Proving $mathbb{P}(xi_1+ xi_2+…+xi_n=1)=(sum_{i=1}^{n}lambda_i)Delta + mathcal{R}Delta^2$












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Let $xi_1, xi_2,...,xi_n$ be independent Bernoulli random variables in $(Omega,mathcal{P}(Omega),mathbb{P})$ and $$mathbb{P}(xi_i=0)=1-lambda_iDelta$$ and $$mathbb{P}(xi_i=1)=lambda_iDelta.$$ Here $lambda_1,lambda_2,...,lambda_n$ are positive parameters and $0< Delta <frac{1}{2max{lambda_1,lambda_2,...,lambda_n}}$.



We need to show $$mathbb{P}(xi_1+ xi_2+...+xi_n=1)=(sum_{i=1}^{n}lambda_i)Delta + mathcal{R}Delta^2,$$ where $|mathcal{R}| leq 2(sum_{i=1}^{n}lambda_i^2+(sum_{i=1}^{n}lambda_i)^2)$



How I should start and what to do? I really have no idea. Can somebody explain this to me?






probability probability-theory random-variables bernoulli-numbers







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  • $begingroup$
    Since the $xi_i$ are Bernoulli distributed, their sum is $1$ iff exactly one of the $xi_i$ is $1$ and the rest is $0$.
    $endgroup$
    – Tki Deneb
    Dec 12 '18 at 13:07


















0












$begingroup$


Let $xi_1, xi_2,...,xi_n$ be independent Bernoulli random variables in $(Omega,mathcal{P}(Omega),mathbb{P})$ and $$mathbb{P}(xi_i=0)=1-lambda_iDelta$$ and $$mathbb{P}(xi_i=1)=lambda_iDelta.$$ Here $lambda_1,lambda_2,...,lambda_n$ are positive parameters and $0< Delta <frac{1}{2max{lambda_1,lambda_2,...,lambda_n}}$.



We need to show $$mathbb{P}(xi_1+ xi_2+...+xi_n=1)=(sum_{i=1}^{n}lambda_i)Delta + mathcal{R}Delta^2,$$ where $|mathcal{R}| leq 2(sum_{i=1}^{n}lambda_i^2+(sum_{i=1}^{n}lambda_i)^2)$



How I should start and what to do? I really have no idea. Can somebody explain this to me?






probability probability-theory random-variables bernoulli-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Since the $xi_i$ are Bernoulli distributed, their sum is $1$ iff exactly one of the $xi_i$ is $1$ and the rest is $0$.
    $endgroup$
    – Tki Deneb
    Dec 12 '18 at 13:07
















0












0








0





$begingroup$


Let $xi_1, xi_2,...,xi_n$ be independent Bernoulli random variables in $(Omega,mathcal{P}(Omega),mathbb{P})$ and $$mathbb{P}(xi_i=0)=1-lambda_iDelta$$ and $$mathbb{P}(xi_i=1)=lambda_iDelta.$$ Here $lambda_1,lambda_2,...,lambda_n$ are positive parameters and $0< Delta <frac{1}{2max{lambda_1,lambda_2,...,lambda_n}}$.



We need to show $$mathbb{P}(xi_1+ xi_2+...+xi_n=1)=(sum_{i=1}^{n}lambda_i)Delta + mathcal{R}Delta^2,$$ where $|mathcal{R}| leq 2(sum_{i=1}^{n}lambda_i^2+(sum_{i=1}^{n}lambda_i)^2)$



How I should start and what to do? I really have no idea. Can somebody explain this to me?






probability probability-theory random-variables bernoulli-numbers







share|cite|improve this question











$endgroup$




Let $xi_1, xi_2,...,xi_n$ be independent Bernoulli random variables in $(Omega,mathcal{P}(Omega),mathbb{P})$ and $$mathbb{P}(xi_i=0)=1-lambda_iDelta$$ and $$mathbb{P}(xi_i=1)=lambda_iDelta.$$ Here $lambda_1,lambda_2,...,lambda_n$ are positive parameters and $0< Delta <frac{1}{2max{lambda_1,lambda_2,...,lambda_n}}$.



