Confidence Intervals for the Coefficient of Variation of the Delta-Lognormal Distribution

Abstract

This paper proposes confidence intervals for the coefficient of variation of the delta-lognormal distribution which are based on the generalized confidence interval (GCI) method, it is compared with the modified Fletcher method. The coverage probabilities and the expected lengths are used for assessing the performance of these confidence interval. Simulation results show that the GCI method perform best in term of the coverage probability and expected length.

A Appendix: Variance of \(\hat{\varvec{\xi } }\)

By Eq. (11), we have

$$\begin{aligned} \begin{aligned} \hat{\xi }&=\frac{1}{2}\ln \left[ \frac{\exp \left( \hat{\sigma }^{2} \right) -\hat{\delta }}{\hat{\delta }} \right] \\&=\frac{1}{2}\left[ \ln \left( \exp \left( \hat{\sigma }^{2} \right) -\hat{\delta }\right) -\ln \left( \hat{\delta } \right) \right] \\&=\frac{1}{2}\left\{ \hat{\sigma }^{2}+\ln \left[ 1-\frac{\hat{\delta }}{\exp \left( \hat{\sigma }^{2} \right) } \right] -\ln \left( \hat{\delta } \right) \right\} . \end{aligned} \end{aligned}$$

Since $\hat{\delta }\in \left[ 0,1 \right]$ and $\hat{\sigma }^{2}>0$, we can compute the limit

$$\begin{aligned} \lim _{\exp \left( \hat{\sigma }^{2} \right) \rightarrow \infty }\left[ \frac{1}{2}\left\{ \hat{\sigma }^{2}+\ln \left[ 1-\frac{\hat{\delta }}{\exp \left( \hat{\sigma }^{2} \right) } \right] -\ln \left( \hat{\delta } \right) \right\} \right] =\frac{1}{2}\left( \hat{\sigma }^{2}-\ln \left( \hat{\delta } \right) \right) . \end{aligned}$$

Now, we will find the variance of $\hat{\xi }$. Let $\mathbf T =T_{1},T_{2},...,T_{n}$ be the random variables defined as

$$\begin{aligned} T_{i}= {\left\{ \begin{array}{ll} 1 &{};X_{i}>0 \\ 0 &{};X_{i}=0. \end{array}\right. } \end{aligned}$$

Let $n_{1}$ have a truncated binomial distribution, $n_{1}=\sum _{i=1}^{n}T_{i}$, and we consider repeated sampling under the constraint $n_{1}>1$, as described in [7]. Then,

$$\begin{aligned} \begin{aligned} P\left( n_{1}=y \right)&=\frac{1}{1-\left( 1-\delta \right) ^{n}-n\delta \left( 1-\delta \right) ^{n-1}}\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y} \\&=\frac{1}{1-\left\{ \left( 1-\delta \right) ^{n-1} \left[ 1+\left( n-1 \right) \delta \right] \right\} }\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y} \\&=\frac{1}{1-ab}\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y} \quad ;y=2,3,...,n \\ \end{aligned} \end{aligned}$$

where $a=\left( 1-\delta \right) ^{n-1}$ and $b=1+\left( n-1 \right) \delta$. The variance of $\hat{\xi }$ is given by

$$\begin{aligned} V\left( \hat{\xi } \right) =V_\mathbf{T }\left[ E\left( \hat{\xi } \mid \mathbf T \right) \right] +E_\mathbf{T }\left[ V\left( \hat{\xi } \mid \mathbf T \right) \right] . \end{aligned}$$

Consider,

$$\begin{aligned} \begin{aligned} E\left( \hat{\xi } \mid \mathbf T \right)&=E\left[ \frac{1}{2}\left( \hat{\sigma }^{2}-\ln \left( \hat{\delta } \right) \right) \mid \mathbf T \right] \\&=\frac{1}{2}\left[ E\left( \hat{\sigma }^{2} \mid \mathbf T \right) - E\left( \ln \left( \hat{\delta } \right) \mid \mathbf T \right) \right] \\&=\frac{1}{2}\left( \sigma ^{2}-\ln \left( \hat{\delta } \right) \right) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} V\left( \hat{\xi } \mid \mathbf T \right)&=V\left[ \frac{1}{2}\left( \hat{\sigma }^{2}-\ln \left( \hat{\delta } \right) \right) \mid \mathbf T \right] \\&=\frac{1}{4}\left[ V\left( \hat{\sigma }^{2} \mid \mathbf T \right) + V\left( \ln \left( \hat{\delta } \right) \mid \mathbf T \right) \right] \\&=\frac{\sigma ^{4}}{2\left( n_{1}-1 \right) }. \end{aligned} \end{aligned}$$

