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Suppose $f, g: \mathbb R \to \mathbb R$ are continuous, non negative functions with $f \leq g$.

Fix some $T > 0$, and denote by $X^f$ the stochastic integral $\int_{[0, T]} f(s) \ dW_s$, where $W_s$ is a standard Brownian motion. Similarly write $X^g$ for the corresponding integral of $g$.

Write $F^f$ for the cumulative distribution function of $|X^f|$, that is $F^f (x) := P(|X^f| < x)$. Similarly write $F^g$ for the corresponding cumulative distribution function of $|X^g|$.

Question: Is it true that $F^f \geq F^g$?

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  • $\begingroup$ Don't you want the opposite inequality? Otherwise take $f=0$, so $F^f\equiv 1$, so for any nontrivial $g$ the inequality fails. $\endgroup$
    – m7e
    Commented Jun 27, 2021 at 10:04
  • $\begingroup$ Ah you’re right. Messed up the order there. $\endgroup$
    – Nate River
    Commented Jun 27, 2021 at 10:23

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$X^f$ and $X^g$ are Gaussian random variables with zero mean and variances $$\mathbb{E}(|X^f|^2) = \int_0^T|f(t)|^2dt \leq \int_0^T|g(t)|^2dt = \mathbb{E}(|X^g|^2).$$

Thus, we know the probability density functions of both random variables. Comparison is up to the reader.

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