0

If $(X_t)_{t\in [0,T]}$ is cadlag and has finite variation, is it true that the total variation $t\mapsto \operatorname{Var}(X)_t$ is cadlag? ($\operatorname{Var}$ denotes the total variation)

This question is related. Erik has seen the proof concerning right-continuity. I want to see it since I couldn't figure it out myself. Can someone help? We're basically using this definition.

colt_browning
  • 2,700
Analysis
  • 2,450

1 Answers1

0

$t\mapsto\operatorname{Var}(X)_t$ is nondecreasing so it has left and right limits everywhere.

You can easily show with its definition that $$ \operatorname{Var}(X)_{t+}-\operatorname{Var}(X)_t=\vert X_{t+}-X_t\vert, $$ from which you trivially deduce that if $X$ is càd then so is $\operatorname{Var}(X)$.

Will
  • 6,842
  • Thanks, I forgot about this property of montone functions. What does $\operatorname{Var}(X)_{t+}$ mean? Is it like $t+\varepsilon$? And "how trivial" is the last deduction? – Analysis Nov 04 '23 at 22:32
  • I use the notation $Y_{t+}=\lim_{h\to0}Y_{t+\vert h\vert}$. And the last deduction is trivial because X càd implies $X_{t+}=X_t$ and therefore $\operatorname{Var}(X)_{t+}=\operatorname{Var}(X)_t$ according to the formula. – Will Nov 05 '23 at 08:27