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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.

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    You want to show that $P(|X|\geq t)\leq \frac{E X^4}{t^2}$, right?2012-05-09
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    @DavideGiraudo : I think you mean $P(|X| \ge t) \le E X^4/t^4$.2012-05-09
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    @RobertIsrael Right, so in this case, maybe what the OP wants is that $P(|X|\geq t)=P(X^4\geq t^4)$.2012-05-09
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    This is a maximial inequality, so actually u want to show that P( max 1t) < E(x_1 + ...+x_n)/t^42012-05-09
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    I don't think this attempt at humor is appropriate here.2012-05-09
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    @erik: do you mean $P(\max_{1 t) < E(X_1^4 + \ldots + X_n^4)/t^4$?2012-05-09
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    Instead of flagging, I downvoted for the extraneous information info. in the title.2012-05-09
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    ... and are the $X_i$ assumed to be independent and identically distributed?2012-05-09
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    @Peter: I won't restore it, but I had no objection to the original title and consider the objections rather silly. If there was a legitimate objection, it's that the question didn't adequately justify the removed part.2012-05-09
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    @Mahmud: And I cancelled you out.2012-05-09
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    @RobertIsrael:NO, the X_i are not assumed to be independent here. Also, intuition tells me this is a special case of Doob's maximial inequality for martingales, except in this case in discrete-time. Also, this can be generalized not just to power ^4, but to any even power (in fact, odd power too, but the odd moments of symmetric r.v.'s is 0). It does have a beauty to it though.2012-05-09
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    Robert, yes I do mean your \Tex-ed version.2012-05-10
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    erik: if the $X_i$ are not independent, you might tell us what you think Kolmogorov's Inequality is. Perhaps you intended them to be independent, but not identically distributed.2012-05-10
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    @Henry: I am so sorry- I meant they are independent, but not identically distributed. Apologies.2012-05-10
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    How is the first part of the current title related to the question?2012-05-10
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    @erik: what is this, third grade?2012-05-10
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    @erik, **Please** edit your question to remove the unnecessary verbiage: it adds *nothing* to the question, and much less to its title.2012-05-11

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