Can I write $E[X_T \vert \mathcal{F}_{t \wedge T}] \mathbb{1}_{T \geq t}=E[X_T \vert \mathcal{F}_t] \mathbb{1}_{T \geq t}$. Here $X$ is a right continuous uniformly integrable martingale, T is a stopping time and $(\mathcal{F}_t)_{t \geq 0}$ is the filtration wrt which $X$ is a martingale I am using it to prove a theorem but I cannot show that its true.
Can I write $E[X_T \vert \mathcal{F}_{t \wedge T}] \mathbb{1}_{T \geq t}=E[X_T \vert \mathcal{F}_t] \mathbb{1}_{T \geq t}$
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stochastic-processes
stochastic-calculus
stopping-times
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1Please define those terms. – 2017-01-31
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0@GrahamKemp Done . I will be careful next time. Thank you for the feedback – 2017-01-31
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0Hint: $E[X_T\mid \mathcal F_t] = X_t$. – 2017-02-01
1 Answers
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You can. Its proof depends some properties of stopping times: For example, if $S,T$ are stopping times, then $\{S\ge T\},\{S>T\}\in \mathscr{F}_{S\wedge T}=\mathscr{F}_S\wedge\mathscr{F}_T$. Using these properties, we have $\mathsf{E}[X_T|\mathscr{F}_t]1_{[T\ge t]}\in\mathscr{F}_{T\wedge t}$, and $$ \begin{aligned} \mathsf{E}[X_T|\mathscr{F}_t]1_{[T\ge t]}&=\mathsf{E}[\mathsf{E}[X_T|\mathscr{F}_t]1_{[T\ge t]}|\mathscr{F}_{T\wedge t}]\\ (\{T\ge t\}\in\mathscr{F}_{T\wedge t})\qquad\qquad &=\mathsf{E}[\mathsf{E}[X_T|\mathscr{F}_t]|\mathscr{F}_{T\wedge t}]1_{[T\ge t]} \\ (\mathscr{F}_{T\wedge t}\subset \mathscr{F}_t)\qquad\qquad&=\mathsf{E}[X_T|\mathscr{F}_{T\wedge t}]1_{[T\ge t]}. \end{aligned}$$