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I have a variable X that is distributed log-normally.

I let Y = lnX ~ N($\mu$, $\sigma^2$) and I've been given that $\sigma$=0.3, $\bar{y}$ = 0.12 and n = 40.

So I find a confidence interval for the mean of the log-transformed data like this:

$(\bar{y}-z_{1-\alpha/2}\times\frac{\sigma}{\sqrt n}, \bar{y}+z_{1-\alpha/2}\times\frac{\sigma}{\sqrt n})\\ (0.12-1.96\times\frac{0.3}{\sqrt 40}, 0.12+1.96\times\frac{0.3}{\sqrt 40})\\ (0.027, 0.213)$

To get the 95% confidence interval for E(X) (the original variable) I just raise e to the power of the endpoints of the interval I just calculated.

so the interval would be $(e^{0.027}, e^{0.213})=\\$ $(1.03, 1.24)$

Is this correct?

Thanks any help appreciated

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  • $\begingroup$ Looks good to me. $\endgroup$ – msitt May 17 '17 at 14:35
  • $\begingroup$ The problem is that $E(X)$ is not $e^{\mu}$. In fact for a log-normal, $e^{\mu}$ is the median. See ww2.amstat.org/publications/jse/v13n1/olsson.html and stats.stackexchange.com/questions/33382/… $\endgroup$ – Just_to_Answer May 17 '17 at 14:45
  • $\begingroup$ @Kane: The mean of lognormal is $\exp(\mu + (1/2) \sigma ^2)$. Your confidence interval before exponentiation is only for $\mu$. You should take a serious look at the links. $\endgroup$ – Just_to_Answer May 18 '17 at 2:11
  • $\begingroup$ @Just_to_Answer I actually used those links to create my answer. I know that the mean of lognormal is $e^{\mu + \frac{1}{2}\sigma^{2}}$. I guess i don't understand how that changes the answer $\endgroup$ – Kane May 18 '17 at 2:18
  • $\begingroup$ @Kane: There are 2 issues. The bigger issue is what I mention in the comment above that mean of lognormal is not exponentiated mean of the corresponding normal. The other smaller issue is the preservation under exponentiating the confidence interval of the log transformed variables. $\endgroup$ – Just_to_Answer May 18 '17 at 2:18
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Quoting https://ww2.amstat.org/publications/jse/v13n1/olsson.html using $\theta=EX$

It would seem natural to use the following "naïve" approach for calculating a confidence interval for $\theta$. A confidence interval for $\mu$ is calculated using standard methods. The limits of the confidence interval are back-transformed to give the limits in a confidence interval for $\theta$. [...an example illustrated...] This illustrates the fact that the naïve method gives a biased estimator of $\theta$

The method outlined by OP is this biased method.

The paper offers a few alternatives including Cox Method. If, as in OP's case $\sigma^2$ is known, a modification should be made to the Cox method. This modification would be to add $(1/2)\sigma^2$ to both ends of the confidence interval for $\mu$ and then take anti-log's.

The end of the paper also provides simulation studies to assess the coverage of the confidence intervals under each method.

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Quantiles and percentiles are preserved under increasing transformations like $\exp(x)$ and $\ln(x)$ so the 5th and 95th percentile of $X$ will still be the 5th and 95th percentile when transformed through $\exp$. (the preservation property is mentioned on Wikipedia).

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  • $\begingroup$ But unfortunately, $E(X)$ is not $e^{\mu}$. Please see ww2.amstat.org/publications/jse/v13n1/olsson.html $\endgroup$ – Just_to_Answer May 17 '17 at 14:42
  • $\begingroup$ To be emphatic, regardless of the exp/ln functions preserve percentiles or not - the confidence interval before exponentiation is a confidence interval for $\mu$. The mean of lognormal is $\exp(\mu + (1/2) \sigma ^2)$. NOT $\exp(\mu)$. So this answer is wrong. $\endgroup$ – Just_to_Answer May 18 '17 at 2:06
  • $\begingroup$ @Just_to_Answer could you please reply to my question showing helping me to understand where my understanding is incorrect? $\endgroup$ – Kane May 18 '17 at 2:09
  • $\begingroup$ @Just_to_Answer you're quite right, this does not answer OP's actual question at all. I accidentally mistook the question as one about percentiles of lognormal variables. And agreed, the mean is not a percentile (unlike the median) and is therefore not preserved, preventing the application of the above principle. $\endgroup$ – Bonnevie May 30 '17 at 12:15

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