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Confidence for the lognormal…

An intermediate exercise in Mathematical statistics is to construct a (100-\alpha)% confidence interval for the mean of a random variable distributed according to the log normal law (X\tilde{\ }LN\left( \mu,\sigma _{{}}^{2}\right).)

The mean is

E\left( X \right)=\exp \left( \mu +\frac{\sigma _{{}}^{2}}{2} \right).


\log \theta =\mu +\frac{\sigma _{{}}^{2}}{2}.

The MVUE is

T\left( Y \right)=\bar{Y}+\frac{s_{{}}^{2}}{2}

, where Y=\log X,{{s}^{2}}=\frac{1}{n-1}\sum{{{\left( Y-\bar{Y} \right)}^{2}}}.

The variance of T\left( Y \right) is

Var\left\{ T\left( Y \right) \right\}=\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}.

Then evidently

\frac{\bar{Y}+\frac{s_{{}}^{2}}{2}-\log \theta }{\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}}}\xrightarrow{d}Z

, where Z\tilde{\ }N\left( 0,1 \right).

To sum thinks up the confidence interval for  mean of X fixing  \alpha=.05 is

\left( \exp \left\{ \bar{Y}+\frac{s_{{}}^{2}}{2}-1.96\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}} \right\},\exp \left\{ \bar{Y}+\frac{s_{{}}^{2}}{2}+1.96\sqrt{\frac{s_{{}}^{2}}{n}+\frac{s_{{}}^{4}}{2\left( n-1 \right)}} \right\} \right).

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