Thus, by delta method we have

$$\begin{aligned} \begin{aligned} V\left( \hat{\xi } \right)&=V_\mathbf{T }\left[ \frac{1}{2}\left( \sigma ^{2}-\ln \left( \hat{\delta } \right) \right) \right] +E_\mathbf{T }\left[ \frac{\sigma ^{4}}{2\left( n_{1} -1\right) } \right] \\&=\frac{1}{4}\left[ V_\mathbf{T }\left( \sigma ^{2} \right) +V_\mathbf{T }\left( \ln \left( \hat{\delta } \right) \right) \right] +\frac{1}{2}E_\mathbf{T }\left[ \frac{\sigma ^{4}}{n_{1}-1} \right] \\&=\frac{1}{4}V_\mathbf{T }\left( \ln \left( \hat{\delta } \right) \right) +\frac{\sigma ^{4}}{2}E_\mathbf{T }\left[ \frac{1}{n_{1}-1} \right] \\&\approx \frac{1}{4}\left\{ \frac{V_\mathbf{T }\left( \hat{\delta } \right) }{\left[ E_\mathbf{T }\left( \hat{\delta } \right) \right] ^{2}} \right\} +\frac{\sigma ^{4}}{2}E_\mathbf{T }\left( \frac{1}{n_{1}-1} \right) \\&=\frac{1}{4}\left\{ \frac{V_\mathbf{T }\left( n_{1} \right) }{\left[ E_\mathbf{T }\left( n_{1} \right) \right] ^{2}} \right\} +\frac{\sigma ^{4}}{2}E_\mathbf{T }\left( \frac{1}{n_{1}-1} \right) \\&=\frac{1}{4}\left\{ \frac{E_\mathbf{T }\left( n_{1}^{2} \right) -\left[ E_\mathbf{T }\left( n_{1} \right) \right] ^{2}}{\left[ E_\mathbf{T }\left( n_{1} \right) \right] ^{2}} \right\} +\frac{\sigma ^{4}}{2}E_\mathbf{T }\left( \frac{1}{n_{1}-1} \right) \\&=\frac{1}{4}\left\{ \frac{E_\mathbf{T }\left( n_{1}^{2} \right) }{\left[ E_\mathbf{T }\left( n_{1} \right) \right] ^{2}}-1 \right\} +\frac{\sigma ^{4}}{2}E_\mathbf{T }\left( \frac{1}{n_{1}-1} \right) . \end{aligned} \end{aligned}$$

Now, we find

$$\begin{aligned} \begin{aligned} E_\mathbf{T }\left( n_{1} \right)&=\frac{1}{1-ab}\sum _{y=2}^{n}y\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y}\\&=\frac{1}{1-ab}\left\{ \sum _{y=0}^{n}y\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y} -\sum _{y=0}^{1}y\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y}\right\} \\&=\frac{n\delta \left( 1-a \right) }{1-ab} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} E_\mathbf{T }\left( n_{1}^{2} \right)&=\frac{1}{1-ab}\sum _{y=2}^{n}y^{2}\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y}\\&=\frac{1}{1-ab}\left\{ \sum _{y=0}^{n}y^{2}\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y} -\sum _{y=0}^{1}y^{2}\left( {\begin{array}{c}n\\ y\end{array}}\right) \delta ^{y}\left( 1-\delta \right) ^{n-y}\right\} \\&=\frac{n\delta \left( b-a \right) }{1-ab}. \end{aligned} \end{aligned}$$

Hence we have

$$\begin{aligned} \begin{aligned} V\left( \hat{\xi } \right)&=\frac{1}{4}\left\{ \frac{\frac{n\delta \left( b-a \right) }{1-ab}}{\left[ \frac{n\delta \left( 1-a \right) }{1-ab} \right] ^{2}} -1\right\} +\frac{\sigma ^{4}}{2}E_\mathbf{T }\left( \frac{1}{n_{1}-1} \right) \\&=\frac{1}{4}\left\{ \frac{\left( b-a \right) \left( 1-ab \right) }{n\delta \left( 1-a \right) ^{2} } -1\right\} +\frac{\sigma ^{4}}{2}E_\mathbf{T }\left( \frac{1}{n_{1}-1} \right) . \end{aligned} \end{aligned}$$

By normal theory results for the sample variance, we can obtain unbiased estimate for $\sigma ^{4}E_\mathbf{T }\left( \frac{1}{n_{1}-1} \right)$,

$$\begin{aligned} E\left( \frac{\hat{\sigma }^{4}}{n_{1}-1} \right) =E_\mathbf{T }\left\{ E\left( \frac{\hat{\sigma }^{4}}{n_{1}-1} \mid \mathbf T \right) \right\} =\sigma ^{4}E_\mathbf{T }\left( \frac{1}{n_{1}-1} \right) . \end{aligned}$$

Therefore, an approximately unbiased estimate of $V\left( \hat{\xi }\right)$ is

$$\begin{aligned} \hat{V}\left( \hat{\xi } \right) \approx \frac{\left( \hat{b}-\hat{a} \right) \left( 1-\hat{a}\hat{b} \right) -n_{1}\left( 1-\hat{a} \right) ^{2}}{4n_{1}\left( 1-\hat{a} \right) ^{2}}+\frac{\hat{\sigma }^{4} }{2\left( n_{1}-1 \right) } \end{aligned}$$

where $\hat{a}=\left( 1-\hat{\delta } \right) ^{n-1}$ and $\hat{b}=1+\left( n-1 \right) \hat{\delta }$.