How I can resolve this problem ?
Set $X \sim \operatorname{Geo}(p)$, $0 \leqslant p \leqslant 1$.
Evaluate $\operatorname{E}(e^{-X})$.
I can't use $e^{-X}$ in this distribution.
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With $f_X(x)=p(1-p)^{x-1}$ we have:
$$\begin{align} \mathsf{E}(e^{-X})&=\sum_{x=1}^{\infty} e^{-x}p(1-p)^{x-1}\\ &=pe^{-1}\sum_{x=1}^{\infty} e^{-(x-1)}(1-p)^{x-1}\\ &=pe^{-1}\sum_{x=1}^{\infty} (e^{-1}(1-p))^{x-1}\\ &=\frac{pe^{-1}}{1-e^{-1}(1-p)} \end{align}$$
where the last term is calculated using the Geometric sum formula.
msm
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Sometimes $\operatorname{Geo}(p)$ means a certain distribution of random variables taking values in the set $\{1,2,3,\ldots\}$, and sometimes it means the set $\{0,1,2,3,\ldots\}$ instead. In the former case the expected value is just $1$ unit bigger than in the latter.
I'll assume it's $\{0,1,2,3,\ldots\}$. You have $\Pr(X=x) = pq^x$. \begin{align} \operatorname{E}(e^{-X}) & = \sum_{x=0}^\infty e^{-x} \Pr(X=x) = \sum_{x=0}^\infty e^{-x} pq^x = p \sum_{x=0}^\infty (qe^{-1})^x \\[10pt] & = p \sum_{x=0}^\infty r^x =p\times \left( \begin{array}{l} \text{the sum of a geometric series with first} \\ \text{term 1 and common ratio }r=qe^{-1} \end{array} \right). \end{align} If you know how to find the sum of a geometric series then you can finish this.
