# One-sided Bound on Sum of Fourth Moments

I'm interested in methods for proving one-sided bounds of the form $$\mathbb{P}[\frac{1}{n}\sum_{i=1}^n X^4_i \geq 3+t]\leq Ce^{-nt}$$ where $X_i$ are standard normal random variables. I've run a few basic random experiments, it seems that this should be possible for t relatively large (but independent of $n$). Note that I don't care about concentration, just a loose upper bound.

Is this possible?

• Did you mean $\geq 3+t$? As it is written, the inequality doesn't make sense: consider very large $t$ for example. The LHS goes to $1$ whereas the RHS goes down to $0$. – Kim Jong Un Jul 23 '15 at 0:20
• Also as $n \to \infty$, since $\mathbb E[X_i^4] = 3$, the left side goes to $1$ for any fixed $t > 0$. $\mathbb P[n^{-1} \sum_{i=1}^n X_i^4 \ge 3 + t]$ does decay exponentially by Cramér's theorem from Large Deviations theory. – Robert Israel Jul 23 '15 at 0:40
• Hi @RobertIsrael, would you mind elaborating on the exponential decay? I'm not familiar with Cramer's Theorem, and the few things I found online don't state the theorem in a clear enough form (to me at least). – squattyroo Jul 23 '15 at 1:17
• @RobertIsrael The random variables $X^4$ have no finite exponential moments hence Cramér needs at least some tweaking before being an option. – Did Jul 23 '15 at 1:19
• The result is false as $t\to\infty$ even for $n=1$. – user940 Jul 23 '15 at 2:46