I am trying to show that whenever we have a sequence of random variables that are symmetric distribued about the origin and have finite fourth moment, then Kolmogorov's inequality will be bounded above by the fourth moment/ t^4.
Originally, we have that the bound is second moment (variance)/ t^2.
My approach has been to use binomial expansion and set bounds, but I am unable to do so. Please give me some advise.
Ok- I think I have made progress on this now and I have a solution. Please give me your comments in the "answers", since a lot of space in the comments has been wasted on unnecessary censorship ram-i-fications. Here it goes:
First of all, we will do a binomial expansion of $(x_1 + x_2 + x_3+...+x_n)^4$ into $(x_1 + x_2 +...x_k) + (x_(k+1)+...+x_n)$. So I evaluate $E(x_1 + x_2 +...+x_n)^4$ in this way. The general idea is then to see the odd moments cancel out, and I have to show that the cross terms are either zero or just positive. Then, I can have that $E(x_1 + x_2 +...+x_n)^4 \geq E(x_1 +...+x_k)^4$. Then, we use Jensen's with our convex function x^2 on positive reals to finish job since we already know Kolmogorov's Inequality is true. Hence, we have refined it.