This question can be stated analytically. To learn more, see our tips on writing great answers. The sum of independent variables each following binomial distributions $B(N_i,p_i)$ is also binomial if all $p_i = p$ are equal (in this case the sum follows $B(\sum_i N_i, p)$. PostgreSQL - CAST vs :: operator on LATERAL table function, Looking for a function that approximates a parabola, Using of the rocket propellant for engine cooling, Generic word for firearms with long barrels. Instead of describing the distribution with a list $p_1,p_1,\dots$ (repeated $N_1$ times), $p_2,p_2,\dots$ (repeated $N_2$ times), ..., where each success probability $p_i$ appears repeated $N_i$ times, one can also describe this distribution with the parameter pairs $(N_1,p_1), (N_2,p_2),\dots$. Using 1. and 2. we have MathOverflow is a question and answer site for professional mathematicians. random variables, Independence depth of linearly dependent random variables. : Let $X,Y$ be independent integer-valued rvs, then \begin{align*} Is the mode of a Poisson Binomial distribution next to the mean? By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. The Poisson-Binomial with parameters q k is the distribution of the sum of Bernoulli variables with success probabilities q k. Since a binomial B ( N i, p i) is the sum of N i Bernoulli variables with success probability p i, it follows that the sum of binomials considered above is equal to the Poisson-Binomial, … Were any IBM mainframes ever run multiuser? Now to your question above. I was simply curious if there was some other motivating factor. Is there a name for the Poisson-Binomial distribution when described in this form? MathJax reference. I don't know why people are voting to close. Then Lovecraft (?) $$\sum_{x}\mathbb{P}(X_k-X_{m-k}=x)^2=\sum_x\mathbb{P}(X_k+X_{m-k}=x)^2=\sum_x\mathbb{P}(X_m=x)^2$$ End Proof. so that ultimately The distribution of a sum S of independent binomial random variables, each with different success probabilities, is discussed. $$ \begin{align*} \mathbb{P}(X_k+Y_{2m-k}=\ell)\leq \mathbb{P}(X_m+Y_m=m)\end{align*}For the right hand side above we have (using 1. below) It only takes a minute to sign up. The convolution of two binomial distributions, one with parameters mand p and the other with parameters nand p, is a binomial distribution with parameters An idea (I don'the know how to pursue all the way): express $f_{n, c}$ through beta functions and prove that it is concave. Here is a (surprising) proof using Cauchy-Schwarz and "rearrangement". “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…, Comparing $X+Y$ and $X-Y$ for independent random variables with values in an abelian locally compact group. Then, by symmetry, the maximum is attained in the middle. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Did Star Trek ever tackle slavery as a theme in one of its episodes? Proof: (a) apply Cauchy-Schwarz to rev 2020.11.24.38066, Sorry, we no longer support Internet Explorer, The best answers are voted up and rise to the top, MathOverflow works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, $$ \sum_z\mathbb{P}(X-Y=z)^2=\mathbb{P}(X-Y=X^\prime-Y^\prime)=\mathbb{P}(X-X^\prime=Y-Y^\prime)\\ But this sum also follows the Poisson-Binomial distribution. $$\mathbb{P}(X_k+Y_{n-k}=z)^2\leq \big(\sum_x \mathbb{P}(X_m=x)^2\big)\big(\sum_y\mathbb{P}(Y_m=y)^2\big)=\big(\sum_x \mathbb{P}(X_m=x)^2\big)^2\;,$$ Using public key cryptography with multiple recipients. The result seems to be true, but I don't see any working monotonicity patterns. \begin{align} 1.\; &Y_k \mbox{ is distributed as}\; k-X^\prime_k, \mbox{where $X_k^\prime$ is distributed as $X_k$,}\\&\mbox{ and independent of $X_k$}\\ (Note that I have no hard evidence for this. Two approximations are examined, one based on a method of Kolmogorov, and another based on fitting a distribution from the Pearson family. Why I can't download packages from Sitecore? I am not demanding that it is necessary. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. \Pr(X_k+Y_{n-k}=l)\leq\Pr(X_{n/2}+Y_{n/2}=n/2). (a)\; &\mbox{ for any } z: \;\mathbb{P}(X+Y=z)^2\leq \big(\sum_x\mathbb{P}(X=x)^2\big)\,\big(\sum_y \mathbb{P}(Y=y)^2\big)\\ $$ Finally using part (b) of the lemma on the first factor gives Thanks for contributing an answer to MathOverflow! What is the cost of health care in the US? \end{align} 7.1. I think it could help if you let us know how this conjecture arose. In probability theory and statistics, the sum of independent binomial random variables is itself a binomial random variable if all the component variables share the same success probability. \sum_z\mathbb{P}(X+Y=z)^2=\mathbb{P}(X+Y=X^\prime+Y^\prime)=\mathbb{P}(X-X^\prime=Y^\prime-Y) rev 2020.11.24.38066, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. as desired. If the p i are distinct, the sum follows the more general Poisson-Binomial distribution. Indeed a very surprising proof. f_{n,c}(k,l)=c^{l+k}\sum_{i=\max(0,k+l-n)}^{\min(k,l)}\binom{k}{i}\binom{n-k}{l-i}c^{-2i}. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Lemma (a)\; &\mbox{ for any } z: \;\mathbb{P}(X+Y=z)^2\leq \big(\sum_x\mathbb{P}(X=x)^2\big)\,\big(\sum_y \mathbb{P}(Y=y)^2\big)\\ "The sum of independent non-identically distributed binomial random variables."