We need to show $$mathbb{P}(xi_1+ xi_2+...+xi_n=1)=(sum_{i=1}^{n}lambda_i)Delta + mathcal{R}Delta^2,$$ where $|mathcal{R}| leq 2(sum_{i=1}^{n}lambda_i^2+(sum_{i=1}^{n}lambda_i)^2)$



How I should start and what to do? I really have no idea. Can somebody explain this to me?






probability probability-theory random-variables bernoulli-numbers




probability probability-theory random-variables bernoulli-numbers






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edited Feb 3 at 0:56









J. W. Tanner

4,0611320




4,0611320










asked Dec 11 '18 at 17:23









AtstovasAtstovas

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  • $begingroup$
    Since the $xi_i$ are Bernoulli distributed, their sum is $1$ iff exactly one of the $xi_i$ is $1$ and the rest is $0$.
    $endgroup$
    – Tki Deneb
    Dec 12 '18 at 13:07




















  • $begingroup$
    Since the $xi_i$ are Bernoulli distributed, their sum is $1$ iff exactly one of the $xi_i$ is $1$ and the rest is $0$.
    $endgroup$
    – Tki Deneb
    Dec 12 '18 at 13:07


















$begingroup$
Since the $xi_i$ are Bernoulli distributed, their sum is $1$ iff exactly one of the $xi_i$ is $1$ and the rest is $0$.
$endgroup$
– Tki Deneb
Dec 12 '18 at 13:07






$begingroup$
Since the $xi_i$ are Bernoulli distributed, their sum is $1$ iff exactly one of the $xi_i$ is $1$ and the rest is $0$.
$endgroup$
– Tki Deneb
Dec 12 '18 at 13:07












1 Answer
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$begingroup$

We have
begin{align*}
Pleft(sum_{i=1}^{n}xi_i=1right) &= sum_{i=1}^{n}P(xi_i=1, {xi_j = 0: jneq i}) \
&=sum_{i=1}^nlambda_iDeltaleft(prod_{jneq i}(1-lambda_j Delta)right) \
&= sum_{i=1}^nlambda_iDeltaleft(1 - Delta sum_{jneq i}lambda_j + Delta^2sum_{j_1, j_2neq i}lambda_{j_1}lambda_{j_2} - cdotsright) \
&= sum_{i=1}^nlambda_iDelta - Delta^2 left{2sum_{i=1}^{n}sum_{j = 1}^{i-1}lambda_ilambda_jright} + cdots
end{align*}
Note that the term in curly brackets equals
begin{align*}
left(sum_{i=1}^{n}lambda_iright)^2 - sum_{i=1}^{n}lambda_i^2
end{align*}
which is clearly less in absolute value than $2(sum_{i=1}^{n}lambda_i^2 + left(sum_{i=1}^{n}lambda_iright)^2)$. Since
begin{align*}
0 < Delta < frac{1}{2max(lambda_1, cdots, lambda_n)}
end{align*}
The alternating series of sums are decreasing, so therefore the largest the sum can be is less than the remainder bound.






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    1 Answer
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    $begingroup$

