y_0]\lt0$, then, by the law of large numbers, $y_0+\cdots+y_k\to-\infty$ hence $x_k\to0$ almost surely, for every $x_0$. On the other hand, what you call the non-stochastic process is solved by $x_{k+1}=x_kE[2^{y_k}]$ hence $x_k=x_0a^k$ with $a=E[2^{y_0}]$. Thus $x_k\to\infty$ if $x_0\ne0$ and $a\gt1$. Both conditions are met simultaneously for every $\frac13\lt p\lt\frac12$. Then the point $0$ is asymptotically stable in probability and asymptotically unstable for the averaged process.

Another example is $f(x,y)=x^2y$ for every $x$ in $\mathbb R$ and $y$ in $\{-1,+1\}$. Then the stochastic process $x_{k+1}=f(x_k,y_k)$ is solved by $x_{k+1}=\pm (x_0)^{2^k}$ hence, for every $|x_0|\geqslant1$, the process is asymptotically unstable in probability. On the other hand, if $p=\frac12$, $E[y_0]=0$ hence the averaged process is such that $x_k=0$ for every $k\geqslant1$. Thus, for every $x_0$, the averaged process is asymptotically stable.

  • $\begingroup$ Thanks for your answer. I guess that $\mathbb{E}[2^{y_0}] > 1$ because there is the a point mass in $1$, namely $m(\{1\}) = p $. Now, $\mathbb{E}[2^{y_0}] = 2p + \frac{1}{2}(1-p)$, where $1-p>p$ to get $\mathbb{E}[y]<0$. Is this right? Moreover, I guess another counterexample can be shown without point masses on $m$. Say we take $m$ so that $\mathbb{E}[y] < 0$, but $\mathbb{E}[2^y]>1$. Is that right? $\endgroup$ – user693 Apr 10 '13 at 18:35
  • $\begingroup$ All of these are correct, starting at "Now". $\endgroup$ – Did Apr 10 '13 at 18:44
  • $\begingroup$ Would be nice to show, if it is the case, that also the converse does not hold. Namely, show an example such that $x_{k+1} = \mathbb{E}[ f(x_k, y_k ) ]$ is asymptotically stable, but the stochastic process $x_{k+1} = f(x_k, y_k )$ is not asymptotically stable in probability. $\endgroup$ – user693 Apr 10 '13 at 19:42
  • $\begingroup$ Let me add that there are no $p \in [0,1]$ such that $x_{k+1} = \mathbb{E}[f(x_k,y_k)]$ is AS, but the process $x_{k+1} = f(x_k,y_k)$ is A. Unstable in probability. $\endgroup$ – user693 Apr 11 '13 at 10:08
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    $\begingroup$ See Edit. $ $ $ $ $\endgroup$ – Did Apr 11 '13 at 11:12

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