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

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Discussion Overview

The discussion centers on the expectation E(log(x-a)) where x is a log-normally distributed variable. Participants are seeking an analytical solution or a good numerical approximation for this expectation, under the condition that x-a>0.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant defines a log-normally distributed variable, stating that if X is log-normally distributed, then log X is normally distributed, leading to the expectation E[log X] = μ_Z.
  • Another participant clarifies that they are specifically interested in E[log(x-a)], not E[log(x)], indicating a different focus in the expectation being sought.
  • A later reply acknowledges a misunderstanding of the original question and admits a lack of knowledge regarding the expectation E[log(x-a)].

Areas of Agreement / Disagreement

Participants do not reach a consensus, and the discussion remains unresolved regarding the expectation E(log(x-a)).

Contextual Notes

The discussion does not provide specific assumptions or definitions regarding the parameters of the log-normal distribution or the variable a, which may affect the analysis.

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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 [itex]X[/itex] is said to be log-normally distributed if [itex]\log X[/itex] is normally distributed (I know, it's a weird naming convention). In other words, [itex]X= e^Z[/itex], where [itex]Z\sim \mathcal N(\mu_Z,\sigma_Z^2)[/itex], a normal random variable. So then [itex]\mathbb E [\log X]= E[Z] = \mu_Z[/itex].
 
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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