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 Mar31 comment Expectation of a Poisson Process Right up top, unless I'm missing something, first one. Can't miss it. They use X and Y there rather than W Mar31 accepted Prove for a simple graph that $n-1 \leq m \leq \frac{n(n-1)}{2}$ Mar30 comment Prove for a simple graph that $n-1 \leq m \leq \frac{n(n-1)}{2}$ exactly, that's what I'm not sure about. I can show that the other extreme is true as well by removing any $3$ edges. Removing any more than that makes the graph disconnected and the inequality would fail. However, I'm not exactly sure that showing this for the extreme cases would be considered a "proof" Mar30 comment Prove for a simple graph that $n-1 \leq m \leq \frac{n(n-1)}{2}$ 0oo, I didn't know I could use MathJax in titles too. thanks for the fix and teaching me something new Mar30 asked Prove for a simple graph that $n-1 \leq m \leq \frac{n(n-1)}{2}$ Mar30 comment Expectation of a Poisson Process I did, I'm not quite understanding. as far as I can tell from wikipedia, the inner function would be a function of $x$ en.wikipedia.org/wiki/Law_of_total_expectation Mar30 comment Expectation of a Poisson Process @Did Maybe it's a notation thing, this subject has a lot of different notations, but the inner expectation should be a function of $x$ and the outer one should turn it into a number, correct? Mar27 revised Expectation of a parallel system added 3 characters in body Mar27 comment Expectation of a parallel system What is $H_n$, the heavyside step function? Mar27 revised Expectation of a parallel system added 61 characters in body Mar27 asked Expectation of a parallel system Mar27 comment Conditioning on a random variable of course that only works if $\Lambda$ and $X$ are independent Mar27 comment Conditioning on a random variable No, thank you! Between both answers I think I've got this well understood now. I hate to have to pick just one of them as best because they were kinda complementary in helping me out Mar27 comment Conditioning on a random variable Multiplying leftmost and rightmost sides by $Pr(\Lambda = \lambda)$ (and dropping the middle term from the equation), we have $$Pr(X \geq 3 |\Lambda = \lambda)Pr(\Lambda = \lambda) = Pr(X \geq 3)$$ Rearranging, I have what you have: $$Pr(X \geq 3) = Pr(X \geq 3 |\Lambda = \lambda)Pr(\Lambda = \lambda)$$ And now we do this for all $\lambda$ Mar27 comment Conditioning on a random variable Not a problem at all; repetition is always good as long as it clears things up. So I think I see it now. So essentially we're fixing $\Lambda$ and asking ourselves what the desired probability would be for every value of $\lambda$. But we also need to be aware of the probability that $\Lambda$ takes on a certain value. In terms of conditional probability: $$Pr(X \geq 3 |\Lambda = \lambda) = \frac{Pr(X \geq 3, \Lambda = \lambda)}{Pr(\Lambda = \lambda)} = \frac{Pr(X \geq 3) Pr(\Lambda = \lambda)}{Pr(\Lambda = \lambda)}$$ multiplying both sides by $Pr(\Lambda = \lambda$) we have...next post Mar27 comment Conditioning on a random variable But regarding the $ds$ stuff, so what you're saying is that the probability that $\Lambda$ is on the set (0,5) is $$\frac{(s+ds)-s}{5}$$ and the $s$ cancels out so we're left with $$\frac{ds}{5}$$ But after that I'm lost because you made the variable we're integrating over $x$ and I don't see any $x$'s in the integration. Typo perhaps? Mar27 comment Conditioning on a random variable Sorry, deleated comment to fix $LaTeX{}$ issues, so they're out of order now, but anyway. So what you're saying is that we multiply the density and mass functions together because by the definition of conditional probability: $$Pr(X = k | \Lambda = \lambda) = \frac{Pr(X = k , \Lambda = \lambda)}{Pr(\Lambda = \lambda)}$$ Mar27 asked Conditioning on a random variable Mar24 comment Expectation of a Poisson Process Ah, now it's clear. Exactly what I wanted; once you said it wasn't really recursion, then it ecame obvious to just factor out an $E[W]$ and solve. Mar24 revised Expectation of a Poisson Process added 213 characters in body