checkRd: (-1) ddst-package.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$W_k=[1/sqrt(n) sum_{i=1}^n l(Z_i)]I^{-1}[1/sqrt(n) sum_{i=1}^n l(Z_i)]'$},
       |                           ^
checkRd: (-1) ddst-package.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$W_k=[1/sqrt(n) sum_{i=1}^n l(Z_i)]I^{-1}[1/sqrt(n) sum_{i=1}^n l(Z_i)]'$},
       |                                            ^
checkRd: (-1) ddst-package.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$W_k=[1/sqrt(n) sum_{i=1}^n l(Z_i)]I^{-1}[1/sqrt(n) sum_{i=1}^n l(Z_i)]'$},
       |                                                               ^
checkRd: (-1) ddst-package.Rd:31: Lost braces; missing escapes or markup?
    31 | where \emph{$l(Z_i)$}, i=1,...,n, is \emph{k}-dimensional (row) score vector, the symbol \emph{'} denotes transposition while \emph{$I=Cov_{theta_0}[l(Z_1)]'[l(Z_1)]$}. Following Neyman's idea of modelling underlying distributions one gets \emph{$l(Z_i)=(phi_1(F(Z_i)),...,phi_k(F(Z_i)))$} and \emph{I} being the identity matrix, where \emph{$phi_j$}'s, j >= 1,  are zero mean orthonormal functions on [0,1], while \emph{F} is the completely specified null distribution function.
       |                                                                                                                                            ^
checkRd: (-1) ddst-package.Rd:35: Lost braces; missing escapes or markup?
    35 | \emph{$W_k^{*}(tilde gamma)=[1/sqrt(n) sum_{i=1}^n l^*(Z_i;tilde gamma)][I^*(tilde gamma)]^{-1}[1/sqrt(n) sum_{i=1}^n l^*(Z_i;tilde gamma)]'$},
       |            ^
checkRd: (-1) ddst-package.Rd:35: Lost braces; missing escapes or markup?
    35 | \emph{$W_k^{*}(tilde gamma)=[1/sqrt(n) sum_{i=1}^n l^*(Z_i;tilde gamma)][I^*(tilde gamma)]^{-1}[1/sqrt(n) sum_{i=1}^n l^*(Z_i;tilde gamma)]'$},
       |                                            ^
checkRd: (-1) ddst-package.Rd:35: Lost braces; missing escapes or markup?
    35 | \emph{$W_k^{*}(tilde gamma)=[1/sqrt(n) sum_{i=1}^n l^*(Z_i;tilde gamma)][I^*(tilde gamma)]^{-1}[1/sqrt(n) sum_{i=1}^n l^*(Z_i;tilde gamma)]'$},
       |                                                                                            ^
checkRd: (-1) ddst-package.Rd:35: Lost braces; missing escapes or markup?
    35 | \emph{$W_k^{*}(tilde gamma)=[1/sqrt(n) sum_{i=1}^n l^*(Z_i;tilde gamma)][I^*(tilde gamma)]^{-1}[1/sqrt(n) sum_{i=1}^n l^*(Z_i;tilde gamma)]'$},
       |                                                                                                               ^
checkRd: (-1) ddst-package.Rd:36: Lost braces; missing escapes or markup?
    36 | where \emph{$tilde gamma$} is an appropriate estimator of \emph{$gamma$} while \emph{$I^*(gamma)=Cov_{theta_0}[l^*(Z_1;gamma)]'[l^*(Z_1;gamma)]$}. More details can be found in Janic and Ledwina (2008), Kallenberg and Ledwina (1997 a,b) as well as Inglot and Ledwina (2006 a,b). 
