big-Oh notation...

[melnick]

well, for one of the ways, you can assume the X_n AREN'T stochastically
bounded (as it's often called) and construct a sequence whose limsup is
equal to 1..... (not sure of the details, though)

"Canuck2378" <canuc...@aol.com> wrote in message
news:20040421200329...@mb-m27.aol.com...
> Say that {X_n} a sequence of random variables satisfies X_n = O(1) if,
given e
> > 0, there exists M_e and an integer N_e such that
>
> P { |X_n| > M_e } < e
>
> for all n >= N_e.
>
> Also, suppose f is strictly increasing on the positive reals, with f(0) =
0,
> lim(t --> oo) f(t) = 1.
>
> Show X_n = O(1) if and only if limsup E f(|X_n|) < 1.
>
> I can easily show that, if the big-Oh condition is satisfied, then limsup
E
> f(|X_n|) <=1, but I don't know how to eliminate the possibility of
equality
> holding. Any ideas? Hopefully this'll illuminate how to show the converse,
> which I'm entirely in the dark on....

[J. Woodward]

"melnick" <blah...@com.com> wrote in message news:<c67vt8$a83$1...@news.wss.yale.edu>...

Assuming the stochastic boundedness doesn't hold, then there exists an
e > 0 such that P { |X_n| > n } < e infinitely often. Call the
sequence of n's for which this happens {n(k)}....I think you can
construct a sequence now that has limsup precisely equal to 1....