$$ \mathbb{P}(X_k+Y_{2m-k}=\ell)=\mathbb{P}(X_{2m-k}+Y_k=2m-\ell)\;,$$ so we may w.l.o.g. Concentration bounds on weighted sum of i.i.d. 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. (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)\\ so that ultimately The convolution of two binomial distributions, one with parameters mand p and the other with parameters nand p, is a binomial distribution with parameters $$. MathJax reference. "The sum of independent non-identically distributed binomial random variables." as desired. Thanks for contributing an answer to MathOverflow! Can everyone with my passport data see my American arrival/departure record (form I-94)? $$ By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. \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 $X_k$}\\ Is is possible to document or add comments to a scriptin file? $Y_{m+j}$ are distributed as $X_m+X_j$ resp. $$ (Note that I have no hard evidence for this. I don't know why people are voting to close. \end{align}, $\sum_l \mathbb{P}(X_m=l)^2=\sum_l \mathbb{P}(Y_m=l)^2$, $$ \mathbb{P}(X_k+Y_{2m-k}=\ell)=\mathbb{P}(X_{2m-k}+Y_k=2m-\ell)\;,$$, $$\mathbb{P}(X_k+Y_{2m-k}=\ell)=\mathbb{P}(X_k-X_{m-k}+Y_m=\ell+k-m)$$, $$\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)$$, $$\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$$, $$\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\;,$$. $Y_m+Y_j$, where}\\ Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 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. If the p i are distinct, the sum follows the more general Poisson-Binomial distribution. $\mathbb{P}(X+Y=z)=\sum_x \mathbb{P}(X=x)\mathbb{P}(Y=z-x)$, (b) let $(X^\prime,Y^\prime)$ be distributed as $(X,Y)$, and independent of $(X,Y)$. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If you let X = X A + X B be the random variable which is the sum of your two binomials, then P ( X = k) is the summation over all the ways that you get X A = k A and X B = k B where k A + k B = k. It is easy to write down this summation formula if you know the formulas … 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$. I was simply curious if there was some other motivating factor. If the $p_i$ are distinct, the sum follows the more general Poisson-Binomial distribution. The Poisson-Binomial with parameters $q_k$ is the distribution of the sum of Bernoulli variables with success probabilities $q_k$. \begin{align*} \mathbb{P}(X_m+Y_m=m)=\mathbb{P}(X_m=X_m^\prime)=\sum_l \mathbb{P}(X_m=l)^2\end{align*} The moment generating function of a Binomial(n,p) random variable is $(1-p+pe^t)^n$. Is there really a need for another term? This question can be stated analytically. Now to your question above. Then Why I can't download packages from Sitecore? Why is it easier to carry a person while spinning than not spinning? The distribution of a sum S of independent binomial random variables, each with different success probabilities, is discussed. 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, $$ 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, with parameters $p_1,p_1,\dots$ (repeated $N_1$ times), $p_2,p_2,\dots$ (repeated $N_2$ times), ..., being the sum of $\sum_i N_i$ Bernoulli variables. Is the mode of a Poisson Binomial distribution next to the mean? 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. 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, … To transform the left hand side, observe that well known properties of the binomial distribution give: Setting $c=(1-p)/p$, define: \sum_z\mathbb{P}(X+Y=z)^2=\mathbb{P}(X+Y=X^\prime+Y^\prime)=\mathbb{P}(X-X^\prime=Y^\prime-Y) $$. The result seems to be true, but I don't see any working monotonicity patterns. I am not demanding that it is necessary. To learn more, see our tips on writing great answers. Did Star Trek ever tackle slavery as a theme in one of its episodes? Lemma Sorry. For a fixed $k$ the maximizing $l$ is the expectation up to rounding. Use MathJax to format equations. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Since $Y-Y^\prime$ and $Y^\prime-Y$ are identically distributed, and independent of $(X,X^\prime)$ the right hand sides above are equal. Proof: (a) apply Cauchy-Schwarz to \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) Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Lovecraft (?) 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}.

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