How is the second last line obtained from the line before it?
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$y$ varies from $x$ to $\infty$.
So, let $z= y-x \Rightarrow z$ varies from $0$ to $\infty$
$$\sum_{z=0}^{\infty}\frac{((1-p)\lambda)^z}{z!} = e^{(1-p)\lambda}$$
This is the Sigma notation of $${e^x}=\sum_{k=0}^\infty\frac{x^k}{k!}$$
19aksh
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1And this is the maclaurin expansion of the exponential function, $$ e^{x} = \sum_{k=0}^{\infty} \frac{x^k}{k!} $$ – Mikal Mar 27 '20 at 13:57
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@Mikal I was adding that one :) – 19aksh Mar 27 '20 at 13:59
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1Ah didn't mean to steal your thunder. – Mikal Mar 27 '20 at 14:01