Calculation of expected value.












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Let N be a random variable with the following distribution: $$ P(N=n)=(n+1)left(frac{3}{4}right)^{2}left(frac{1}{4}right)^{n}, n=0,1,2,...$$ Let $xi_{1},xi_{2},...$ be a sequence of independent random variables with the same Bernoulli distribution: $$P(xi_{i}=1)=p, P(xi_{i}=0)=q, p>0, p+q=1.$$ Assume that a sequence $(xi_{i})_{i=1}^{infty}$ is independent of random variable N. Let: $$eta(omega)=
left{ begin{array}{ll}
displaystylesum_{i=1}^{N(omega)}xi_{i}(omega) & textrm{when $N(omega)>0$}\
0 & textrm{when $N(omega)=0$,}\
end{array} right.
$$ and assume that $zeta=N-eta$.
Find expected value $Eleft(frac{eta}{zeta +1}right)$.



I tried to use conditional expectation: $$Ebigg(frac{eta}{zeta +1}bigg)=Ebigg(Ebigg(frac{eta}{zeta +1}|Nbigg)bigg)$$ and calculate: $$Ebigg(frac{eta}{zeta +1}|N=nbigg)=Ebigg(frac{sum_{i=1}^{n}xi_{i}(omega)}{n-sum_{i=1}^{n}xi_{i}(omega)+1}bigg)$$ but I have no idea if it's right and what to do next.






expected-value







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    Let N be a random variable with the following distribution: $$ P(N=n)=(n+1)left(frac{3}{4}right)^{2}left(frac{1}{4}right)^{n}, n=0,1,2,...$$ Let $xi_{1},xi_{2},...$ be a sequence of independent random variables with the same Bernoulli distribution: $$P(xi_{i}=1)=p, P(xi_{i}=0)=q, p>0, p+q=1.$$ Assume that a sequence $(xi_{i})_{i=1}^{infty}$ is independent of random variable N. Let: $$eta(omega)=
    left{ begin{array}{ll}
    displaystylesum_{i=1}^{N(omega)}xi_{i}(omega) & textrm{when $N(omega)>0$}\
    0 & textrm{when $N(omega)=0$,}\
    end{array} right.
    $$     and assume that $zeta=N-eta$.
    Find expected value $Eleft(frac{eta}{zeta +1}right)$.



    I tried to use conditional expectation: $$Ebigg(frac{eta}{zeta +1}bigg)=Ebigg(Ebigg(frac{eta}{zeta +1}|Nbigg)bigg)$$ and calculate: $$Ebigg(frac{eta}{zeta +1}|N=nbigg)=Ebigg(frac{sum_{i=1}^{n}xi_{i}(omega)}{n-sum_{i=1}^{n}xi_{i}(omega)+1}bigg)$$ but I have no idea if it's right and what to do next.






    expected-value







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      Let N be a random variable with the following distribution: $$ P(N=n)=(n+1)left(frac{3}{4}right)^{2}left(frac{1}{4}right)^{n}, n=0,1,2,...$$ Let $xi_{1},xi_{2},...$ be a sequence of independent random variables with the same Bernoulli distribution: $$P(xi_{i}=1)=p, P(xi_{i}=0)=q, p>0, p+q=1.$$ Assume that a sequence $(xi_{i})_{i=1}^{infty}$ is independent of random variable N. Let: $$eta(omega)=
      left{ begin{array}{ll}
      displaystylesum_{i=1}^{N(omega)}xi_{i}(omega) & textrm{when $N(omega)>0$}\
      0 & textrm{when $N(omega)=0$,}\
      end{array} right.
      $$       and assume that $zeta=N-eta$.
      Find expected value $Eleft(frac{eta}{zeta +1}right)$.



      I tried to use conditional expectation: $$Ebigg(frac{eta}{zeta +1}bigg)=Ebigg(Ebigg(frac{eta}{zeta +1}|Nbigg)bigg)$$ and calculate: $$Ebigg(frac{eta}{zeta +1}|N=nbigg)=Ebigg(frac{sum_{i=1}^{n}xi_{i}(omega)}{n-sum_{i=1}^{n}xi_{i}(omega)+1}bigg)$$ but I have no idea if it's right and what to do next.






      expected-value







      share|cite|improve this question















      Let N be a random variable with the following distribution: $$ P(N=n)=(n+1)left(frac{3}{4}right)^{2}left(frac{1}{4}right)^{n}, n=0,1,2,...$$ Let $xi_{1},xi_{2},...$ be a sequence of independent random variables with the same Bernoulli distribution: $$P(xi_{i}=1)=p, P(xi_{i}=0)=q, p>0, p+q=1.$$ Assume that a sequence $(xi_{i})_{i=1}^{infty}$ is independent of random variable N. Let: $$eta(omega)=
      left{ begin{array}{ll}
      displaystylesum_{i=1}^{N(omega)}xi_{i}(omega) & textrm{when $N(omega)>0$}\
      0 & textrm{when $N(omega)=0$,}\
      end{array} right.
      $$       and assume that $zeta=N-eta$.
      Find expected value $Eleft(frac{eta}{zeta +1}right)$.



      I tried to use conditional expectation: $$Ebigg(frac{eta}{zeta +1}bigg)=Ebigg(Ebigg(frac{eta}{zeta +1}|Nbigg)bigg)$$ and calculate: $$Ebigg(frac{eta}{zeta +1}|N=nbigg)=Ebigg(frac{sum_{i=1}^{n}xi_{i}(omega)}{n-sum_{i=1}^{n}xi_{i}(omega)+1}bigg)$$ but I have no idea if it's right and what to do next.






      expected-value




      expected-value






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          Under condition $N=n$ random variable $eta$ has binomial distribution
          with parameters $n$ and $p$.