       |                                                                                                      ^
checkRd: (-1) ddst-package.Rd:40: Lost braces
    40 | \emph{$T = min{1 <= k <= d: W_k-pi(k,n,c) >= W_j-pi(j,n,c), j=1,...,d}$}
       |               ^
checkRd: (-1) ddst-package.Rd:45: Lost braces
    45 | $T^* = min{1 <= k <= d: W_k^*(tilde gamma)-pi^*(k,n,c) >= W_j^*(tilde gamma)-pi^*(j,n,c), j=1,...,d}$}.
       |           ^
checkRd: (-1) ddst-package.Rd:49: Lost braces
    49 | \emph{$pi(j,n,c)={jlog n,  if  max{1 <= k <= d}|Y_k| <= sqrt(c log(n)), 2j,  if max{1 <= k <= d}|Y_k|>sqrt(c log(n)). }$}
       |                  ^
checkRd: (-1) ddst-package.Rd:49: Lost braces
    49 | \emph{$pi(j,n,c)={jlog n,  if  max{1 <= k <= d}|Y_k| <= sqrt(c log(n)), 2j,  if max{1 <= k <= d}|Y_k|>sqrt(c log(n)). }$}
       |                                   ^
checkRd: (-1) ddst-package.Rd:49: Lost braces
    49 | \emph{$pi(j,n,c)={jlog n,  if  max{1 <= k <= d}|Y_k| <= sqrt(c log(n)), 2j,  if max{1 <= k <= d}|Y_k|>sqrt(c log(n)). }$}
       |                                                                                    ^
checkRd: (-1) ddst-package.Rd:54: Lost braces
    54 | $pi^*(j,n,c)={jlog n,  if max{1 <= k <= d}|Y_k^*| <= sqrt(c log(n)),2j  if max(1 <= k <= d)|Y_k^*| > sqrt(c log(n))}$}.
       |              ^
checkRd: (-1) ddst-package.Rd:54: Lost braces
    54 | $pi^*(j,n,c)={jlog n,  if max{1 <= k <= d}|Y_k^*| <= sqrt(c log(n)),2j  if max(1 <= k <= d)|Y_k^*| > sqrt(c log(n))}$}.
       |                              ^
checkRd: (-1) ddst-package.Rd:58: Lost braces; missing escapes or markup?
    58 | \emph{$(Y_1,...,Y_k)=[1/sqrt(n) sum_{i=1}^n l(Z_i)]I^{-1/2}$} 
       |                                     ^
checkRd: (-1) ddst-package.Rd:58: Lost braces; missing escapes or markup?
    58 | \emph{$(Y_1,...,Y_k)=[1/sqrt(n) sum_{i=1}^n l(Z_i)]I^{-1/2}$} 
       |                                                      ^
checkRd: (-1) ddst-package.Rd:62: Lost braces; missing escapes or markup?
    62 | \emph{$(Y_1^*,...,Y_k^*)=[1/sqrt(n) sum_{i=1}^n l^*(Z_i; tilde gamma)][I^*(tilde gamma)]^{-1/2}$}.
       |                                         ^
checkRd: (-1) ddst-package.Rd:62: Lost braces; missing escapes or markup?
    62 | \emph{$(Y_1^*,...,Y_k^*)=[1/sqrt(n) sum_{i=1}^n l^*(Z_i; tilde gamma)][I^*(tilde gamma)]^{-1/2}$}.
       |                                                                                          ^
checkRd: (-1) ddst-package.Rd:65: Lost braces; missing escapes or markup?
    65 | and \emph{$W_{T^*} = W_{T^*}(tilde gamma)$}, respectively. For details see Inglot and Ledwina (2006 a,b,c). 
       |              ^
checkRd: (-1) ddst-package.Rd:65: Lost braces; missing escapes or markup?
    65 | and \emph{$W_{T^*} = W_{T^*}(tilde gamma)$}, respectively. For details see Inglot and Ledwina (2006 a,b,c). 
       |                        ^
checkRd: (-1) ddst-package.Rd:67: Lost braces; missing escapes or markup?
