search for: tau3

...;- na.omit(mydta) # theta is the parameter vector ologit.lf <- function(theta, y, X) { n <- nrow(X) k <- ncol(X) # b is the coefficient vector for independent variables b <- theta[1:k] # tau1 is cut-point 1 tau1 <- theta [k+1] # tau2 is cut-point 2 tau2 <- theta [k+2] # tau3 is cut-point 1 tau3 <- theta [k+3] if (y == 1){ p <- (1/(1+exp( - tau1 + X %*% b))) } if (y == 2) { p <- (1/(1+exp( - tau2 + X %*% b))) - (1/(1+exp( - tau1 + X %*% b))) } if (y == 3) { p <- (1/(1+exp( - tau3 + X %*% b))) - (1/(1+exp( - tau2 + X %*% b))) } if...

...3()? Lmom package info just gives: pelgam(lmom), lelln3(lmom), where lmom is a numeric vector containing the L-moments of the distribution or of a data sample. # Draw an L-moment ratio diagram and add a line for the # Weibull distribution lmrd() weimom <- sapply( seq(0, 0.9, by=0.01), function(tau3) lmrwei(pelwei(c(0,1,tau3)), nmom=4) ) lmrdlines(t(weimom), col='darkgreen', lwd=2) source: http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=lmom:lmrdpoints Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Adding-gamma-and-3-parameter-log-normal-distributions-to-L...

...2.reg <- lm(z.diff ~ -1 + z.diff.lag) phi3.reg <- lm(z.diff ~ z.diff.lag) phi2 <- anova(phi2.reg, result)$F[2] phi3 <- anova(phi3.reg, result)$F[2] teststat <- as.matrix(t(c(tau, phi2, phi3))) colnames(teststat) <- c('tau3', 'phi2', 'phi3') } } else { if (type == "none") { result <- lm(z.diff ~ z.lag.1 - 1) tau <- coef(summary(result))[1, 3] teststat <- as.matrix(tau) colnames(teststat) <- 'tau1'...

Respected Sir I have to select a probability distribution using L-moment Ratio Diagram. I am not understanding how to plot sample TAU3 and TUA4 on L-moment Ratio Diagram. Please Guide me Best Regards -- AMINA SHAHZADI Department of Statistics GC University Lahore, Pakistan. Email: amnakhan493@gmail.com amna_989@hotmail.com amna_989@yahoo.com [[alternative HTML version deleted]]

...phi3.reg <- lm(z.diff ~ 1) phi2 <- anova(phi2.reg, result)$F[2] phi3 <- anova(phi3.reg, result)$F[2] tau <- coef(summary(result))[2, 3] teststat <- as.matrix(t(c(tau, phi2, phi3))) colnames(teststat) <- c("tau3", "phi2", "phi3") [[alternative HTML version deleted]]

...n conclude that x is mean-reverting. Am I correct? 2/ I would like to see critical values also, so I tried with ur.df > summary(ur.df(x, "trend", lag=0)) <snip> Value of test-statistic is: -31.8629 338.4156 507.6231 Critical values for test statistics: 1pct 5pct 10pct tau3 -3.96 -3.41 -3.12 phi2 6.09 4.68 4.03 phi3 8.27 6.25 5.34 Here if I understand correctly, as my first critical value is significantly less than the 1% critical value I reject the null hypothesis that x has a unit root, so x is stationary and then mean reverting. Thanks, -Arnaud

...3.459 NA NA NA Residual standard error: NaN on 0 degrees of freedom Multiple R-squared: 1, Adjusted R-squared: NaN F-statistic: NaN on 2 and 0 DF, p-value: NA Value of test-statistic is: NaN NaN NaN Critical values for test statistics: 1pct 5pct 10pct tau3 -4.38 -3.60 -3.24 phi2 8.21 5.68 4.67 phi3 10.61 7.24 5.91 " Can anyone tell me what is wrong with the codes???The code used is given below. Thanx t=1:length(x) plot(t,x) trend = lm(x~t) abline(lm(x~t)) summary(trend) library(urca) x = ts(x, start=1, end = length(x), frequency=1) x.ct...