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If follows a standard normal distribution, then

Assume that are iid. Denote . In the case where the distribution of is standard normal, then The bound is not bad in but not very nice in .

Let be iid standard gaussian random variables. Then, by Jensen's inequality, for all , Taking the and dividing by , we get: Choosing such that yields for , With a similar argument, we can show that Finally, a Markov inequality yields for all which is better than the previous bound \eqref{eq:bound:max} wrt but dramatic wrt …

Let be a non-negative random variable on a probability space and assume that there exists a constant such that for all , Then,

The proof is completed by noting that

It may seem a bit convoluted to bound using a bound of . I tried using a direct proof. The bound is less sharp because on the second line, we only apply a rough bound on the survival function of a Gaussian distribution. And not surprisingly, the resulting bound is less sharp that the previous one because: .

Let with the convention that . Then, We first rewrite the rhs using that is also a -martingale. To see this last property, write , which implies Now, the rhs of \eqref{eq:kolm:one} can be written using

Define with the convention that . Then,

• We prove (i). From \eqref{eq:fond}, we have

where we used in the second inequality that is non-negative and in the third inequality that is a supermartingale. The proof then follows by letting goes to infinity.

• We now turn to the proof of (ii). Using \eqref{eq:fond},

where we used in the last inequality that is non-negative. The proof is completed.