Find max likelihood estimator for a 2 variable function
$begingroup$
I have been given a problem in which I am given a set of independent random variables, X1, X2, ... Xn with the distribution:
$f_x(x ; b, m) = frac{1}{2b}e^{-|x-m|/b}$ for $-∞ < x < ∞$
and I am tasked with finding the maxiumum likelihood estimators for b and m given the sample: $-2.1,: -1.5,: -.6,: .1,: 1.2,: 1.7,: 2.3,: 3.1,: 4.4$
I followed the standard way of solving these questions as taught by my professor by first taking the product of the distribution over the set of variables:
$prod_{i=1}^{n} (frac{1}{2b}e^{-|x_i-m|/b})$ = $frac{1}{(2b)^n}e^{-1/bsum_{i=1}^{n}|x_i-m|}$
Taking the natural log of this equation gives:
$-nln{2b}:-:frac{1}{b}sum_{i=1}^{n}|x_i-m|$
So then I must take the derivative in respect to m and in respect to f, set each of these equations to zero and solve for m and f, respectively. I have found the equations for each of these as follows:
for f: $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$
for m: $frac{-1}{b}sum_{i=1}^{n}sign(m-x_i) = 0$
With the equation for m having the sum of the signum function. After this I am stuck as I have never worked with an estimator with 2 variables nor have I ever had to actually solve an equation for a set of real values. I am also unsure of how I would work out the sum of the signum function so I could solve for m as I have never worked with this function either. If anyone could help solve the equation or point me in the right direction that would be great.
probability statistics estimation maximum-likelihood
asked Dec 1 '18 at 19:31
mkohlermkohler
82
$endgroup$
add a comment |
$begingroup$
I have been given a problem in which I am given a set of independent random variables, X1, X2, ... Xn with the distribution:
$f_x(x ; b, m) = frac{1}{2b}e^{-|x-m|/b}$ for $-∞ < x < ∞$
and I am tasked with finding the maxiumum likelihood estimators for b and m given the sample: $-2.1,: -1.5,: -.6,: .1,: 1.2,: 1.7,: 2.3,: 3.1,: 4.4$
I followed the standard way of solving these questions as taught by my professor by first taking the product of the distribution over the set of variables:
$prod_{i=1}^{n} (frac{1}{2b}e^{-|x_i-m|/b})$ = $frac{1}{(2b)^n}e^{-1/bsum_{i=1}^{n}|x_i-m|}$
Taking the natural log of this equation gives:
$-nln{2b}:-:frac{1}{b}sum_{i=1}^{n}|x_i-m|$
So then I must take the derivative in respect to m and in respect to f, set each of these equations to zero and solve for m and f, respectively. I have found the equations for each of these as follows:
for f: $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$
for m: $frac{-1}{b}sum_{i=1}^{n}sign(m-x_i) = 0$
With the equation for m having the sum of the signum function. After this I am stuck as I have never worked with an estimator with 2 variables nor have I ever had to actually solve an equation for a set of real values. I am also unsure of how I would work out the sum of the signum function so I could solve for m as I have never worked with this function either. If anyone could help solve the equation or point me in the right direction that would be great.
probability statistics estimation maximum-likelihood
asked Dec 1 '18 at 19:31
mkohlermkohler
82
$endgroup$
add a comment |
$begingroup$
I have been given a problem in which I am given a set of independent random variables, X1, X2, ... Xn with the distribution:
$f_x(x ; b, m) = frac{1}{2b}e^{-|x-m|/b}$ for $-∞ < x < ∞$
and I am tasked with finding the maxiumum likelihood estimators for b and m given the sample: $-2.1,: -1.5,: -.6,: .1,: 1.2,: 1.7,: 2.3,: 3.1,: 4.4$
I followed the standard way of solving these questions as taught by my professor by first taking the product of the distribution over the set of variables:
$prod_{i=1}^{n} (frac{1}{2b}e^{-|x_i-m|/b})$ = $frac{1}{(2b)^n}e^{-1/bsum_{i=1}^{n}|x_i-m|}$
Taking the natural log of this equation gives:
$-nln{2b}:-:frac{1}{b}sum_{i=1}^{n}|x_i-m|$
So then I must take the derivative in respect to m and in respect to f, set each of these equations to zero and solve for m and f, respectively. I have found the equations for each of these as follows:
for f: $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$
for m: $frac{-1}{b}sum_{i=1}^{n}sign(m-x_i) = 0$
With the equation for m having the sum of the signum function. After this I am stuck as I have never worked with an estimator with 2 variables nor have I ever had to actually solve an equation for a set of real values. I am also unsure of how I would work out the sum of the signum function so I could solve for m as I have never worked with this function either. If anyone could help solve the equation or point me in the right direction that would be great.