    67 | The choice of \emph{c} in \emph{T} and \emph{$T^*$} is decisive to finite sample behaviour of the selection rules and pertaining statistics \emph{$W_T$} and \emph{$W_{T^*}(tilde gamma)$}. In particular, under large \emph{c}'s the rules behave similarly as Schwarz's (1978) BIC while for \emph{c=0} they mimic Akaike's (1973) AIC. For moderate sample sizes, values \emph{c in (2,2.5)} guarantee, under `smooth' departures, only slightly smaller power as in case BIC were used and simultaneously give much higher power than BIC under multimodal alternatives. In genral, large \emph{c's} are recommended if changes in location, scale, skewness and kurtosis are in principle aimed to be detected. For evidence and discussion see Inglot and Ledwina (2006 c). 
       |                                                                                                                                                                       ^
checkRd: (-1) ddst-package.Rd:69: Lost braces; missing escapes or markup?
    69 | It \emph{c>0} then the limiting null distribution of \emph{$W_T$} and \emph{$W_{T^*}(tilde gamma)$} is central chi-squared with one degree of freedom. In our implementation, for given \emph{n}, both critical values and \emph{p}-values are computed by MC method.
       |                                                                                ^
checkRd: (-1) ddst-package.Rd:71: Lost braces; missing escapes or markup?
    71 | Empirical distributions of \emph{T} and \emph{$T^*$} as well as \emph{$W_T$} and \emph{$W_{T^*}(tilde gamma)$} are not essentially influenced by the choice of reasonably large \emph{d}'s, provided that sample size is at least moderate.
       |                                                                                           ^
checkRd: (-1) ddst.exp.test.Rd:27: Lost braces; missing escapes or markup?
    27 | Modelling alternatives similarly as in Kallenberg and Ledwina (1997 a,b), e.g., and estimating \emph{$gamma$} by \emph{$tilde gamma= 1/n sum_{i=1}^n Z_i$} yields the efficient score 
       |                                                                                                                                              ^
checkRd: (-1) ddst.exp.test.Rd:30: Lost braces; missing escapes or markup?
    30 | The matrix \emph{$[I^*(tilde gamma)]^{-1}$} does not  depend on \emph{$tilde gamma$} and is calculated for succeding dimensions \emph{k} using some recurrent relations for Legendre's polynomials and computed in a numerical way in case of cosine basis. In the implementation the default value of \emph{c} in \emph{$T^*$} is set to be 100. 
       |                                      ^
checkRd: (-1) ddst.extr.test.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$gamma=(gamma_1,gamma_2)$} is estimated by \emph{$tilde gamma=(tilde gamma_1,tilde gamma_2)$}, where \emph{$tilde gamma_1=-1/n sum_{i=1}^n Z_i + varepsilon  G$}, where \emph{$varepsilon approx 0.577216 $} is the Euler constant and \emph{$ G = tilde gamma_2 = [n(n-1) ln2]^{-1}sum_{1<= j < i <= n}(Z_{n:i}^o - Z_{n:j}^o) $} while \emph{$Z_{n:1}^o <= ... <= Z_{n:n}^o$}
       |                                                                                                                                          ^
checkRd: (-1) ddst.extr.test.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$gamma=(gamma_1,gamma_2)$} is estimated by \emph{$tilde gamma=(tilde gamma_1,tilde gamma_2)$}, where \emph{$tilde gamma_1=-1/n sum_{i=1}^n Z_i + varepsilon  G$}, where \emph{$varepsilon approx 0.577216 $} is the Euler constant and \emph{$ G = tilde gamma_2 = [n(n-1) ln2]^{-1}sum_{1<= j < i <= n}(Z_{n:i}^o - Z_{n:j}^o) $} while \emph{$Z_{n:1}^o <= ... <= Z_{n:n}^o$}
       |                                                                                                                                                                                                                                                                                       ^
