R: How to compare the quality of estimators (method of moments vs max likelihood)
0
It's my first question in here, so please do not be very strict on me.
I need to compare the quality of estimators. I have the following data and codes. What I need to understand is whether the only way to understand the better estimation is to see graphics, or whether there is a more precise way to do it. Thanks in advance!
WT<-c(75, 265, 225, 402, 35, 105, 411, 346, 159, 229, 62, 256, 431, 177, 56, 144, 354, 178, 386, 294)
hist(WT,breaks=10,freq=F)
h<-hist(WT,breaks=quantile(WT,seq(0,1,0.1)),main="WT distribution")
cumfreq2<-cumsum(h$counts)/length(WT)
plot(h$breaks,c(0,cumfreq2),"l",main="Distribution
function",xlab="WT",ylab="Probabilite")+
lines(h$breaks,punif(h$breaks,min(WT),max(WT)),col="red")
#Estimate the parameter of this law by the method of moments.
u<-mean(WT)
v<-sqrt(sum((WT-u)^2)/length(WT)) # ce n'est pas le sqrt(var?) sqrt(var(WT))
a<-u-sqrt(3)*v #pourquoi 3
b<-u+sqrt(3)*v
#Estimate this parameter by the maximum likelihood method.
teta2<-max(WT)
#Compare the quality of these estimators by cumulative frequency graph
plot(h$breaks,c(0,cumfreq2),"l",main="Distribution
function",xlab="WT",ylab="Probabilite")+
lines(h$breaks,punif(h$breaks,min=a,max=b),col="red")
uni<-function(x){
if (x<=0){
y<-0
}else if (x<teta2) {
y<-x/teta2
}else {
y<-1
}
return(y)
}
lines(h$breaks,uni(h$breaks),col="blue")
r statistics
edited Nov 22 '18 at 0:29
Christopher Bradshaw
91311230
asked Nov 21 '18 at 23:25
Azat AleksanyanAzat Aleksanyan
11
add a comment |
0
It's my first question in here, so please do not be very strict on me.
I need to compare the quality of estimators. I have the following data and codes. What I need to understand is whether the only way to understand the better estimation is to see graphics, or whether there is a more precise way to do it. Thanks in advance!
WT<-c(75, 265, 225, 402, 35, 105, 411, 346, 159, 229, 62, 256, 431, 177, 56, 144, 354, 178, 386, 294)
hist(WT,breaks=10,freq=F)
h<-hist(WT,breaks=quantile(WT,seq(0,1,0.1)),main="WT distribution")
cumfreq2<-cumsum(h$counts)/length(WT)
plot(h$breaks,c(0,cumfreq2),"l",main="Distribution
function",xlab="WT",ylab="Probabilite")+
lines(h$breaks,punif(h$breaks,min(WT),max(WT)),col="red")
#Estimate the parameter of this law by the method of moments.
u<-mean(WT)
v<-sqrt(sum((WT-u)^2)/length(WT)) # ce n'est pas le sqrt(var?) sqrt(var(WT))
a<-u-sqrt(3)*v #pourquoi 3
b<-u+sqrt(3)*v
#Estimate this parameter by the maximum likelihood method.
teta2<-max(WT)
#Compare the quality of these estimators by cumulative frequency graph
plot(h$breaks,c(0,cumfreq2),"l",main="Distribution
function",xlab="WT",ylab="Probabilite")+
lines(h$breaks,punif(h$breaks,min=a,max=b),col="red")
uni<-function(x){
if (x<=0){
y<-0
}else if (x<teta2) {
y<-x/teta2
}else {
y<-1
}
return(y)
}
lines(h$breaks,uni(h$breaks),col="blue")
r statistics
edited Nov 22 '18 at 0:29
Christopher Bradshaw
91311230
asked Nov 21 '18 at 23:25
Azat AleksanyanAzat Aleksanyan
11
What is your metric - rmse? can you compare based on that? (I am uncertain - just asking)
– javadba
Nov 21 '18 at 23:41
Dear @javadba , frankly speaking I do not understand your question, what metric you want?
– Azat Aleksanyan
Nov 22 '18 at 8:48
add a comment |
0
0
0
1
It's my first question in here, so please do not be very strict on me.
I need to compare the quality of estimators. I have the following data and codes. What I need to understand is whether the only way to understand the better estimation is to see graphics, or whether there is a more precise way to do it. Thanks in advance!