probability statistics estimation maximum-likelihood
asked Dec 1 '18 at 19:31
mkohlermkohler
82
$endgroup$
I have been given a problem in which I am given a set of independent random variables, X1, X2, ... Xn with the distribution:
$f_x(x ; b, m) = frac{1}{2b}e^{-|x-m|/b}$ for $-∞ < x < ∞$
and I am tasked with finding the maxiumum likelihood estimators for b and m given the sample: $-2.1,: -1.5,: -.6,: .1,: 1.2,: 1.7,: 2.3,: 3.1,: 4.4$
I followed the standard way of solving these questions as taught by my professor by first taking the product of the distribution over the set of variables:
$prod_{i=1}^{n} (frac{1}{2b}e^{-|x_i-m|/b})$ = $frac{1}{(2b)^n}e^{-1/bsum_{i=1}^{n}|x_i-m|}$
Taking the natural log of this equation gives:
$-nln{2b}:-:frac{1}{b}sum_{i=1}^{n}|x_i-m|$
So then I must take the derivative in respect to m and in respect to f, set each of these equations to zero and solve for m and f, respectively. I have found the equations for each of these as follows:
for f: $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$
for m: $frac{-1}{b}sum_{i=1}^{n}sign(m-x_i) = 0$
With the equation for m having the sum of the signum function. After this I am stuck as I have never worked with an estimator with 2 variables nor have I ever had to actually solve an equation for a set of real values. I am also unsure of how I would work out the sum of the signum function so I could solve for m as I have never worked with this function either. If anyone could help solve the equation or point me in the right direction that would be great.
probability statistics estimation maximum-likelihood
probability statistics estimation maximum-likelihood
asked Dec 1 '18 at 19:31
mkohlermkohler
82
asked Dec 1 '18 at 19:31
mkohlermkohler
82
asked Dec 1 '18 at 19:31
mkohlermkohler
82
asked Dec 1 '18 at 19:31
mkohlermkohler
82
asked Dec 1 '18 at 19:31
mkohlermkohler
82
82
add a comment |
add a comment |
1 Answer
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Your distribution is a Laplace distribution
In your equation for $m$, you will get $sum_{i=1}^{n}text{sign}(m-x_i) = 0$ when ($n$ is even) half the signs are $+1$ and half $-1$ or when ($n$ odd) one is $0$ and half the rest $+1$ and half $-1$. This will make MLE $m$ the median
Knowing the MLE for $m$, this gives a fixed value for $sum_{i=1}^{n}|x_i-m|$ and so you can solve $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$ to get the MLE estimate $b=frac1n sum_{i=1}^{n}|x_i-m|$, which is the mean absolute deviation about the median
answered Dec 1 '18 at 19:57
HenryHenry
100k480166
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$begingroup$
Your distribution is a Laplace distribution
In your equation for $m$, you will get $sum_{i=1}^{n}text{sign}(m-x_i) = 0$ when ($n$ is even) half the signs are $+1$ and half $-1$ or when ($n$ odd) one is $0$ and half the rest $+1$ and half $-1$. This will make MLE $m$ the median
Knowing the MLE for $m$, this gives a fixed value for $sum_{i=1}^{n}|x_i-m|$ and so you can solve $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$ to get the MLE estimate $b=frac1n sum_{i=1}^{n}|x_i-m|$, which is the mean absolute deviation about the median
answered Dec 1 '18 at 19:57
HenryHenry
100k480166
$endgroup$
add a comment |
$begingroup$
Your distribution is a Laplace distribution
In your equation for $m$, you will get $sum_{i=1}^{n}text{sign}(m-x_i) = 0$ when ($n$ is even) half the signs are $+1$ and half $-1$ or when ($n$ odd) one is $0$ and half the rest $+1$ and half $-1$. This will make MLE $m$ the median
Knowing the MLE for $m$, this gives a fixed value for $sum_{i=1}^{n}|x_i-m|$ and so you can solve $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$ to get the MLE estimate $b=frac1n sum_{i=1}^{n}|x_i-m|$, which is the mean absolute deviation about the median
answered Dec 1 '18 at 19:57
HenryHenry
100k480166
$endgroup$
add a comment |
$begingroup$
Your distribution is a Laplace distribution
In your equation for $m$, you will get $sum_{i=1}^{n}text{sign}(m-x_i) = 0$ when ($n$ is even) half the signs are $+1$ and half $-1$ or when ($n$ odd) one is $0$ and half the rest $+1$ and half $-1$. This will make MLE $m$ the median
Knowing the MLE for $m$, this gives a fixed value for $sum_{i=1}^{n}|x_i-m|$ and so you can solve $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$ to get the MLE estimate $b=frac1n sum_{i=1}^{n}|x_i-m|$, which is the mean absolute deviation about the median
answered Dec 1 '18 at 19:57
HenryHenry
100k480166
$endgroup$
Your distribution is a Laplace distribution
In your equation for $m$, you will get $sum_{i=1}^{n}text{sign}(m-x_i) = 0$ when ($n$ is even) half the signs are $+1$ and half $-1$ or when ($n$ odd) one is $0$ and half the rest $+1$ and half $-1$. This will make MLE $m$ the median
Knowing the MLE for $m$, this gives a fixed value for $sum_{i=1}^{n}|x_i-m|$ and so you can solve $frac{-n}{b}:+:frac{1}{b^2}sum_{i=1}^{n}|x_i-m| = 0$ to get the MLE estimate $b=frac1n sum_{i=1}^{n}|x_i-m|$, which is the mean absolute deviation about the median
answered Dec 1 '18 at 19:57
HenryHenry
100k480166
answered Dec 1 '18 at 19:57
HenryHenry
100k480166
answered Dec 1 '18 at 19:57
HenryHenry
100k480166
answered Dec 1 '18 at 19:57
HenryHenry
100k480166
100k480166
add a comment |
add a comment |
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