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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: $$\eta(\omega)= \left\{ \begin{array}{ll} \displaystyle\sum_{i=1}^{N(\omega)}\xi_{i}(\omega) & \textrm{when $N(\omega)>0$}\\ 0 & \textrm{when $N(\omega)=0$,}\\ \end{array} \right. $$ and assume that $\zeta=N-\eta$. Find expected value $E\left(\frac{\eta}{\zeta +1}\right)$.

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

A.Fue
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1 Answers1

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

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

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

I leave the rest to you.

drhab
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