checkRd: (-1) ddst.extr.test.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$gamma=(gamma_1,gamma_2)$} is estimated by \emph{$tilde gamma=(tilde gamma_1,tilde gamma_2)$}, where \emph{$tilde gamma_1=-1/n sum_{i=1}^n Z_i + varepsilon  G$}, where \emph{$varepsilon approx 0.577216 $} is the Euler constant and \emph{$ G = tilde gamma_2 = [n(n-1) ln2]^{-1}sum_{1<= j < i <= n}(Z_{n:i}^o - Z_{n:j}^o) $} while \emph{$Z_{n:1}^o <= ... <= Z_{n:n}^o$}
       |                                                                                                                                                                                                                                                                                               ^
checkRd: (-1) ddst.extr.test.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$gamma=(gamma_1,gamma_2)$} is estimated by \emph{$tilde gamma=(tilde gamma_1,tilde gamma_2)$}, where \emph{$tilde gamma_1=-1/n sum_{i=1}^n Z_i + varepsilon  G$}, where \emph{$varepsilon approx 0.577216 $} is the Euler constant and \emph{$ G = tilde gamma_2 = [n(n-1) ln2]^{-1}sum_{1<= j < i <= n}(Z_{n:i}^o - Z_{n:j}^o) $} while \emph{$Z_{n:1}^o <= ... <= Z_{n:n}^o$}
       |                                                                                                                                                                                                                                                                                                                  ^
checkRd: (-1) ddst.extr.test.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$gamma=(gamma_1,gamma_2)$} is estimated by \emph{$tilde gamma=(tilde gamma_1,tilde gamma_2)$}, where \emph{$tilde gamma_1=-1/n sum_{i=1}^n Z_i + varepsilon  G$}, where \emph{$varepsilon approx 0.577216 $} is the Euler constant and \emph{$ G = tilde gamma_2 = [n(n-1) ln2]^{-1}sum_{1<= j < i <= n}(Z_{n:i}^o - Z_{n:j}^o) $} while \emph{$Z_{n:1}^o <= ... <= Z_{n:n}^o$}
       |                                                                                                                                                                                                                                                                                                                              ^
checkRd: (-1) ddst.extr.test.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$gamma=(gamma_1,gamma_2)$} is estimated by \emph{$tilde gamma=(tilde gamma_1,tilde gamma_2)$}, where \emph{$tilde gamma_1=-1/n sum_{i=1}^n Z_i + varepsilon  G$}, where \emph{$varepsilon approx 0.577216 $} is the Euler constant and \emph{$ G = tilde gamma_2 = [n(n-1) ln2]^{-1}sum_{1<= j < i <= n}(Z_{n:i}^o - Z_{n:j}^o) $} while \emph{$Z_{n:1}^o <= ... <= Z_{n:n}^o$}
       |                                                                                                                                                                                                                                                                                                                                                         ^
checkRd: (-1) ddst.extr.test.Rd:29: Lost braces; missing escapes or markup?
    29 | \emph{$gamma=(gamma_1,gamma_2)$} is estimated by \emph{$tilde gamma=(tilde gamma_1,tilde gamma_2)$}, where \emph{$tilde gamma_1=-1/n sum_{i=1}^n Z_i + varepsilon  G$}, where \emph{$varepsilon approx 0.577216 $} is the Euler constant and \emph{$ G = tilde gamma_2 = [n(n-1) ln2]^{-1}sum_{1<= j < i <= n}(Z_{n:i}^o - Z_{n:j}^o) $} while \emph{$Z_{n:1}^o <= ... <= Z_{n:n}^o$}
       |                                                                                                                                                                                                                                                                                                                                                                             ^
checkRd: (-1) ddst.extr.test.Rd:33: Lost braces; missing escapes or markup?
