Timeline for How can we show that the total variation distance of $X_s$ and $Y_s$ is bounded by the distance of $(X_t)_{t\ge s}$ and $(Y_t)_{t\ge s}$?
Current License: CC BY-SA 4.0
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| Jan 19, 2019 at 9:37 | comment | added | 0xbadf00d | @RW Yes, correctly. The left-hand side is the total variation distance on $\mathcal B(\mathbb R)$ and the right-hand side the total variation distance on $\mathcal B(\mathbb R)^{[s,\:\infty)}$. | |
| Jan 19, 2019 at 2:00 | comment | added | Algernon | Note that the $\sigma$-algebra $\sigma(X_s)$ of events depending only on the state at time $s$ is included in the $\sigma$-algebra $\sigma((X_{s+t})_{t\geq 0})$ of events depending on the states from time $s$ onward. What you want to show thus follows directly from the definition of total variation distance. | |
| Jan 19, 2019 at 1:10 | comment | added | R W | I don't understand your notation. Is the LHS of (1) the total variation distance between the time $s$ distributions and the RHS the distance between the measures on the space of sample paths? | |
| Jan 18, 2019 at 23:13 | history | asked | 0xbadf00d | CC BY-SA 4.0 |