Question 15.3: Consider the i.i.d. r.v.’s X1, . . . , X20 with continuous d......

Consider the i.i.d. r.v.’s X_{1},\,\ldots,\,X_{20} with continuous d.f. F, which has unique x_{0.50},\mathcal{x}_{0.25},\mathrm{~and~}x_{0.75}, and let \,Y_{1},\,\ldots,\,Y_{20} be the corresponding order statistics. Then consider several confidence intervals for x_{0.50},x_{0.25},\mathrm{and}\;x_{0.75}, and calculate the respective coverage probabilities.

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Using formula (11),
P(Y_{i}\leq x_{p}\leq Y_{j})=\sum\limits_{k=i}^{n}{\binom{n}{k}}p^{k}(1-p)^{n-k}-P(Y_{j}\leq x_{p})

~~=\sum\limits_{k=i}^{n}{\binom{n}{k}}p^{k}(1-p)^{n-k}-\sum\limits_{k=j}^{n}{\binom{n}{k}}p^{k}(1-p)^{n-k}

~~=\sum\limits_{k=i}^{j-1}{\binom{n}{k}}p^{k}(1-p)^{n-k}.           (11)

we obtain the coverage probabilities listed in Table 15.1 for several confidence intervals for the median x_{0.50} and the first quartile x_{0.25}. For the calculation of coverage probabilities for confidence intervals for the third quartile x_{0.75}, we employ the following formula, which allows us to use the Binomial tables; namely,
\sum\limits_{k=i}^{j-1}{\binom{20}{k}}(0.75)^{k}(0.25)^{20-k}=\sum\limits_{r=20-j+1}^{20-i}{\binom{20}{r}}(0.25)^{r}(0.75)^{20-r}.

Table 15.1
Quantile Confidence Interval Coverage Probability
x_{0.50} (Y_{9},Y_{12}) 0.3364
(Y_{8},Y_{13}) 0.6167
(Y_{7},Y_{14}) 0.8107
(Y_{6},Y_{15}) 0.9216
x_{0.25} (Y_{6},\,Y_{6}) 0.2024
(Y_{3},Y_{7}) 0.5606
(Y_{2},\,Y_{8}) 0.8069
(Y_{1},\,Y_{9}) 0.9348
x_{0.75} (Y_{15},Y_{17}) 0.2024
(Y_{14},Y_{18}) 0.5606
(Y_{13},Y_{19}) 0.8069
(Y_{12},Y_{20}) 0.9348

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