Finding mistake in likelihood function and likelihood
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pmf =$e^{{(-(y-a)/b)-e^{-((y-a)/b)}}}$
liklihood function = $e^{-((sumlimits_{n=1}^{infty} y_n)-na/b)}*e^{-e^{(a/b)}*(sumlimits_{n=1}^{infty} e^{(y_n/b)}})$
log likelihood function in r = +-((sum(y)-length(y)p[1])/p[2])-exp((p[1]/p[2])(sum(exp(-y/p[2]))))
where p[1]=a and p[2]=b
I suspect that my likelihood function is incorrect because when I optim in R it gives an impossible interval.
statistics statistical-inference
asked Nov 16 at 21:55
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pmf =$e^{{(-(y-a)/b)-e^{-((y-a)/b)}}}$
liklihood function = $e^{-((sumlimits_{n=1}^{infty} y_n)-na/b)}*e^{-e^{(a/b)}*(sumlimits_{n=1}^{infty} e^{(y_n/b)}})$
log likelihood function in r = +-((sum(y)-length(y)p[1])/p[2])-exp((p[1]/p[2])(sum(exp(-y/p[2]))))
where p[1]=a and p[2]=b
I suspect that my likelihood function is incorrect because when I optim in R it gives an impossible interval.
statistics statistical-inference
asked Nov 16 at 21:55
Extra mint
124
Please write the pmf ( which I actually think is a pdf) clearly using MathJax, mentioning the domain of $y$.
– StubbornAtom
Nov 17 at 9:48
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
pmf =$e^{{(-(y-a)/b)-e^{-((y-a)/b)}}}$
liklihood function = $e^{-((sumlimits_{n=1}^{infty} y_n)-na/b)}*e^{-e^{(a/b)}*(sumlimits_{n=1}^{infty} e^{(y_n/b)}})$
log likelihood function in r = +-((sum(y)-length(y)p[1])/p[2])-exp((p[1]/p[2])(sum(exp(-y/p[2]))))
where p[1]=a and p[2]=b
I suspect that my likelihood function is incorrect because when I optim in R it gives an impossible interval.
statistics statistical-inference
asked Nov 16 at 21:55
Extra mint
124
pmf =$e^{{(-(y-a)/b)-e^{-((y-a)/b)}}}$
liklihood function = $e^{-((sumlimits_{n=1}^{infty} y_n)-na/b)}*e^{-e^{(a/b)}*(sumlimits_{n=1}^{infty} e^{(y_n/b)}})$
log likelihood function in r = +-((sum(y)-length(y)p[1])/p[2])-exp((p[1]/p[2])(sum(exp(-y/p[2]))))
where p[1]=a and p[2]=b
I suspect that my likelihood function is incorrect because when I optim in R it gives an impossible interval.
statistics statistical-inference
statistics statistical-inference
asked Nov 16 at 21:55
Extra mint
124
asked Nov 16 at 21:55
Extra mint
124
asked Nov 16 at 21:55
Extra mint
124
asked Nov 16 at 21:55
Extra mint
124
asked Nov 16 at 21:55
Extra mint
124
124
Please write the pmf ( which I actually think is a pdf) clearly using MathJax, mentioning the domain of $y$.
– StubbornAtom
Nov 17 at 9:48
add a comment |
Please write the pmf ( which I actually think is a pdf) clearly using MathJax, mentioning the domain of $y$.
– StubbornAtom
Nov 17 at 9:48
Please write the pmf ( which I actually think is a pdf) clearly using MathJax, mentioning the domain of $y$.
– StubbornAtom
Nov 17 at 9:48
Please write the pmf ( which I actually think is a pdf) clearly using MathJax, mentioning the domain of $y$.
– StubbornAtom
Nov 17 at 9:48
add a comment |
1 Answer
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0
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I think you are talking about the Gumbel distribution whose probability density function (pdf) is of the form
$$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
, where I assume the parameter $(a,b)$ to be unknown.
This is not a pmf but a pdf (it is a continuous distribution) and you can see that the normalising constant $b$ is missing in your equations.
If $Y_1,Y_2,ldots,Y_n$ are independent and identically distributed random variables having the above distribution, then the likelihood function given the sample $(y_1,y_2,ldots,y_n)inmathbb R^n$ is
begin{align}
L(a,b)&=prod_{i=1}^n f(y_i,;a,b)
\&=frac{1}{b^n}expleft[-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}right]quad,,ainmathbb R,b>0
end{align}
So the log-likelihood is
begin{align}
ell(a,b)&=-nln b-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}
\&=-nln b-frac{n}{b}(bar y-a)-e^{a/b}sum_{i=1}^n e^{-y_i/b} qquad[,bar y=text{ sample mean }]
end{align}
answered Nov 17 at 17:24
StubbornAtom
4,88411137
Thank you , I managed to estimate my a ,b now
– Extra mint
Nov 18 at 14:17
Consider 'accepting' answers to your questions if they fully address your queries.
– StubbornAtom
Nov 18 at 15:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I think you are talking about the Gumbel distribution whose probability density function (pdf) is of the form
$$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
, where I assume the parameter $(a,b)$ to be unknown.
This is not a pmf but a pdf (it is a continuous distribution) and you can see that the normalising constant $b$ is missing in your equations.
