Does This Probabilistic Inequality Hold for IID Random Variables?

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The discussion centers on the validity of a probabilistic inequality for non-negative independent and identically-distributed (IID) random variables. Specifically, it questions whether P(x1+x2+x3+x4<2δ) is less than or equal to 2P(x1<δ) for any δ>0. Suggestions include analyzing cases based on the value of x1 and considering the implications of changing the constant from 2 to 3 in the inequality. The conversation also explores the scenario where the random variables are Bernoulli, raising further questions about the inequality's applicability. Overall, the complexity of proving the inequality under different conditions is emphasized.
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Suppose x_1,x_2,x_3,x_4 are non-negative Independent and identically-distributed random variables, is it true that <br /> P\left(x_{1}+x_{2}+x_{3}+x_{4}&lt;2\delta\right)\leq2P\left(x_{1}&lt;\delta\right) for any \delta&gt;0?

Any answer or suggestion will be highly appreciated!
 
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This might well hold without the independence assumption. Use x1+x2+x3+x4>=x1+x2 then consider the cases x1<d and x1>=d separately.
 
Thanks. But then is it true that P\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}&lt;3\delta\right)\leq2P\left(x_{1}&lt;\delta\right) for any \delta&gt;0 ?
 
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forumfann said:
Thanks. But then is it true that P\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}&lt;3\delta\right)\leq2P\left(x_{1}&lt;\delta\right) for any \delta&gt;0 ?

This is not easy. Change the 2 to 3 and it is certainly true (using same method as before). What if the variables are Bernoulli, does the inequality hold?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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