$$\sum\limits_{x=0}^{\infty} p(1-p)^x \cdot \mathrm{e}^{-\alpha x} \overbrace{=}^{(1)}\, (1-p) \sum\limits_{x=0}^{\infty} p^x \mathrm{e}^{-\alpha x}$$
Why one can put $(1-p)$ before the sum although there is the $x$ as an exponent. Any hints?
Calculation of infinite sum
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$\begingroup$
calculus
summation
stochastic-calculus
moment-generating-functions
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0Is it MGF of geometric distribution? I cannot see how it is so... have you wrongly interchanged $p$ and $1-p$ in the lhs? – 2017-02-19
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0Yeah it is! I get this with an solution of worksheet and I can't see how it works. I mean the PDF of geometric distribution is $p(1-p)^x$. So maybe there is a mistake? – 2017-02-19
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2It looks like a mistake... Can you provide a context of it? By all means, $\sum pq^x e^{-\alpha x} =p/ (1-q e^{-\alpha}) \neq q/ (1-p e^{-\alpha}) =\sum qp^x e^{-\alpha x}$ in general. – 2017-02-19
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0I had to compute the MGF with the laplace transformation, $\mathcal{L}_x(\alpha)=\mathbb{E}[\mathrm{e}^{-\alpha x}]$ and with this one get the sum above. Yeah right, that's what I thought too! – 2017-02-19