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My question is how to calculate the probability density function (pdf) of consequent, but otherwise independent events, defined as follows (in case of 3 events):

$A$ is an independent event with pdf $f(x)=\lambda_1 e^{-\lambda_1x} $.

$P(B|A')=0$, $B|A$ is event with pdf $f(x)=\lambda_2 e^{-\lambda_2x} $.

$P(C|B')=0$, $P(C|B)$ is event with pdf $f(x)=\lambda_3 e^{-\lambda_3x} $.

I understand that for independent events I would need to apply convolution, but I could not find the way to solve the problem for consecutive events.

Note (in case I defined the problem unclearly): once the previous event has occured, the next event becomes an independent event with a defined pdf. Each event can happen only once.

Anna401
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1 Answers1

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From Bayes' theorem, P(B)=P(B|A)P(A)/P(A|B). P(A|B) = 1, => P(B)=P(A)*P(B|A) I assume that I need P(B), not P(A+B), because P(A|B)=1. Then P(C)=P(B)*P(C|B), and so on.

Though I am a bit surprised that the solution is the same as for independent events.

Anna
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