What Is the Expectation E(log(x-a)) for a Log-Normally Distributed Variable x?

AI Thread Summary
The discussion focuses on finding the expectation E(log(x-a)) for a log-normally distributed variable x, with the condition that x-a > 0. It clarifies that a log-normally distributed variable x can be expressed as X = e^Z, where Z follows a normal distribution. Participants express confusion about the specific expectation being sought, indicating a lack of clarity in the initial inquiry. There is a request for either an analytical solution or a good numerical approximation for the expectation. Overall, the conversation highlights the complexity of deriving the expectation for log-transformed variables in this context.
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What is the expectation, E(log(x-a)), when x is log normally distributed? Also x-a>0. I am looking for analytical solution or good numerical approximation.

Thanks
 
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The random variable X is said to be log-normally distributed if \log X is normally distributed (I know, it's a weird naming convention). In other words, X= e^Z, where Z\sim \mathcal N(\mu_Z,\sigma_Z^2), a normal random variable. So then \mathbb E [\log X]= E[Z] = \mu_Z.
 
Yes of course, but I am looking for E[log (x-a)] not E[log(x)].

Thanks.
 
Oh, yikes. I misread. Sorry for my useless answer.

I have no clue about your question.
 
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