 # Consider the following stochastic differential equation: d Y t </msub> = Jaylene Duarte 2022-05-15 Answered
Consider the following stochastic differential equation:
$d{Y}_{t}={Z}_{t}d{W}_{t}$
and terminal condition ${Y}_{T}=b,$ for which holds: $E\left[|b{|}^{2}\right]<\mathrm{\infty }$ Furthermore b is adapted to the filtration generated by the Browian motion only at terminal time $T$.
${Z}_{t}$ is a predictable square integrable process. So the right hand side ${Z}_{t}d{W}_{t}$ is martingale.
Why is the solution ${Y}_{t}$ adapted to the underlying filtration If I rewrite the equation, I get:
${Y}_{t}={Y}_{T}-{\int }_{t}^{T}{Z}_{s}d{W}_{s}=b-{\int }_{t}^{T}{Z}_{s}d{W}_{s}$
Since $b$ is only mb w.r.t to terminal time $T$, ${Y}_{t}$ cannot be adapted.
Where did I do a mistake?
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While it's true that $b$ and ${\int }_{t}^{T}{Z}_{s}d{W}_{s}$ are not ${\mathcal{F}}_{t}$ measurable, $b-{\int }_{t}^{T}{Z}_{s}d{W}_{s}$ is ${\mathcal{F}}_{t}$ measurable. That's because
$b-{\int }_{t}^{T}{Z}_{s}d{W}_{s}={Y}_{T}-{\int }_{t}^{T}{Z}_{s}d{W}_{s}={\int }_{0}^{T}{Z}_{s}d{W}_{s}-{\int }_{t}^{T}{Z}_{s}d{W}_{s}={\int }_{0}^{t}{Z}_{s}d{W}_{s}.$
If you're confused about why there is a solution to this backwards stochastic differential equation, note that ${Y}_{t}:=\mathbb{E}\left[b|{\mathcal{F}}_{t}\right]$ is a martingale satisfying ${Y}_{T}=b$, so by the martingale representation theorem there exists an adapted process $Z$ such that $d{Y}_{t}={Z}_{t}d{W}_{t}$.

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