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Let $X_t$ be a stochastic process on $[0, 1]$ that is almost surely Hölder continuous of order $\alpha > 0$, and almost surely uniformly bounded by some deterministic constant. It is not hard to see that the function $t \to \mathbb E[X_t]$ need not be $\alpha$-Hölder continuous. However,

Question: Is it true that $\mathbb E[X_t]$ is Hölder continuous of order $\beta$ for all $\beta < \alpha$?

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  • $\begingroup$ math.stackexchange.com/questions/1260388/… $\endgroup$
    – user479223
    Commented Jun 15 at 7:32
  • $\begingroup$ @user479223 Hm, the average of uniformly bounded continuous functions (by a deterministic constant) is always at least continuous. The example given in the post is not uniformly bounded. $\endgroup$
    – Nate River
    Commented Jun 15 at 7:35
  • $\begingroup$ @user479223 See for example the answer here. $\endgroup$
    – Nate River
    Commented Jun 15 at 7:36
  • $\begingroup$ Although, my intuition says that this is in general not true, and only true if you assume some probablistic bounds on the Hölder constant of $X_t$. $\endgroup$
    – Nate River
    Commented Jun 15 at 7:51

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No. Take $X_t = (Zt^\alpha)\wedge 1$ for $Z$ a positive random variable with $P(Z>K) \sim 1/\log K$ for large $K$. Then $$ E(X_t) \ge t^\alpha E(Z 1_{Z \le t^{-\alpha}}) = t^\alpha\int_0^{t^{-\alpha}} (P(Z\ge s) - P(Z \ge t^{-\alpha}))ds. $$ It's now not too hard to see that this integral is bounded below by $1/(\log t)^3$ for small $t$...

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    $\begingroup$ Very instructive counterexample, thanks. What kind of bounds do you think are needed on $F(x) := \mathbb P(\|X_t\|_{C^{0, \alpha}} \geq x)$ are needed to make the statement true? $\endgroup$
    – Nate River
    Commented Jun 15 at 8:03
  • $\begingroup$ Also, I have not worked it out in full yet, but to help me work out the computation for the integral, where does the power of $4$ in $1/(\log t)^4$ come from? $\endgroup$
    – Nate River
    Commented Jun 15 at 8:05
  • $\begingroup$ @NateRiver That was just a brutal bound obtained by replacing the integral by $t^{-\alpha}/(\log t) (P(Z\ge t^{-\alpha}/\log t) - P(Z\ge t^{-\alpha}))$ , it's probably not the "right" power. Come to think of it, one actually gets a cube there. $\endgroup$
    – Martin Hairer
    Commented Jun 15 at 8:45
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    $\begingroup$ @NateRiver The statement is of course trivial if you assume that the Hölder norm has finite expectation, and this is also pretty much optimal as you can see from variations of my counterexample. $\endgroup$
    – Martin Hairer
    Commented Jun 15 at 8:49
  • $\begingroup$ Ah I see about the bound. Yes it is the optimality I am interested in; for example I want to see if weak $L^p$ bounds on $\mathbb P(\|X_t\|_{C^{0, \alpha}} \geq x)$ would suffice. But I can probably work it out from here thanks to your guidance. Since this is an exercise I am giving myself, I should work it out myself anyway... $\endgroup$
    – Nate River
    Commented Jun 15 at 8:51

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