Does $Y$ $\mathcal{F_t}$-adapted and $X$ adapted to $Y$ imply that $X$ is $\mathcal{F_t}$-adapted

Question

Suppose $(Y_t)_{t \geq0}$ is stochastic process which is adapted to a filtration $(\mathcal{F_t})_{t\geq0}$ (meaning $Y_t$ is $\mathcal{F_t}$-measurable for every $t\geq0$).

Suppose another stochastic process $(X_t)_{t\geq0} $ is adapted to the canonical filtration of $(Y_t)_{t \geq0}$ (namely $\mathcal{F_t}^Y:= \sigma\left(\{Y_s^{-1}(B)\colon s\leq t, B \in \mathcal{B}(\mathbb{R}\}\right)$.

Does it hold that $X$ is adapted to $(\mathcal{F_t})_{t\geq0}$?

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This question came up when I studied the existence and uniqueness of solution Stochastic Differential Equations with respect to Brownian Motion. Here we speak of a "strong solution" which is as solution that is both adapted to the underlying filtration as well as the canonical filtration of Brownian Motion.

I have thought about it for a little while now without being able to figure out whether it is true or not. Am I missing something obvious?

Answer 1

Yes, $X$ is adapted to $(\mathcal{F}_t)_t$, because $\mathcal{F}^{Y}_t\subset \mathcal{F}_t$ necessarily for each $t$, otherwise $Y$ could not be adapted to $(\mathcal{F}_t)_t$.

(In general $(\mathcal{F}^Y_t)_t$ is known as the natural filtration of $Y$.)