Hi folks, could you help me with this simple question?

Let $\displaystyle X$ be a binomial random variable. How to show that $\displaystyle Pr(X > E[X]) \le 1/2$

thanks!

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- Nov 10th 2009, 11:17 AMandre.vignattiBinomial Random Variable probability
Hi folks, could you help me with this simple question?

Let $\displaystyle X$ be a binomial random variable. How to show that $\displaystyle Pr(X > E[X]) \le 1/2$

thanks! - Nov 10th 2009, 12:38 PMLaurent
The major difficulty of this question comes from the fact that its statement is false.

If $\displaystyle X$ has parameter $\displaystyle (n,1/2)$, then this is obvious by symmetry, but otherwise it has no reason to be true.

Consider a binomial variable $\displaystyle X$ of parameter $\displaystyle (n,p)$. Then $\displaystyle Y=n-X$ is binomial of parameter $\displaystyle (n,1-p)$. And $\displaystyle P(X>E[X])=P(Y<E[Y])$, $\displaystyle P(X<E[X])=P(Y>E[Y])$, so if your result was true, we would have (applying it to X and Y): $\displaystyle P(X<E[X])\leq 1/2$ and $\displaystyle P(X>E[X])\leq 1/2$. However, if the expected value $\displaystyle E[X]=np$ is not an integer, these terms sum to one, hence they would have to be equal to 1/2 : $\displaystyle P(X<E[X])=P(X>E[X])=1/2$. Consider for instance a binomial r.v. of parameter $\displaystyle (n,p)$ such that $\displaystyle n-1<np<n$, then $\displaystyle P(X>E[X])=P(X=n)=p^n$ is usually different from 1/2... - Nov 10th 2009, 03:02 PMandre.vignatti
Nice answer, but I forgot to tell that the mean is an integer! (Happy) I.e., $\displaystyle Pr(X=E[X]) \neq 0$

Anyone have the answer? - Nov 11th 2009, 05:34 AMLaurent
Did you mean that the statement of the question is simple, or that you were told there is a simple proof?

What this question means, is that the median of a binomial distribution with integer mean is equal to the mean.

I had to check by computer to convince myself this was true, and it seemed to be (up to $\displaystyle n=1000$). And indeed it is: you can find a proof in this article ("Monotone Convergence of Binomial Probabilities and a Generalization of Ramanujan's Equation" by Kumar Jogdeo and S. M. Samuels in the Annals of Mathematical Statistics). This result (theorem 3.2) is one of a few more general ones; so if you extract the part of the proof that leads to thm 3.2, this is probably not very long. I admit I did not read it through; feel free to post the outline of the proof below if you like (I'd read it for sure). - Nov 11th 2009, 07:04 AMmatheagle
I had to check if that was Steve Samuels from Purdue.

It was.

I was a student there when he and his wife were professors there. - Nov 11th 2009, 07:37 PMandre.vignatti
Using this observation, I trying another way to proof it (that paper solves the question, but is very sketchy, its difficult for me..). So, I looked at Wikipedia

(Binomial distribution article) and it says that the median is one of $\displaystyle \{\lceil np\rceil, \lfloor np\rfloor\}$, citing a paper which the result comes from (The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions). The abstract is simply "*We show that for the binomial and Poisson distributions, the distance between the mean and the median is always less than ln 2.*"

Thus, I'm trying to use the result of this paper to get the result claimed in Wikipedia, but i cannot figure it out. Any ideas? - Nov 12th 2009, 02:55 AMLaurent
I saw this article as well. The implication is because $\displaystyle \ln 2<1$ and the median is an integer: if the mean is also an integer, it has to be equal to the median. And if not, the median is either the least integer greater than the mean, or the greatest integer less than the mean (the only integers possibly at distance at most 1 from the mean). That's what the Wikipedia says.

This article is not available for free online; if you're interested, PM me your e-mail. Let me warn you however: it is probably not easier to read than the other one (and it seems to use a result from an article I couldn't get on internet, even from university). That's why I didn't mention it.

By the way, this article also quotes an older article, "Mean, Median and Mode in Binomial Distributions" by R. Kaas and J.M. Buhrman, in Statistica Neerlandica, which proves that the mean and median differ by at most $\displaystyle \max(p,1-p)$, which is enough to conclude (in the same way) that they're equal for integer means. However, I couldn't get this article either.

Let me come back to my question about the word "simple": is it a question you were asked by a teacher? or is it just out of curiosity? - Nov 12th 2009, 07:09 AMandre.vignatti
Yes, but the median of a binomial random variable is always integer? Why?

I used this result in a paper that i wrote to a conference (here), but now I'm submiting to a journal, and one of the referees ask this. At the time I wrote the paper to the conference, I asked this question to the other author of my paper, and he said that it is true, but he does not give me the proof, i just believe on him. A lot of referees read the paper, and no one complains about that part, until now. So I asked again to the co-author, and he cannot give me the proof. As the question (not the proof) is simple, I tried this forum. - Nov 12th 2009, 08:40 AMLaurent
I'd say: since the binomial is integer-valued, either the median is an integer, or it is not well-defined (like when $\displaystyle p=1/2$ and $\displaystyle n$ is odd (which gives a non-integral mean)). And since the article by Hamza deals with "the" median, it must mean that he proves also that it is well defined (except in the aforementioned case, where we "decide" consensually that the mean is $\displaystyle n/2$)

The median is not well-defined iff there is $\displaystyle x$ such that $\displaystyle P(X>x)=P(X<x)=1/2$. This is quite a coincidence, that's why I guess the symmetric case $\displaystyle p=1/2$ (and $\displaystyle n$ odd) is the only one where this happens. A rigorous proof could be rather cumbersome, I guess... - Nov 12th 2009, 04:45 PMandre.vignatti
Laurent,

thank you very much for your answers, it help me a lot!