ks.test左截断威布尔

Question

分布我接受我要遵循的值是左截断的威布尔分布。我知道使用$('.new-venue-item').each(function() { var $map = $(this).find('.map'), lat = parseFloat($map.data('lat')), lng = parseFloat($map.data('lng')); // Hook up the map: var mapOptions = { scrollwheel: false, streetViewControl: false, mapTypeControl: false, panControl: false, zoomControl: false, draggable: false, zoomLevel: 14, center: { lat: lat, lng: lng } }; //new google.maps.Map( document.getElementById('map8160'), mapOptions ); $(this).data('map', new google.maps.Map($map[0], mapOptions)); }); 命令的此分布的参数a，形状和比例：

ptrunc

所以我希望require(truncdist); ptrunc(x,"weibull",a=a,scale=b,shape=c) 命令（见下文）使用描述的左截断weibull分布而不是＆＃34;正常weibull＆＃34;。

ks.test

所以我知道，在这种情况下，没有必要进行左侧截断。但在其他人看来会如此。

myvalues<-c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4); 
a<-7;
scale<-36.37516;
shape<-9.437013; 

但

ks.test(myvalues,"pweibull",scale=b,shape=c) #for normal weibull

给出了错误的结果。

Answer 1

首先，ptrunc应替换为rtrunc。 ptrunc给出概率值的向量。但是documentation of ks.test我们需要一个样本，这就是rtrunc给我们的。如果a的参数rtrunc设置为-Inf，则不会截断，a=-Inf的结果与a=7的结果相同：

library(truncdist)

myvalues <- c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4)

a <- 7
scale<-36.37516
shape <- 9.437013

set.seed(1)
y1 <- rtrunc(myvalues,"weibull",a=-Inf,scale=scale,shape=shape)

set.seed(1)
y2 <- rtrunc(myvalues,"weibull",a=a,scale=scale,shape=shape)

set.seed(1)
ks0 <- ks.test( myvalues, "pweibull",scale=scale,shape=shape  )

set.seed(1)
ks1 <- ks.test( myvalues, y1 )

set.seed(1)
ks2 <- ks.test( myvalues, y2 )

> ks1

    Two-sample Kolmogorov-Smirnov test

data:  myvalues and y1
D = 0.21429, p-value = 0.2898
alternative hypothesis: two-sided

> ks2

    Two-sample Kolmogorov-Smirnov test

data:  myvalues and y2
D = 0.21429, p-value = 0.2898
alternative hypothesis: two-sided

但ks.test( myvalues, "pweibull",scale=scale,shape=shape )的结果仍然不同：

> ks0

    One-sample Kolmogorov-Smirnov test

data:  myvalues
D = 0.15612, p-value = 0.2576
alternative hypothesis: two-sided

原因是myvalues太小了。如果我们在rtrunc（不是ks.test）的调用中将其设置得更大，则ks0，ks1和ks2几乎相同：

library(truncdist)

myvalues <- c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4)

myManyValues <- c(outer((0:9999)/100000,myvalues,"+"))

a <- 7
scale<-36.37516
shape <- 9.437013

set.seed(1)
y1 <- rtrunc(myManyValues,"weibull",a=-Inf,scale=scale,shape=shape)

set.seed(1)
y2 <- rtrunc(myManyValues,"weibull",a=a,scale=scale,shape=shape)

set.seed(1)
ks0 <- ks.test( myvalues, "pweibull",scale=scale,shape=shape  )

set.seed(1)
ks1 <- ks.test( myvalues, y1 )

set.seed(1)
ks2 <- ks.test( myvalues, y2 )

> ks0

    One-sample Kolmogorov-Smirnov test

data:  myvalues
D = 0.15612, p-value = 0.2576
alternative hypothesis: two-sided

> ks1

    Two-sample Kolmogorov-Smirnov test

data:  myvalues and y1
D = 0.15655, p-value = 0.2548
alternative hypothesis: two-sided

> ks2

    Two-sample Kolmogorov-Smirnov test

data:  myvalues and y2
D = 0.15655, p-value = 0.2548
alternative hypothesis: two-sided

现在让我们看看当做截断分发时会发生什么：

library(truncdist)

myvalues <- c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4)

myManyValues <- c(outer((0:9999)/100000,myvalues,"+"))

a <- 29
scale<-36.37516
shape <- 9.437013

set.seed(1)
y1 <- rtrunc(myManyValues,"weibull",a=-Inf,scale=scale,shape=shape)

set.seed(1)
y2 <- rtrunc(myManyValues,"weibull",a=a,scale=scale,shape=shape)

set.seed(1)
ks0 <- ks.test( myvalues, "pweibull",scale=scale,shape=shape  )

set.seed(1)
ks1 <- ks.test( myvalues, y1 )

set.seed(1)
ks2 <- ks.test( myvalues, y2 )

> ks0

    One-sample Kolmogorov-Smirnov test

data:  myvalues
D = 0.15612, p-value = 0.2576
alternative hypothesis: two-sided

> ks1

    Two-sample Kolmogorov-Smirnov test

data:  myvalues and y1
D = 0.15655, p-value = 0.2548
alternative hypothesis: two-sided

> ks2

    Two-sample Kolmogorov-Smirnov test

data:  myvalues and y2
D = 0.2059, p-value = 0.05683
alternative hypothesis: two-sided

Answer 2

您正在错误地使用ptrunc函数（我假设），需要输入一系列分位数。下面我根据你的尺度和形状参数计算威布尔的平均值和标准差，然后从上下5个标准差进行采样，得到一个比较集。

require(truncdist);

myvalues <- c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4); 
a <- 7;
scale <- 36.37516;
shape <- 9.437013;

# Calculate standard deviation of the weibull
weib_mean <- scale * gamma(1 + 1/shape)
weib_sd <- sqrt((scale^2) * (gamma(1 + 2/shape) - (gamma(1 + 1/shape))^2))

# Get a sample
quant <- seq(weib_mean - 5 * weib_sd, weib_mean + 5 * weib_sd, length.out = 1E5)
weibull_samp <- ptrunc(quant, "weibull", a = a, scale = scale, shape = shape)

# Take a look
plot(weibull_samp ~ quant)

# Use with test
> ks.test(sort(myvalues), weibull_samp)
       Two-sample Kolmogorov-Smirnov test

data:  sort(myvalues) and weibull_samp
D = 1, p-value < 2.2e-16
alternative hypothesis: two-sided