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#1
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Hi,
Using the direct calculation method, I am not seeing how E(N²) is calculated as: 0.5(5² + 5) + 0.5(1² + 1) = 16 In my mind, this would've been: 0.5(5²) + 0.5(1²) = 13 I do see how the double expectation theorem works though. What am I not understanding? |
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#4
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Yes, that is exactly what kind of problem this is.
Given 50% Good drivers, 50% Bad drivers. LambdaG = 1, LambdaB = 5 Find E(N^2). It then shows: E(N^2) = .5(Nbad^2) + .5(Ngood^2) E(N^2) = .5(5^2 + 5) + .5(1^2 + 1) What formula does this come from, or what is the logic behind this? |
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#5
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Line one is just the standard conditional expectation applied to N^2:
E(anything) = E(anything | A) * prob(A) = E(anything | ~A) * prob(~ A) Line 2 is just substituting in E(N^2|good) and E(N^2|bad) |
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#6
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Quote:
Line 1 should read:E(anything) = E(anything | A) * prob(A) + E(anything | ~A) * prob(~ A) I still don't understand how line 2 works, I'm feeling ![]() How do you get to 0.5(5² + 5) + 0.5(1² + 1) = 16 ? |
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#7
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If the distribution of Nbad is Poisson with mean 5, then
Var(Nbad) = E(Nbad^2) - (E(Nbad))^2; definition of variance 5 = E(Nbad^2) - 5^2 since Poisson with mean = variance =5 E(Nbad)^2 = 5^2 + 5 Similarly for Ngood. .5's are P(good) and P(not good) = P(bad) |
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#8
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:: slaps forehead ::
duh! Thanks for clearing that up! |
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#9
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![]() ![]() Awesome, thank you very much!!! ![]() ![]() ![]() |
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