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$$\begin{align*} P(X=k)&={k-1 \choose r-1}(1-p)^{k-r}p^{r}={k-1 \choose r-1}\left(\frac12\right)^k \\ \\ E(X) &= \frac rp = 2r\\ \\ E(Y) &= \sum_{k=r}^{\infty} kP(Y=k) \\ &= rP(Y=r) + (r+1)P(Y=r+1) + \sum_{k=r+2}^{\infty} kP(Y=k) \\ &= rP(X=r) + (r+1)(P(X=r+1)+P(X=r+2)) + \sum_{k=r+2}^{\infty} kP(X=k+1) \\ &= rP(X=r) + (r+1)P(X=r+1)+ (r+1)P(X=r+2) + \sum_{k=r+3}^{\infty} (k-1)P(X=k) \\ &= rP(X=r) + (r+1)P(X=r+1)+ \sum_{k=r+2}^{\infty} (k-1)P(X=k) \\ &= rP(X=r) + (r+1)P(X=r+1)+ \sum_{k=r+2}^{\infty} kP(X=k) - \sum_{k=r+2}^{\infty} P(X=k)\\ &= \sum_{k=r}^{\infty} kP(X=k) - (1 - P(X=r) - P(X=r+1))\\ &= E(X) - 1 + {r-1 \choose r-1}\left(\frac12\right)^r + {r \choose r-1}\left(\frac12\right)^{r+1}\\ &= 2r - 1 + \left(\frac12\right)^r + r\left(\frac12\right)^{r+1}\\ &= 2r - 1 + \frac{2+r}{2^{r+1}} \end{align*}$$
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Sun, 08 Dec 2019 17:14 GMT