Sum of independent random variables with different distributions











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Can we find the distribution of the sum of random variables with different pmf and different possible values?For example let X be a Poisson random variable with parameter $lambda$ and Y an independent Bernoulli random variable with parameter $p$. Then is the probability mass function of X + Y
$$p_{X+Y}(n)=sum_{k=1}^n P(X=n-k)P(Y=k) = e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p = pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}?$$










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    Can we find the distribution of the sum of random variables with different pmf and different possible values?For example let X be a Poisson random variable with parameter $lambda$ and Y an independent Bernoulli random variable with parameter $p$. Then is the probability mass function of X + Y
    $$p_{X+Y}(n)=sum_{k=1}^n P(X=n-k)P(Y=k) = e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p = pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}?$$










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      up vote
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      down vote

      favorite











      Can we find the distribution of the sum of random variables with different pmf and different possible values?For example let X be a Poisson random variable with parameter $lambda$ and Y an independent Bernoulli random variable with parameter $p$. Then is the probability mass function of X + Y
      $$p_{X+Y}(n)=sum_{k=1}^n P(X=n-k)P(Y=k) = e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p = pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}?$$










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      Can we find the distribution of the sum of random variables with different pmf and different possible values?For example let X be a Poisson random variable with parameter $lambda$ and Y an independent Bernoulli random variable with parameter $p$. Then is the probability mass function of X + Y
      $$p_{X+Y}(n)=sum_{k=1}^n P(X=n-k)P(Y=k) = e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p = pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}?$$







      probability proof-verification






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      asked Nov 22 at 14:27









      dxdydz

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      1169






















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          if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$



          if $n>0$, then we have



          begin{align}
          p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
          &= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
          end{align}






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            1 Answer
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            1 Answer
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            active

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            active

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            active

            oldest

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            up vote
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            down vote



            accepted










            if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$



            if $n>0$, then we have



            begin{align}
            p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
            &= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
            end{align}






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              up vote
              1
              down vote



              accepted










              if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$



              if $n>0$, then we have



              begin{align}
              p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
              &= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
              end{align}






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                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$



                if $n>0$, then we have



                begin{align}
                p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
                &= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
                end{align}






                share|cite|improve this answer












                if $n=0$, then we have $$p_{{X+Y}}(0)=exp(-lambda)(1-p)$$



                if $n>0$, then we have



                begin{align}
                p_{X+Y}(n) &= sum_{k=0}^1 P(X=n-k) P(Y=k)\
                &= e^{-lambda}frac{lambda^n}{n!}(1-p)+e^{-lambda}frac{lambda^{n-1}}{(n-1)!}p \&= pe^{-lambda}(frac{lambda^{n-1}n-lambda^n}{n!})+e^{-lambda}frac{lambda^{n}}{n!}
                end{align}







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                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 14:34









                Siong Thye Goh

                95.2k1462115




                95.2k1462115






























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