Find max likelihood estimator for a 2 variable function












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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







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    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Dec 1 '18 at 19:31









      mkohlermkohler

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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






          share|cite|improve this answer









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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






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              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






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                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






                share|cite|improve this answer









                $endgroup$



                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







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 '18 at 19:57









                HenryHenry

                100k480166




                100k480166






























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