Compare cumulative distribution functions of binomial distribution

CoolMath

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Feb 18, 2021
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Hello,

I'm currently looking into the the balls into bins problem. More specifically, I compare the case that we have [MATH]x[/MATH] balls placed into [MATH]y[/MATH] bins with the case that we have .[MATH]2x[/MATH] balls placed into [MATH]2y[/MATH] bins ([MATH]x\geq y[/MATH]). Both cases follow the binomial distribution with density functions [MATH]F(k, x, 1/y)[/MATH] and [MATH]F(k, 2x, 1/(2y))[/MATH]. In average, each bin gets [MATH]x/y[/MATH] elements.

I'm now wondering whether the following inequality folds for [MATH]k\geq x/y[/MATH]:
[MATH]F(k, x, 1/y)\geq F(k, 2x, 1/(2y))[/MATH]I.e., the probability that a bin gets more than [MATH]k[/MATH] balls for the "[MATH]x[/MATH] balls placed into [MATH]y[/MATH] bins" case is lower than the probability that a bin gets more than [MATH]k[/MATH] balls in the "[MATH]2x[/MATH] balls placed into [MATH]2y[/MATH] bins" case.

Intuitively, I think that this proposition is true. I also evaluated the inequality for different [MATH]k[/MATH], [MATH]x[/MATH], and [MATH]y[/MATH] with Mathematica which always returned true.

Unfortunately, I do not come up with a more formal argumentation.

Do you have any idea?

Best
 
Actually, it should be the cumulative distribution function [MATH]F(k, x, 1/y)[/MATH] instead of the density function [MATH]F(k, x, 1/y)[/MATH].
 
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