0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ See Edit. $ $ $ $ $\endgroup$ – Did Apr 11 '13 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.