Having a little trouble understanding this notation from lecture. Hopefully someone can help clear it up, because I wasn't able to find this notation online. What exactly is I{A,w}?
Define: I{A, w} = 1 if w ∈ A and 0 if w ∈ 0.
Define: P(A) = E[ I{A,w} ] (expectation)
Prove: Given disjoint A, B, then P(A ∪ B) = P(A) + P(B). Here's the proof from lecture:
I(A ∪ B, w) = I{A, w} + I{B, w} if A, B disjoint. Let's call this equation EQN.
Take 3 cases:
a) w ∉ A and w ∉ B: LHS of EQN = 0, RHS of EQN = 0
b) w ∈ A and w ∉ B: LHS of EQN = 1, RHS of EQN = 1
c) w ∉ A and w ∈ B: LHS of EQN = 1, RHS of EQN = 1
Take the expectation of both sides of EQN and use
the fact that E(X_1 + E_2) = E(X_1) + E(X_2).
I think my confusion lies in what I{A, w} is. Would the solution look something like this?
E[I{A ∪ B, w}] = E[I{A, w} + I{B, w}]
E[I{A ∪ B, w}] = E[I{A, w}] + E[I{B, w}]
P(A ∪ B) = P(A) + P(B)
Also, if A ⊂ B, then I{A, w} ≤ I{B, w}. How would I go about proving this? Hopefully understanding the above will give me insight.
Thanks.
$I$ is the indicator function, usually defined for a set A, parametrized with $w$
$$ I_A\{w\} = \begin{cases} 1 &\text{if } w \in A, \\ 0 &\text{if } w \notin A. \end{cases} $$
Following your notation it's easy to prove that for disjoint sets $$I\{A ∪ B, w\} = I\{A, w\} + I\{B, w\}$$
and the rest follows by taking expectations and linearity.