Fuzzy hypothesis testing with vague data using likelihood ratio test

Abstract

Hypothesis testing is one of the most significant facets of statistical inference, which like other situations in the real world is definitely affected by uncertain conditions. The aim of this paper is to develop hypothesis testing based on likelihood ratio test in fuzzy environment, where it is supposed that both hypotheses under study and sample data are fuzzy. The main idea is to employ Zadeh’s extension principle. In this regard, a pair of non-linear programming problems is exploited toward obtaining membership function of likelihood ratio test statistic. Afterwards, the membership function is compared with critical value of the test in order to assess acceptability of the fuzzy null hypothesis under consideration. In this step, two distinct procedures are applied. In the first procedure, a ranking method for fuzzy numbers is utilized to make an absolute decision about acceptability of fuzzy null hypothesis. From a different point of view, in the second procedure, membership degrees of fuzzy null hypothesis acceptance and rejection are first derived using resolution identity and then, a relative decision is made on fuzzy null hypothesis acceptance or rejection based on some arbitrary decision rules. Flexibility of the proposed approach in testing fuzzy hypothesis with vague data is presented using some numerical examples.

Appendices

Appendix A

This appendix gives the likelihood ratio statistic in crisp environment when one is going to evaluate the null hypothesis ‘H: $\mu = \mu_{0}$’ against the alternative hypothesis ‘K: $\mu \ne \mu_{0}$’. This statistic relates to the Example 1.

Suppose $X_{1} , \ldots ,X_{n}$ is a crisp random sample taken from a normally distributed population with the unknown variance $\sigma^{2}$. In this case, likelihood ratio function is as follows:

$$ L(\mu,\sigma^{2} )=\prod\limits_{i = 1}^{n} {f(x_{i} ;\mu, \sigma^{2})}=(\sigma \sqrt {2n})^{ - n}.\,\exp \left(-\sum\limits_{i = 1}^{n} {{{(x_{i} - \mu )^{2} }/{2\sigma^{2}}}}\right) $$

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Assume $\theta = (\mu,\sigma^{2} )$. In this case, the above function should be maximized to obtain the likelihood ratio statistic $\Uplambda$ when $\theta \in \Upomega_{h}$ and $\theta \in \Upomega$ giving $L(x_{1} , \ldots ,x_{n} ;\theta^{\prime})$ and $L(x_{1} , \ldots ,x_{n} ;\theta^{\prime\prime})$, respectively. Since there is just one value for $\mu$ when $\theta \in \Upomega_{h}$, its likelihood ratio estimation is $\mu_{0}$. This estimation is also equal to $\bar{X}$ when $\theta \in \Upomega$. On the other hand, likelihood ratio estimation of $\sigma^{2}$ is equal to $\sum\nolimits_{i = 1}^{n} {{{(x_{i} - \hat{\mu })^{2} }/n}}$ where $\hat{\mu }$ is likelihood ratio estimation of $\mu$. Therefore, likelihood ratio estimation of $\sigma^{2}$ is equal to $\sum\nolimits_{i = 1}^{n} {{{(X_{i} - \mu_{0} )^{2} }/n}}$ when $\theta \in \Upomega_{h}$ and is equal to $\sum\nolimits_{i = 1}^{n} {{{(X_{i} - \bar{X})^{2} }/n}}$ when $\theta \in \Upomega$. Hence, we have

$$ L(\theta^{\prime};x_{1} , \ldots ,x_{n} ) = \left(\sqrt {2\sum\nolimits_{i = 1}^{n} {(X_{i} - \mu_{0} )^{2} } } \right)^{ - n} .\exp ({{ - n} \mathord{\left/ {\vphantom {{ - n} {2)}}} \right. \kern-\nulldelimiterspace} {2)}} $$

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$$ L(\theta^{\prime\prime};x_{1} , \ldots ,x_{n} ) = \left(\sqrt {2\sum\nolimits_{i = 1}^{n} {(X_{i} - \bar{X})^{2} } } \right)^{ - n} .\exp ({{ - n} \mathord{\left/ {\vphantom {{ - n} {2)}}} \right. \kern-\nulldelimiterspace} {2)}} $$

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Consequently, based on the Eq. (1), the likelihood ratio test statistic is as follows:

$$ \Uplambda = \left( {{{\sum\nolimits_{i = 1}^{n} {(X_{i} - \bar{X})^{2} } } \mathord{\left/ {\vphantom {{\sum\nolimits_{i = 1}^{n} {(X_{i} - \bar{X})^{2} } } {\sum\nolimits_{i = 1}^{n} {(X_{i} - \mu_{0} )^{2} } }}} \right. \kern-\nulldelimiterspace} {\sum\nolimits_{i = 1}^{n} {(X_{i} - \mu_{0} )^{2} } }}} \right)^{{{n \mathord{\left/ {\vphantom {n 2}} \right. \kern-\nulldelimiterspace} 2}}} $$

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Regarding sample data, sample mean and population mean as fuzzy, Eq. (18) is obtained. □

Appendix B

This Appendix presents a proof of Lemma 1 introduced in the Example 1. Suppose $A = \sum\nolimits_{i = 1}^{n} {(X_{i} - \bar{X}} )^{2} = (n - 1)\,S^{2}$, where $S^{2}$ is the sample variance. Therefore,

$$ \begin{aligned}\Uplambda^{\frac{2}{n}}&=A/(A + n(\bar{X} - \mu_{0} )^{2} ) \\&=1/(1 + nA^{ - 1} (\bar{X} - \mu_{0} )^{2} )=1/(1 + (n/n - 1)S^{ - 2} (\bar{X} - \mu_{0} )^{2} )\end{aligned} $$

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Since $T = {{\sqrt n (\bar{X} - \mu )} \mathord{\left/ {\vphantom {{\sqrt n (\bar{X} - \mu )} S}} \right. \kern-\nulldelimiterspace} S}$, we have

$$ \Uplambda^{2/n}=(1+T^{2}/(n - 1))^{ - 1} $$

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On the other hand, since $S = \sqrt {\sum\nolimits_{i = 1}^{n} {{{(X_{i} - \bar{X})^{2} } \mathord{\left/ {\vphantom {{(X_{i} - \bar{X})^{2} } {(n - 1)}}} \right. \kern-\nulldelimiterspace} {(n - 1)}}} }$, we have

$$ \Uplambda^{{{2 \mathord{\left/ {\vphantom {2 n}} \right. \kern-\nulldelimiterspace} n}}} = {{\sum\nolimits_{i = 1}^{n} {(X_{i} - \bar{X})^{2} } } \mathord{\left/ {\vphantom {{\sum\nolimits_{i = 1}^{n} {(X_{i} - \bar{X})^{2} } } {\sum\nolimits_{i = 1}^{n} {(X_{i} - \mu )^{2} } }}} \right. \kern-\nulldelimiterspace} {\sum\nolimits_{i = 1}^{n} {(X_{i} - \mu )^{2} } }} = \left( {1 + {{T^{2} } \mathord{\left/ {\vphantom {{T^{2} } {(n - 1)}}} \right. \kern-\nulldelimiterspace} {(n - 1)}}} \right)^{ - 1} $$

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