    We have
    begin{align*}
    Pleft(sum_{i=1}^{n}xi_i=1right) &= sum_{i=1}^{n}P(xi_i=1, {xi_j = 0: jneq i}) \
    &=sum_{i=1}^nlambda_iDeltaleft(prod_{jneq i}(1-lambda_j Delta)right) \
    &= sum_{i=1}^nlambda_iDeltaleft(1 - Delta sum_{jneq i}lambda_j + Delta^2sum_{j_1, j_2neq i}lambda_{j_1}lambda_{j_2} - cdotsright) \
    &= sum_{i=1}^nlambda_iDelta - Delta^2 left{2sum_{i=1}^{n}sum_{j = 1}^{i-1}lambda_ilambda_jright} + cdots
    end{align*}
    Note that the term in curly brackets equals
    begin{align*}
    left(sum_{i=1}^{n}lambda_iright)^2 - sum_{i=1}^{n}lambda_i^2
    end{align*}
    which is clearly less in absolute value than $2(sum_{i=1}^{n}lambda_i^2 + left(sum_{i=1}^{n}lambda_iright)^2)$. Since
    begin{align*}
    0 < Delta < frac{1}{2max(lambda_1, cdots, lambda_n)}
    end{align*}
    The alternating series of sums are decreasing, so therefore the largest the sum can be is less than the remainder bound.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We have
      begin{align*}
      Pleft(sum_{i=1}^{n}xi_i=1right) &= sum_{i=1}^{n}P(xi_i=1, {xi_j = 0: jneq i}) \
      &=sum_{i=1}^nlambda_iDeltaleft(prod_{jneq i}(1-lambda_j Delta)right) \
      &= sum_{i=1}^nlambda_iDeltaleft(1 - Delta sum_{jneq i}lambda_j + Delta^2sum_{j_1, j_2neq i}lambda_{j_1}lambda_{j_2} - cdotsright) \
      &= sum_{i=1}^nlambda_iDelta - Delta^2 left{2sum_{i=1}^{n}sum_{j = 1}^{i-1}lambda_ilambda_jright} + cdots
      end{align*}
      Note that the term in curly brackets equals
      begin{align*}
      left(sum_{i=1}^{n}lambda_iright)^2 - sum_{i=1}^{n}lambda_i^2
      end{align*}
      which is clearly less in absolute value than $2(sum_{i=1}^{n}lambda_i^2 + left(sum_{i=1}^{n}lambda_iright)^2)$. Since
      begin{align*}
      0 < Delta < frac{1}{2max(lambda_1, cdots, lambda_n)}
      end{align*}
      The alternating series of sums are decreasing, so therefore the largest the sum can be is less than the remainder bound.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We have
        begin{align*}
        Pleft(sum_{i=1}^{n}xi_i=1right) &= sum_{i=1}^{n}P(xi_i=1, {xi_j = 0: jneq i}) \
        &=sum_{i=1}^nlambda_iDeltaleft(prod_{jneq i}(1-lambda_j Delta)right) \
        &= sum_{i=1}^nlambda_iDeltaleft(1 - Delta sum_{jneq i}lambda_j + Delta^2sum_{j_1, j_2neq i}lambda_{j_1}lambda_{j_2} - cdotsright) \
        &= sum_{i=1}^nlambda_iDelta - Delta^2 left{2sum_{i=1}^{n}sum_{j = 1}^{i-1}lambda_ilambda_jright} + cdots
        end{align*}
        Note that the term in curly brackets equals
        begin{align*}
        left(sum_{i=1}^{n}lambda_iright)^2 - sum_{i=1}^{n}lambda_i^2
        end{align*}
        which is clearly less in absolute value than $2(sum_{i=1}^{n}lambda_i^2 + left(sum_{i=1}^{n}lambda_iright)^2)$. Since
        begin{align*}
        0 < Delta < frac{1}{2max(lambda_1, cdots, lambda_n)}
        end{align*}
        The alternating series of sums are decreasing, so therefore the largest the sum can be is less than the remainder bound.






        share|cite|improve this answer









        $endgroup$



        We have
        begin{align*}
        Pleft(sum_{i=1}^{n}xi_i=1right) &= sum_{i=1}^{n}P(xi_i=1, {xi_j = 0: jneq i}) \
        &=sum_{i=1}^nlambda_iDeltaleft(prod_{jneq i}(1-lambda_j Delta)right) \
        &= sum_{i=1}^nlambda_iDeltaleft(1 - Delta sum_{jneq i}lambda_j + Delta^2sum_{j_1, j_2neq i}lambda_{j_1}lambda_{j_2} - cdotsright) \
        &= sum_{i=1}^nlambda_iDelta - Delta^2 left{2sum_{i=1}^{n}sum_{j = 1}^{i-1}lambda_ilambda_jright} + cdots
        end{align*}
        Note that the term in curly brackets equals
        begin{align*}
        left(sum_{i=1}^{n}lambda_iright)^2 - sum_{i=1}^{n}lambda_i^2
        end{align*}
        which is clearly less in absolute value than $2(sum_{i=1}^{n}lambda_i^2 + left(sum_{i=1}^{n}lambda_iright)^2)$. Since
        begin{align*}
        0 < Delta < frac{1}{2max(lambda_1, cdots, lambda_n)}
        end{align*}
        The alternating series of sums are decreasing, so therefore the largest the sum can be is less than the remainder bound.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 3 at 1:40









        Tom ChenTom Chen

        1,773715




        1,773715






























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