Discontinuous Distribution?


Discontinuous Distribution?

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gmerton
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Hello,

I'm not sure I agree with the statement: "The likelihood of a loss of size L is L/300,000." This is only true if losses are rounded off to the nearest dollar- which leads to other problems. For example, if you're allowed to lose any amount to the nearest cent, then the sum of p(lossing L) = (1/300,000)*30,000,000 = 100. But of course, the sum of all probabilities must be one. So, okay, let's assume in the problem that we're only allowed discrete dollar values. But if we do this, then any talk of integration is invalid since the density function is discontinuous.

So right now, you're stuck. If you say "the likelihood of a loss of size L is L/300,000" you're implying a discrete distribution, which invalidates integration.

[NEAS: Likelihood is not the same as probability.  If the size of loss distribution is continuous, the probability of a loss of size L is zero, but the likelihood is the pdf.  Likelihoods and probabilities are covered on the actuarial exams in the textbook by Klugman, Panjer, and Willmot.]


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