    33 | The related matrix \emph{$[I^*(tilde gamma)]^{-1}$} does not  depend on \emph{$tilde gamma$} and is calculated for succeding dimensions \emph{k} using some recurrent relations for Legendre's polynomials and numerical methods for cosine functions. In the implementation the default value of \emph{c} in \emph{$T^*$} was fixed to be 100. Hence, \emph{$T^*$} is Schwarz-type model selection rule. The resulting data driven test statistic for extreme value distribution is \emph{$W_{T^*}=W_{T^*}(tilde gamma)$}.
       |                                              ^
checkRd: (-1) ddst.extr.test.Rd:33: Lost braces; missing escapes or markup?
    33 | The related matrix \emph{$[I^*(tilde gamma)]^{-1}$} does not  depend on \emph{$tilde gamma$} and is calculated for succeding dimensions \emph{k} using some recurrent relations for Legendre's polynomials and numerical methods for cosine functions. In the implementation the default value of \emph{c} in \emph{$T^*$} was fixed to be 100. Hence, \emph{$T^*$} is Schwarz-type model selection rule. The resulting data driven test statistic for extreme value distribution is \emph{$W_{T^*}=W_{T^*}(tilde gamma)$}.
       |                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               ^
checkRd: (-1) ddst.extr.test.Rd:33: Lost braces; missing escapes or markup?
    33 | The related matrix \emph{$[I^*(tilde gamma)]^{-1}$} does not  depend on \emph{$tilde gamma$} and is calculated for succeding dimensions \emph{k} using some recurrent relations for Legendre's polynomials and numerical methods for cosine functions. In the implementation the default value of \emph{c} in \emph{$T^*$} was fixed to be 100. Hence, \emph{$T^*$} is Schwarz-type model selection rule. The resulting data driven test statistic for extreme value distribution is \emph{$W_{T^*}=W_{T^*}(tilde gamma)$}.
       |                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       ^
checkRd: (-1) ddst.norm.test.Rd:30: Lost braces; missing escapes or markup?
    30 | \emph{$gamma=(gamma_1,gamma_2)$} is estimated by \emph{$tilde gamma=(tilde gamma_1,tilde gamma_2)$}, where \emph{$tilde gamma_1=1/n sum_{i=1}^n Z_i$} and 
       |                                                                                                                                         ^
checkRd: (-1) ddst.norm.test.Rd:31: Lost braces; missing escapes or markup?
    31 | \emph{$tilde gamma_2 = 1/(n-1) sum_{i=1}^{n-1}(Z_{n:i+1}-Z_{n:i})(H_{i+1}-H_i)$},
       |                                    ^
checkRd: (-1) ddst.norm.test.Rd:31: Lost braces; missing escapes or markup?
    31 | \emph{$tilde gamma_2 = 1/(n-1) sum_{i=1}^{n-1}(Z_{n:i+1}-Z_{n:i})(H_{i+1}-H_i)$},
       |                                          ^
checkRd: (-1) ddst.norm.test.Rd:31: Lost braces; missing escapes or markup?
    31 | \emph{$tilde gamma_2 = 1/(n-1) sum_{i=1}^{n-1}(Z_{n:i+1}-Z_{n:i})(H_{i+1}-H_i)$},
       |                                                  ^
checkRd: (-1) ddst.norm.test.Rd:31: Lost braces; missing escapes or markup?
    31 | \emph{$tilde gamma_2 = 1/(n-1) sum_{i=1}^{n-1}(Z_{n:i+1}-Z_{n:i})(H_{i+1}-H_i)$},
       |                                                            ^
checkRd: (-1) ddst.norm.test.Rd:31: Lost braces; missing escapes or markup?
    31 | \emph{$tilde gamma_2 = 1/(n-1) sum_{i=1}^{n-1}(Z_{n:i+1}-Z_{n:i})(H_{i+1}-H_i)$},
       |                                                                     ^
checkRd: (-1) ddst.norm.test.Rd:32: Lost braces; missing escapes or markup?
