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Here's the problem which confused me:

Claim sizes for an insurance coverage follow a lognormal distribution with mean 1000 and medium 800. Determine the probability that a claim will be greater than 1200.

I have given that for a lognormal distribution, E(X) = $e^{\mu + \sigma^2/2}$

Which means that E(x) is not simply $\mu$. Isn't that the definition of E(X)?

(The solution continues - since the medium is 800, $e^\mu$ = 800 ...)

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up vote 3 down vote accepted

$\mu$ is the mean of $\log(X)$.

If $X$ is a Log-normally-distributed variable, then $Y=\log(X)$ is normally-distributed variable. Historically, it is common to parametrize the distribution of $Y$ it by $\mu$ and $\sigma$ of $X$. They are, however, different than the actual expectation value and standard deviation $Y$. This is because, in general, $$\mathbb{E}(\log(X))\ne \mathbb{E}(X)$$

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Thanks, that makes much more sense! Do you know if that's historically, true for any other distributions, where $\mu$ isn't the mean of the distribution? – Eliyahu Dec 19 '12 at 12:10
With respect to the actual question, the remark to make is rather that $E(\log X)\ne E(X)$. – Did Dec 19 '12 at 12:10
@did Of course. Corrected. – yohBS Dec 19 '12 at 12:13

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