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In the proof of existence/uniqueness of SDE the following inequality is used:

$$E\left[ \left( \int_0^t a(s,\omega) ds \right)^2 \right] \leq t E\left[ \int_0^t a(s,\omega)^2 ds \right]$$

and I cannot really see how it is obtained. Here, $a$ is defined as $$a(s,\omega) = b(s,X_1(s)) - b(s,X_2(s)),$$ where $X_1$ and $X_2$ are stochastic processes, both of which satisfy the SDE $$dX_t = b dt + \sigma dBt,$$ and $b$ is assumed to satisfy:

  1. $|b(t,x)| \leq C(1+|x|)$
  2. $|b(t,x)-b(t,y)| \leq D|x-y|$

for some constants $C,D$. But given neither appears in the inequality, I do not think they are used.

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This inequality uses the (deterministic) Jensen's inequality, i.e.$$ \left( \frac{1}{t} \int_0^t a(s, ω) \,\mathrm{d}s \right)^2 \leqslant \frac{1}{t} \int_0^t (a(s, ω))^2 \,\mathrm{d}s. \quad \forall t > 0,\ ω \in Ω $$ Multiplying by $t^2$ and taking expectations yields$$ E\left( \left( \int_0^t a(s, ω) \,\mathrm{d}s \right)^2 \right) \leqslant t E\left( \int_0^t (a(s, ω))^2 \,\mathrm{d}s \right). $$

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  • $\begingroup$ Thanks! I wasn't aware of that form of Jensen's inequality. Just found the one you refer to one wikipedia page. Many thanks! $\endgroup$ – Xiaomi Dec 17 '18 at 1:13

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