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

[econmath]

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

 

[economicsnerd]

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 $\mathbb E [\log X]= E[Z] = \mu_Z$.

 

[econmath]

Yes of course, but I am looking for E[log (x-a)] not E[log(x)].

Thanks.

 

[economicsnerd]

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

I have no clue about your question.

 

[CellCoree]

for your question. The expectation, E(log(x-a)), of a log normally distributed variable x with a shift parameter a and x-a>0 can be calculated analytically using the following formula:

E(log(x-a)) = ln(a) + (μ + σ^2/2)

Where μ and σ are the mean and standard deviation of the underlying normal distribution for x. This formula is derived from the properties of log normal distributions and can be found in many textbooks on probability and statistics.

Alternatively, a good numerical approximation for the expectation can be obtained using numerical integration techniques. This involves dividing the range of x-a into small intervals and calculating the integral of log(x-a) multiplied by the probability density function of the log normal distribution within each interval. The sum of these integrals will give a good approximation of the expectation.

I hope this helps. Please let me know if you have any further questions.