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Let $X$ be an $L^p$ random variable, where $p\in (0,1)$ and $W_t$ usual Brownian motion (with $W_t$ independent from $X$). I'd like to bound $$\mathbb E|X+W_t|^p$$ purely in terms of $\mathbb E|X|^p$ and $\mathbb E|X+W_1|^p$ (which I can assume to exist).

I was able to show the following, which looks a bit similar, but this does not solve my problem: For $X,Y\in L^p$, I can bound $$\begin{align*}\mathbb E|X + tY|^p &= \mathbb E|(1-t)X + t(X+Y)|^p \leq \mathbb E(\max\{|X|^p, |X+Y|^p\}) \leq \mathbb E[|X|^p + |X+Y|^p] \\ &= \mathbb E|X|^p + \mathbb E|X+Y|^p\end{align*}$$ (first inequality by quasiconvexity of the p-norm). This is essentially the kind of bound I would like to obtain. But of course, $W_t \neq t Y$ for a random variable $Y$, so I cannot directly apply this.

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With $Y:=W_1$ and $t\in[0,1]$, the expectation in question is $$ \begin{aligned} E|X+\sqrt t\,Y|^p&=E|(1-\sqrt t)X+\sqrt t\,(X+Y)|^p \\ &\le(1-\sqrt t)^p E|X|^p+(\sqrt t)^p E|X+Y|^p \\ &\le E|X|^p+E|X+Y|^p \\ &=E|X|^p+E|X+W_1|^p. \\ \end{aligned} $$

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  • $\begingroup$ But $W_t \neq \sqrt t W_1$, so I don't understand why the first term is valid. $\endgroup$
    – Philipp Wacker
    Commented Apr 20, 2021 at 5:32
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    $\begingroup$ $W_t$ has the same distribution as $\sqrt t W_1$. $\endgroup$
    – Anthony Quas
    Commented Apr 20, 2021 at 5:54
  • $\begingroup$ And I can just swap random variables out with other ones having the same distribution? I am worried that I can't do that here because the path of $W_t$ and $\sqrt t W_1$ are very different. $\endgroup$
    – Philipp Wacker
    Commented Apr 20, 2021 at 6:07
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    $\begingroup$ @MercuryBench : (i) The path here is quite irrelevant. The expectation of a given function of two random variables (r.v.'s) (here of $X$ and $W_t$) depends only on the joint distribution of the r.v.'s. In this case, $X$ is independent of each $W_t$, so that the joint distribution of $X$ and $W_t$ is determined by the individual distributions of $X$ and of $W_t$. Also, $W_t$ has the same distribution as $\sqrt t\,W_1$. So, the joint distribution of $X$ and $W_t$ is the same as that of $X$ and $\sqrt t\,W_1$. So, $E|X+W_t|^p=E|X+\sqrt t\,W_1|^p=E|X+\sqrt t\,Y|^p$. $\endgroup$
    – Iosif Pinelis
    Commented Apr 20, 2021 at 11:56
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    $\begingroup$ (ii) The inequality $E|(1-\sqrt t)X+\sqrt t\,(X+Y)|^p\le(1-\sqrt t)^p E|X|^p+(\sqrt t)^p E|X+Y|^p$ follows from the inequality $(a+b)^p\le a^p+b^p$ for $p\in(0,1]$ and nonnegative $a,b$, and standard properties of the expectation. $\endgroup$
    – Iosif Pinelis
    Commented Apr 20, 2021 at 12:00

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