We can consider the sequence space R Z , which becomes a Polish space (complete,...

Savanah Boone

Savanah Boone

Answered

2022-07-04

We can consider the sequence space R Z , which becomes a Polish space (complete, separable metric space) when equipped with the product topology. In the theory of stochastic processes, one is frequently interested in probability measures on this space.
A classical result on measure theory in Polish spaces tells us that any probability measure μ on a Polish space X is tight: given ϵ > 0, then there is a compact set K ϵ X such that μ ( K ϵ ) > 1 ϵ.
Hence any product probability measure on R Z is tight.
Say we fix an i.i.d. probability measure μ on R Z . Precisely, μ ( A i ) = ν ( A i ) for some probability measure ν on R .
Is there a nice, concrete proof that μ is tight (independent of ν)?
The reason I ask is tightness is somewhat counter-intuitive here. For instance, if ν has unbounded support, then, for each M > 0, the set { k Z | x k | M } is infinite almost surely under μ (by the 0-1 law). Thinking along those lines, it's not easy to see K ϵ intuitively.

Answer & Explanation

bap1287dg

bap1287dg

Expert

2022-07-05Added 13 answers

Here is one concrete example of such a set K ϵ . Let M n for n N be numbers such that ν ( [ M n , M n ] c ) < ϵ 2 n . Then, consider the set
K ϵ = { ( a n ) n N : n , | a n | M n } R N .
We have
μ ( K ϵ c ) n = 1 ν ( { | a n | > M n } ) < ϵ n = 1 2 n = ϵ .
Also, K ϵ = n = 1 [ M n , M n ] and so K ϵ is the product of compact sets, and hence compact itself by Tychonoff's theorem.

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