          This leads to: $mathbb{E}left[frac{eta}{zeta+1}mid N=nright]=sum_{k=0}^{n}binom{n}{k}p^{k}q^{n-k}frac{k}{n+1-k}=sum_{k=1}^{n}binom{n}{k-1}p^{k}q^{n-k}=frac{p}{q}sum_{k=0}^{n-1}binom{n}{k}p^{k}q^{n-k}=frac{p}{q}left(1-p^{n}right)$



          So $mathbb{E}left[frac{eta}{zeta+1}mid Nright]=frac{p}{q}left(1-p^{N}right)$
          and $mathbb{E}left[frac{eta}{zeta+1}right]=mathbb{E}left[mathbb{E}left[frac{eta}{zeta+1}mid Nright]right]=mathbb{E}frac{p}{q}left(1-p^{N}right)=frac{p}{q}-frac{p}{q}sum_{n=0}^{infty}left(n+1right)left(frac{3}{4}right)^{2}left(frac{1}{4}pright)^{n}$



          I leave the rest to you.






          share|cite|improve this answer





















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            Under condition $N=n$ random variable $eta$ has binomial distribution
            with parameters $n$ and $p$.



            This leads to: $mathbb{E}left[frac{eta}{zeta+1}mid N=nright]=sum_{k=0}^{n}binom{n}{k}p^{k}q^{n-k}frac{k}{n+1-k}=sum_{k=1}^{n}binom{n}{k-1}p^{k}q^{n-k}=frac{p}{q}sum_{k=0}^{n-1}binom{n}{k}p^{k}q^{n-k}=frac{p}{q}left(1-p^{n}right)$



            So $mathbb{E}left[frac{eta}{zeta+1}mid Nright]=frac{p}{q}left(1-p^{N}right)$
            and $mathbb{E}left[frac{eta}{zeta+1}right]=mathbb{E}left[mathbb{E}left[frac{eta}{zeta+1}mid Nright]right]=mathbb{E}frac{p}{q}left(1-p^{N}right)=frac{p}{q}-frac{p}{q}sum_{n=0}^{infty}left(n+1right)left(frac{3}{4}right)^{2}left(frac{1}{4}pright)^{n}$



            I leave the rest to you.






            share|cite|improve this answer


























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              Under condition $N=n$ random variable $eta$ has binomial distribution
              with parameters $n$ and $p$.



              This leads to: $mathbb{E}left[frac{eta}{zeta+1}mid N=nright]=sum_{k=0}^{n}binom{n}{k}p^{k}q^{n-k}frac{k}{n+1-k}=sum_{k=1}^{n}binom{n}{k-1}p^{k}q^{n-k}=frac{p}{q}sum_{k=0}^{n-1}binom{n}{k}p^{k}q^{n-k}=frac{p}{q}left(1-p^{n}right)$



              So $mathbb{E}left[frac{eta}{zeta+1}mid Nright]=frac{p}{q}left(1-p^{N}right)$
              and $mathbb{E}left[frac{eta}{zeta+1}right]=mathbb{E}left[mathbb{E}left[frac{eta}{zeta+1}mid Nright]right]=mathbb{E}frac{p}{q}left(1-p^{N}right)=frac{p}{q}-frac{p}{q}sum_{n=0}^{infty}left(n+1right)left(frac{3}{4}right)^{2}left(frac{1}{4}pright)^{n}$



              I leave the rest to you.






              share|cite|improve this answer
























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                Under condition $N=n$ random variable $eta$ has binomial distribution
                with parameters $n$ and $p$.



                This leads to: $mathbb{E}left[frac{eta}{zeta+1}mid N=nright]=sum_{k=0}^{n}binom{n}{k}p^{k}q^{n-k}frac{k}{n+1-k}=sum_{k=1}^{n}binom{n}{k-1}p^{k}q^{n-k}=frac{p}{q}sum_{k=0}^{n-1}binom{n}{k}p^{k}q^{n-k}=frac{p}{q}left(1-p^{n}right)$



                So $mathbb{E}left[frac{eta}{zeta+1}mid Nright]=frac{p}{q}left(1-p^{N}right)$
                and $mathbb{E}left[frac{eta}{zeta+1}right]=mathbb{E}left[mathbb{E}left[frac{eta}{zeta+1}mid Nright]right]=mathbb{E}frac{p}{q}left(1-p^{N}right)=frac{p}{q}-frac{p}{q}sum_{n=0}^{infty}left(n+1right)left(frac{3}{4}right)^{2}left(frac{1}{4}pright)^{n}$



                I leave the rest to you.






                share|cite|improve this answer












                Under condition $N=n$ random variable $eta$ has binomial distribution
                with parameters $n$ and $p$.



                This leads to: $mathbb{E}left[frac{eta}{zeta+1}mid N=nright]=sum_{k=0}^{n}binom{n}{k}p^{k}q^{n-k}frac{k}{n+1-k}=sum_{k=1}^{n}binom{n}{k-1}p^{k}q^{n-k}=frac{p}{q}sum_{k=0}^{n-1}binom{n}{k}p^{k}q^{n-k}=frac{p}{q}left(1-p^{n}right)$



                So $mathbb{E}left[frac{eta}{zeta+1}mid Nright]=frac{p}{q}left(1-p^{N}right)$
                and $mathbb{E}left[frac{eta}{zeta+1}right]=mathbb{E}left[mathbb{E}left[frac{eta}{zeta+1}mid Nright]right]=mathbb{E}frac{p}{q}left(1-p^{N}right)=frac{p}{q}-frac{p}{q}sum_{n=0}^{infty}left(n+1right)left(frac{3}{4}right)^{2}left(frac{1}{4}pright)^{n}$



                I leave the rest to you.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered 16 hours ago









                drhab

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                98.1k544129






























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