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I’m now faced with an SDE problem to prove the solution of the SDE: $$dX_t=(X_t-X_t^3)dt+\sigma dW_t, X_0=0$$ Is ergodic and estimate the bound of $EX_t^4$. I started to try to solve the SDE to get the explicit solution. Consider $Y_t=f(X_t)$, then we have:

$$dY_t=(f^{\prime}(X_t)(X_t-X_t^3)+\frac{\sigma}{2}f^{\prime\prime}(X_t))dt+\sigma f^\prime(X_t)dW_t$$

When $f^\prime(x)=exp(\frac{x^2}{\sigma}(\frac{x^2}{2}-1))$, we have:

$$dY_t=\sigma f^\prime (X_t)dW_t$$

And I don’t know how to go on now.

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    $\begingroup$ What's the given definition for ergodicity? For your bound apply Ito to $X^{4}$ and estimate cleverly. In any case I highly doubt that there is a closed solution for the equation. $\endgroup$
    – Tobsn
    Commented Jul 29, 2022 at 20:52
  • $\begingroup$ @Tobsn: For me, I'd take the definition to calculate $\int_0^T X_tdt$ to converge to be the definition. And for the bound, I also do not think there is the sharp estimation. But, whatever, I couldn't even give a bound here... That's why I'm asking this question. $\endgroup$
    – Holden Lyu
    Commented Jul 30, 2022 at 8:54
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    $\begingroup$ That for sure is no conventional definition of ergodicity If anything, you should consider the time avergaed integral. A gave you a hint on how to get bounds for your 4th moment. $\endgroup$
    – Tobsn
    Commented Jul 30, 2022 at 11:22
  • $\begingroup$ @Tobsn : I didn't totally get your hint, do you mean calculate the Ito differential of $X_t^4$? It may leave a term about $X_t^6$ which I don't know how to estimate. $\endgroup$
    – Holden Lyu
    Commented Jul 30, 2022 at 11:56
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    $\begingroup$ the $X^{6}$ will come with a minus term, thus can be dropped in your estimate. $\endgroup$
    – Tobsn
    Commented Jul 30, 2022 at 12:38

1 Answer 1

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As suggested, to find the bound we can apply Ito to $f=X_{t}^{4}$, which, taking the exptectation and remembering that the stochastic part vanishes, yields:

$$ m(t) = \alpha(t)+\int_{0}^{t}4m(s)ds -4 \int_{0}^{t}\mathbb{E}[X_{s}^{6}] ds $$ which we can bound from above

$$ m(t) \le \alpha(t)+\int_{0}^{t}4m(s)ds $$

where $\alpha(t) = \int_{0}^{t}6 \sigma^2\mathbb{E}[X_{s}^{2}]ds$ and $m(t)= \mathbb{E}[X_{t}^{4}]$.

Since $4$ is a non-decreasing function, by the integral version of Gronwall Lemma we get:

$$ m(t) \le e^{4t} \alpha(t) $$

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