# A probabilistic inequality [closed]

Suppose $x_1,x_2,...,x_6$ are non-negative Independent and identically-distributed random variables, is it true that $P(x_1+x_2+x_3+x_4+x_5+x_6 \lt 3\delta) \lt 2P(x_1 \lt \delta)$ for any $\delta \gt 0$?

Any help on this will be highly appreciated.

## closed as too localized by Scott Morrison♦Dec 6 '09 at 8:58

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• Arrgh it looks like this be homework on the high seas. – Harry Gindi Dec 6 '09 at 8:35
• Closed as too localized. Please explain why you're interested, what you've tried already, etc. Also, downvoting all answers below. Don't feed the raccoons. – Scott Morrison Dec 6 '09 at 8:59
• Downvoting the question and the answer below for the same reasons. – Akhil Mathew Dec 6 '09 at 23:37

## 1 Answer

I think it's true. My reasoning goes like this (and I kept making mistakes with the algebra/arithmetic so check carefully)

$P(S_6 \lt 3 \delta) \le P(\hbox{exactly 4 }X_i\hbox{s less than }\delta)+P(\hbox{exactly 5 }X_i\hbox{s less than }\delta)+P(\hbox{all 6 }X_i\hbox{s less than }\delta)$. So if we let $p=P(X_1 \lt \delta)$, then the right hand part of this inequality is

$(p^4)(10p^2-24p+15)$. I then plugged $2p-(p^4)(10p^2-24p+15)$ into R and got a function that looks always non-negetive between 0 and 1 (shouldn't be too bad to show this by derivatives but I'm lazy and have computer power).