If $Y_1,Y_2,ldots,Y_n$ are independent and identically distributed random variables having the above distribution, then the likelihood function given the sample $(y_1,y_2,ldots,y_n)inmathbb R^n$ is
begin{align}
L(a,b)&=prod_{i=1}^n f(y_i,;a,b)
\&=frac{1}{b^n}expleft[-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}right]quad,,ainmathbb R,b>0
end{align}
So the log-likelihood is
begin{align}
ell(a,b)&=-nln b-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}
\&=-nln b-frac{n}{b}(bar y-a)-e^{a/b}sum_{i=1}^n e^{-y_i/b} qquad[,bar y=text{ sample mean }]
end{align}
answered Nov 17 at 17:24
StubbornAtom
4,88411137
Thank you , I managed to estimate my a ,b now
– Extra mint
Nov 18 at 14:17
Consider 'accepting' answers to your questions if they fully address your queries.
– StubbornAtom
Nov 18 at 15:19
add a comment |
up vote
0
down vote
accepted
I think you are talking about the Gumbel distribution whose probability density function (pdf) is of the form
$$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
, where I assume the parameter $(a,b)$ to be unknown.
This is not a pmf but a pdf (it is a continuous distribution) and you can see that the normalising constant $b$ is missing in your equations.
If $Y_1,Y_2,ldots,Y_n$ are independent and identically distributed random variables having the above distribution, then the likelihood function given the sample $(y_1,y_2,ldots,y_n)inmathbb R^n$ is
begin{align}
L(a,b)&=prod_{i=1}^n f(y_i,;a,b)
\&=frac{1}{b^n}expleft[-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}right]quad,,ainmathbb R,b>0
end{align}
So the log-likelihood is
begin{align}
ell(a,b)&=-nln b-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}
\&=-nln b-frac{n}{b}(bar y-a)-e^{a/b}sum_{i=1}^n e^{-y_i/b} qquad[,bar y=text{ sample mean }]
end{align}
answered Nov 17 at 17:24
StubbornAtom
4,88411137
Thank you , I managed to estimate my a ,b now
– Extra mint
Nov 18 at 14:17
Consider 'accepting' answers to your questions if they fully address your queries.
– StubbornAtom
Nov 18 at 15:19
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I think you are talking about the Gumbel distribution whose probability density function (pdf) is of the form
$$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
, where I assume the parameter $(a,b)$ to be unknown.
This is not a pmf but a pdf (it is a continuous distribution) and you can see that the normalising constant $b$ is missing in your equations.
If $Y_1,Y_2,ldots,Y_n$ are independent and identically distributed random variables having the above distribution, then the likelihood function given the sample $(y_1,y_2,ldots,y_n)inmathbb R^n$ is
begin{align}
L(a,b)&=prod_{i=1}^n f(y_i,;a,b)
\&=frac{1}{b^n}expleft[-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}right]quad,,ainmathbb R,b>0
end{align}
So the log-likelihood is
begin{align}
ell(a,b)&=-nln b-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}
\&=-nln b-frac{n}{b}(bar y-a)-e^{a/b}sum_{i=1}^n e^{-y_i/b} qquad[,bar y=text{ sample mean }]
end{align}
answered Nov 17 at 17:24
StubbornAtom
4,88411137
I think you are talking about the Gumbel distribution whose probability density function (pdf) is of the form
$$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
, where I assume the parameter $(a,b)$ to be unknown.
This is not a pmf but a pdf (it is a continuous distribution) and you can see that the normalising constant $b$ is missing in your equations.
If $Y_1,Y_2,ldots,Y_n$ are independent and identically distributed random variables having the above distribution, then the likelihood function given the sample $(y_1,y_2,ldots,y_n)inmathbb R^n$ is
begin{align}
L(a,b)&=prod_{i=1}^n f(y_i,;a,b)
\&=frac{1}{b^n}expleft[-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}right]quad,,ainmathbb R,b>0
end{align}
So the log-likelihood is
begin{align}
ell(a,b)&=-nln b-frac{1}{b}sum_{i=1}^n (y_i-a)-sum_{i=1}^n e^{-(y_i-a)/b}
\&=-nln b-frac{n}{b}(bar y-a)-e^{a/b}sum_{i=1}^n e^{-y_i/b} qquad[,bar y=text{ sample mean }]
end{align}
answered Nov 17 at 17:24
StubbornAtom
4,88411137
answered Nov 17 at 17:24
StubbornAtom
4,88411137
answered Nov 17 at 17:24
StubbornAtom
4,88411137
answered Nov 17 at 17:24
StubbornAtom
4,88411137
4,88411137
Thank you , I managed to estimate my a ,b now
– Extra mint
Nov 18 at 14:17
Consider 'accepting' answers to your questions if they fully address your queries.
– StubbornAtom
Nov 18 at 15:19
add a comment |
Thank you , I managed to estimate my a ,b now
– Extra mint
Nov 18 at 14:17
Consider 'accepting' answers to your questions if they fully address your queries.
– StubbornAtom
Nov 18 at 15:19
Thank you , I managed to estimate my a ,b now
– Extra mint
Nov 18 at 14:17
Thank you , I managed to estimate my a ,b now
– Extra mint
Nov 18 at 14:17
Consider 'accepting' answers to your questions if they fully address your queries.
– StubbornAtom
Nov 18 at 15:19
Consider 'accepting' answers to your questions if they fully address your queries.
– StubbornAtom
Nov 18 at 15:19
add a comment |
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Please write the pmf ( which I actually think is a pdf) clearly using MathJax, mentioning the domain of $y$.
– StubbornAtom
Nov 17 at 9:48