WT<-c(75, 265, 225, 402, 35, 105, 411, 346, 159, 229, 62, 256, 431, 177, 56, 144, 354, 178, 386, 294)
hist(WT,breaks=10,freq=F)
h<-hist(WT,breaks=quantile(WT,seq(0,1,0.1)),main="WT distribution")
cumfreq2<-cumsum(h$counts)/length(WT)
plot(h$breaks,c(0,cumfreq2),"l",main="Distribution
function",xlab="WT",ylab="Probabilite")+
lines(h$breaks,punif(h$breaks,min(WT),max(WT)),col="red")
#Estimate the parameter of this law by the method of moments.
u<-mean(WT)
v<-sqrt(sum((WT-u)^2)/length(WT)) # ce n'est pas le sqrt(var?) sqrt(var(WT))
a<-u-sqrt(3)*v #pourquoi 3
b<-u+sqrt(3)*v
#Estimate this parameter by the maximum likelihood method.
teta2<-max(WT)
#Compare the quality of these estimators by cumulative frequency graph
plot(h$breaks,c(0,cumfreq2),"l",main="Distribution
function",xlab="WT",ylab="Probabilite")+
lines(h$breaks,punif(h$breaks,min=a,max=b),col="red")
uni<-function(x){
if (x<=0){
y<-0
}else if (x<teta2) {
y<-x/teta2
}else {
y<-1
}
return(y)
}
lines(h$breaks,uni(h$breaks),col="blue")
r statistics
edited Nov 22 '18 at 0:29
Christopher Bradshaw
91311230
asked Nov 21 '18 at 23:25
Azat AleksanyanAzat Aleksanyan
11
It's my first question in here, so please do not be very strict on me.
I need to compare the quality of estimators. I have the following data and codes. What I need to understand is whether the only way to understand the better estimation is to see graphics, or whether there is a more precise way to do it. Thanks in advance!
WT<-c(75, 265, 225, 402, 35, 105, 411, 346, 159, 229, 62, 256, 431, 177, 56, 144, 354, 178, 386, 294)
hist(WT,breaks=10,freq=F)
h<-hist(WT,breaks=quantile(WT,seq(0,1,0.1)),main="WT distribution")
cumfreq2<-cumsum(h$counts)/length(WT)
plot(h$breaks,c(0,cumfreq2),"l",main="Distribution
function",xlab="WT",ylab="Probabilite")+
lines(h$breaks,punif(h$breaks,min(WT),max(WT)),col="red")
#Estimate the parameter of this law by the method of moments.
u<-mean(WT)
v<-sqrt(sum((WT-u)^2)/length(WT)) # ce n'est pas le sqrt(var?) sqrt(var(WT))
a<-u-sqrt(3)*v #pourquoi 3
b<-u+sqrt(3)*v
#Estimate this parameter by the maximum likelihood method.
teta2<-max(WT)
#Compare the quality of these estimators by cumulative frequency graph
plot(h$breaks,c(0,cumfreq2),"l",main="Distribution
function",xlab="WT",ylab="Probabilite")+
lines(h$breaks,punif(h$breaks,min=a,max=b),col="red")
uni<-function(x){
if (x<=0){
y<-0
}else if (x<teta2) {
y<-x/teta2
}else {
y<-1
}
return(y)
}
lines(h$breaks,uni(h$breaks),col="blue")
r statistics
r statistics
edited Nov 22 '18 at 0:29
Christopher Bradshaw
91311230
asked Nov 21 '18 at 23:25
Azat AleksanyanAzat Aleksanyan
11
edited Nov 22 '18 at 0:29
Christopher Bradshaw
91311230
asked Nov 21 '18 at 23:25
Azat AleksanyanAzat Aleksanyan
11
edited Nov 22 '18 at 0:29
Christopher Bradshaw
91311230
edited Nov 22 '18 at 0:29
Christopher Bradshaw
91311230
edited Nov 22 '18 at 0:29
Christopher Bradshaw
91311230
91311230
asked Nov 21 '18 at 23:25
Azat AleksanyanAzat Aleksanyan
11
asked Nov 21 '18 at 23:25
Azat AleksanyanAzat Aleksanyan
11
asked Nov 21 '18 at 23:25
Azat AleksanyanAzat Aleksanyan
11
11
What is your metric - rmse? can you compare based on that? (I am uncertain - just asking)
– javadba
Nov 21 '18 at 23:41
Dear @javadba , frankly speaking I do not understand your question, what metric you want?
– Azat Aleksanyan
Nov 22 '18 at 8:48
add a comment |
What is your metric - rmse? can you compare based on that? (I am uncertain - just asking)
– javadba
Nov 21 '18 at 23:41
Dear @javadba , frankly speaking I do not understand your question, what metric you want?
– Azat Aleksanyan
Nov 22 '18 at 8:48
What is your metric - rmse? can you compare based on that? (I am uncertain - just asking)
– javadba
Nov 21 '18 at 23:41
What is your metric - rmse? can you compare based on that? (I am uncertain - just asking)
– javadba
Nov 21 '18 at 23:41
Dear @javadba , frankly speaking I do not understand your question, what metric you want?
– Azat Aleksanyan
Nov 22 '18 at 8:48
Dear @javadba , frankly speaking I do not understand your question, what metric you want?
– Azat Aleksanyan
Nov 22 '18 at 8:48
add a comment |
0
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What is your metric - rmse? can you compare based on that? (I am uncertain - just asking)
– javadba
Nov 21 '18 at 23:41
Dear @javadba , frankly speaking I do not understand your question, what metric you want?
– Azat Aleksanyan
Nov 22 '18 at 8:48