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Jan 18, 2019 at 10:28 vote accept Sanjay
Jan 13, 2019 at 20:23 comment added Gordon Note that $0 \le \frac{Y_u^2}{1+Y_u^2} < 1$. Then $\exp\int_0^t \frac{Y_u^2}{1+Y_u^2} du < \exp(t)$, and $E\left( \exp\int_0^t \frac{Y_u^2}{1+Y_u^2} du\right) \le \exp(t)$.
Jan 13, 2019 at 14:43 answer added Ezy timeline score: 6
Jan 13, 2019 at 14:12 comment added Sanjay $dY_t$ instead of $Y_t$. Thanks for the reminder! I have now corrected.
Jan 13, 2019 at 14:11 history edited Sanjay CC BY-SA 4.0
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Jan 13, 2019 at 12:59 comment added Ezy I don’t understand the first equation. Please correct
Jan 13, 2019 at 12:24 history edited Bob Jansen CC BY-SA 4.0
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Jan 13, 2019 at 10:42 history asked Sanjay CC BY-SA 4.0