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 $(N_1,p_1), (N_2,p_2),\dots$. (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)\\ $Y_m+Y_j$, where}\\ 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. Now to your question above. Is the mode of a Poisson Binomial distribution next to the mean? What is the cost of health care in the US? $Y_m+Y_j$, where}\\ \end{align*} Setting $c=(1-p)/p$, define: Can everyone with my passport data see my American arrival/departure record (form I-94)? It only takes a minute to sign up. $Y_{m+j}$ are distributed as $X_m+X_j$ resp. It would just be an alternative parameterization for the poisson-binomial. 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}. Here is a (surprising) proof using Cauchy-Schwarz and "rearrangement". $$ MathJax reference. 7.1. $$, $$ Is is possible to document or add comments to a scriptin file? where $X_k,X_{m-k}$ and $Y_m$ on the right hand side are independent. I don't know why people are voting to close. $$. It has been my observation over 40 years of being a statistician that statisticians tend not to want to make new names for specific departures of standard procedures and that when a new name is introduced it is many times from someone in a different field. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. I was simply curious if there was some other motivating factor. A mystery! Using part (a) of the lemma (with $X=X_{k}-X_{m-k}$ and $Y=Y_m$) \Pr(X_k+Y_{n-k}=l)\leq\Pr(X_{n/2}+Y_{n/2}=n/2). What distribution does X+Y follow? Fisher information matrix for comparing two treatments, Skewness for a sum of independent weighted bernoulli random variables with different probabilities of success, Proof of the standard error of the distribution between two normal distributions in A/B testing. Why is it easier to carry a person while spinning than not spinning? \end{align} Using 1. gives that $\sum_l \mathbb{P}(X_m=l)^2=\sum_l \mathbb{P}(Y_m=l)^2$ and further that $$\mathbb{P}(X_k+Y_{n-k}=z)^2\leq \big(\sum_x \mathbb{P}(X_m=x)^2\big)\big(\sum_y\mathbb{P}(Y_m=y)^2\big)=\big(\sum_x \mathbb{P}(X_m=x)^2\big)^2\;,$$ \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 $X_k$}\\ Using public key cryptography with multiple recipients. SUMS OF DISCRETE RANDOM VARIABLES 289 For certain special distributions it is possible to flnd an expression for the dis-tribution that results from convoluting the distribution with itself ntimes. But this sum also follows the Poisson-Binomial distribution. Let $n=2m$. Thanks for contributing an answer to MathOverflow! We have to show that To learn more, see our tips on writing great answers. (b)\; &\sum_z\mathbb{P}(X-Y=z)^2=\sum_z\mathbb{P}(X+Y=z)^2\end{align*}, $\mathbb{P}(X+Y=z)=\sum_x \mathbb{P}(X=x)\mathbb{P}(Y=z-x)$, \begin{align*} Use MathJax to format equations. There is a small typo in the third row of proving (b): The middle term is $\mathbb{P}(X+Y = X' + Y')$. 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, … 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 $B(\sum_i N_i, p)$. What happens if someone casts Dissonant Whisper on my halfling? The convolution of two binomial distributions, one with parameters mand p and the other with parameters nand p, is a binomial distribution with parameters 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, $$ gives $$\mathbb{P}(X_k+Y_{2m-k}=z)^2 \leq \big(\sum_x \mathbb{P}(X_k-X_{m-k}=x)^2\big)\big(\sum_y\mathbb{P}(Y_m=y)^2\big)$$ Making statements based on opinion; back them up with references or personal experience. To transform the left hand side, observe that well known properties of the binomial distribution give: By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. MathOverflow is a question and answer site for professional mathematicians.

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