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.