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“…$$ many parameters on level 1js$$ 1\le j\le s $$. Hence, using a result of Raisbeck, 17 the likelihood function features a total of d=j=1sp0.3emn!false(nprefix−jfalse)!=ptrue(j=0nn!false(nprefix−jfalse)!prefix−n!n!prefix−n!false(nprefix−sprefix−1false)!j=0nprefix−sprefix−1false(nprefix−sprefix−1false)!false(nprefix−sprefix−1prefix−jfalse)!true)=ptrue(e0.3emn!prefix−1prefix−n!efalse(nprefix−sprefix−1false)!false(nprefix−sprefix−1false)!true)$$ d=\sum \limits_{j=1}^sp\kern0.3em \frac{n!}{\left(n-j\right)! }=p\left(\sum \limits_{j=0}^n\frac{n!}{\left(n-j\right)!}-\frac{n!}{n!}-\frac{n!}{\left(n-s-1\right)!…”
Section: Likelihood Ratio Testmentioning
confidence: 98%
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“…$$ many parameters on level 1js$$ 1\le j\le s $$. Hence, using a result of Raisbeck, 17 the likelihood function features a total of d=j=1sp0.3emn!false(nprefix−jfalse)!=ptrue(j=0nn!false(nprefix−jfalse)!prefix−n!n!prefix−n!false(nprefix−sprefix−1false)!j=0nprefix−sprefix−1false(nprefix−sprefix−1false)!false(nprefix−sprefix−1prefix−jfalse)!true)=ptrue(e0.3emn!prefix−1prefix−n!efalse(nprefix−sprefix−1false)!false(nprefix−sprefix−1false)!true)$$ d=\sum \limits_{j=1}^sp\kern0.3em \frac{n!}{\left(n-j\right)! }=p\left(\sum \limits_{j=0}^n\frac{n!}{\left(n-j\right)!}-\frac{n!}{n!}-\frac{n!}{\left(n-s-1\right)!…”
Section: Likelihood Ratio Testmentioning
confidence: 98%
“…(n−j)! many parameters on level 1 ≤ j ≤ s. Hence, using a result of Raisbeck, 17 the likelihood function features a total of…”
Section: Likelihood Ratio Testmentioning
confidence: 99%