Skip to main content

New answers tagged

2 votes
Accepted

Proving the Conditional Markov's Inequality

You are correct as far as I can see. You could express "almost surely" as "there exists $N_A$ with $A^c \subset N_A$ and $P(N_A)=0$" instead (where $A^c$ denotes the complement of $...
Flounderer's user avatar
  • 10.9k

Top 50 recent answers are included