    32 | while \emph{$Z_{n:1}<= ... <= Z_{n:n}$} are ordered values of \emph{$Z_1, ..., Z_n$} and \emph{$H_i= phi^{-1}((i-3/8)(n+1/4))$}, cf. Chen and Shapiro (1995). 
       |                ^
checkRd: (-1) ddst.norm.test.Rd:32: Lost braces; missing escapes or markup?
    32 | while \emph{$Z_{n:1}<= ... <= Z_{n:n}$} are ordered values of \emph{$Z_1, ..., Z_n$} and \emph{$H_i= phi^{-1}((i-3/8)(n+1/4))$}, cf. Chen and Shapiro (1995). 
       |                                 ^
checkRd: (-1) ddst.norm.test.Rd:32: Lost braces; missing escapes or markup?
    32 | while \emph{$Z_{n:1}<= ... <= Z_{n:n}$} are ordered values of \emph{$Z_1, ..., Z_n$} and \emph{$H_i= phi^{-1}((i-3/8)(n+1/4))$}, cf. Chen and Shapiro (1995). 
       |                                                                                                          ^
checkRd: (-1) ddst.norm.test.Rd:35: Lost braces; missing escapes or markup?
    35 | The pertaining matrix \emph{$[I^*(tilde gamma)]^{-1}$} does not  depend on \emph{$tilde gamma$} and is calculated for succeding dimensions \emph{k} using some recurrent relations for Legendre's polynomials and is computed in a numerical way in case of cosine basis. In the implementation of \emph{$T^*$} the default value of \emph{c} is set  to be 100. Therefore, in practice, \emph{$T^*$} is Schwarz-type criterion. See Inglot and Ledwina (2006) as well as Janic and Ledwina (2008) for comments. The resulting data driven test statistic for normality is \emph{$W_{T^*}=W_{T^*}(tilde gamma)$}.
       |                                                 ^
checkRd: (-1) ddst.norm.test.Rd:35: Lost braces; missing escapes or markup?
    35 | The pertaining matrix \emph{$[I^*(tilde gamma)]^{-1}$} does not  depend on \emph{$tilde gamma$} and is calculated for succeding dimensions \emph{k} using some recurrent relations for Legendre's polynomials and is computed in a numerical way in case of cosine basis. In the implementation of \emph{$T^*$} the default value of \emph{c} is set  to be 100. Therefore, in practice, \emph{$T^*$} is Schwarz-type criterion. See Inglot and Ledwina (2006) as well as Janic and Ledwina (2008) for comments. The resulting data driven test statistic for normality is \emph{$W_{T^*}=W_{T^*}(tilde gamma)$}.
       |                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                     ^
checkRd: (-1) ddst.norm.test.Rd:35: Lost braces; missing escapes or markup?
    35 | The pertaining matrix \emph{$[I^*(tilde gamma)]^{-1}$} does not  depend on \emph{$tilde gamma$} and is calculated for succeding dimensions \emph{k} using some recurrent relations for Legendre's polynomials and is computed in a numerical way in case of cosine basis. In the implementation of \emph{$T^*$} the default value of \emph{c} is set  to be 100. Therefore, in practice, \emph{$T^*$} is Schwarz-type criterion. See Inglot and Ledwina (2006) as well as Janic and Ledwina (2008) for comments. The resulting data driven test statistic for normality is \emph{$W_{T^*}=W_{T^*}(tilde gamma)$}.
       |                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             ^
checkRd: (-1) ddst.uniform.test.Rd:25: Lost braces; missing escapes or markup?
    25 | $W_k=[1/sqrt(n) sum_{j=1}^k sum_{i=1}^n phi_j(Z_i)]^2$},
       |                     ^
checkRd: (-1) ddst.uniform.test.Rd:25: Lost braces; missing escapes or markup?
    25 | $W_k=[1/sqrt(n) sum_{j=1}^k sum_{i=1}^n phi_j(Z_i)]^2$},
       |                                 ^