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function that is finite but has infinite expectation
- Thread starter vkaul
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- 1/1/09
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Cauchy distribution doesn't have well defined expectation (do you remember Lebesgue integral?) 
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Let \(X\) be r.v. with following distribution \(\mathbb{P}(X=2^n)=c\frac{1}{n^2},n\in\mathbb{N}\) (where \(c\) is a constant such that it all sums up to 1). It is trivial to check that it has a well defined expectation (trivial since r.v. is nonnegative), that is a.e. finite (\(2^n<\infty,\forall n\in\mathbb{N}\)) and that it has infinite expectation (\(2^n\) grows faster than \(n^2\)).
. Let \(X\) be r.v. with following distribution \(\mathbb{P}(X=2^n)=c\frac{1}{n^2},n\in\mathbb{N}\) (where \(c\) is a constant such that it all sums up to 1). It is trivial to check that it has a well defined expectation (trivial since r.v. is nonnegative), that is a.e. finite (\(2^n<\infty,\forall n\in\mathbb{N}\)) and that it has infinite expectation (\(2^n\) grows faster than \(n^2\)).
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