Fisher information of the Rayleigh distribution












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Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(theta) = -E(frac{partial^2 log L}{partial theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(theta) = frac{n}{theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$log L = prod log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$log L = sum_{i}(log x_{i} -frac{2}{theta} - frac{x_{i}^2}{2theta^2}) $$ Then I take the partial derivative with respect to $theta$ two times and obtain $$frac{partial^2 log L}{partial theta^2} = frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} $$ So $I(theta) = -E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $theta$?






probability-distributions expectation







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  • $begingroup$
    how to take an expected value with respect to the $x_i^2$?
    $endgroup$
    – Karl
    Nov 16 '14 at 13:26










  • $begingroup$
    The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
    $endgroup$
    – Sayan
    Nov 16 '14 at 13:44












  • $begingroup$
    I did take them with respect to $x_{i}$'s. I will post the result asap.
    $endgroup$
    – user29163
    Nov 16 '14 at 13:54
















0












$begingroup$


Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(theta) = -E(frac{partial^2 log L}{partial theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(theta) = frac{n}{theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$log L = prod log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$log L = sum_{i}(log x_{i} -frac{2}{theta} - frac{x_{i}^2}{2theta^2}) $$ Then I take the partial derivative with respect to $theta$ two times and obtain $$frac{partial^2 log L}{partial theta^2} = frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} $$ So $I(theta) = -E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $theta$?






probability-distributions expectation







share|cite|improve this question









$endgroup$












  • $begingroup$
    how to take an expected value with respect to the $x_i^2$?
    $endgroup$
    – Karl
    Nov 16 '14 at 13:26










  • $begingroup$
    The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
    $endgroup$
    – Sayan
    Nov 16 '14 at 13:44












  • $begingroup$
    I did take them with respect to $x_{i}$'s. I will post the result asap.
    $endgroup$
    – user29163
    Nov 16 '14 at 13:54














0












0








0





$begingroup$


Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(theta) = -E(frac{partial^2 log L}{partial theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(theta) = frac{n}{theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$log L = prod log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$log L = sum_{i}(log x_{i} -frac{2}{theta} - frac{x_{i}^2}{2theta^2}) $$ Then I take the partial derivative with respect to $theta$ two times and obtain $$frac{partial^2 log L}{partial theta^2} = frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} $$ So $I(theta) = -E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $theta$?






probability-distributions expectation







share|cite|improve this question









$endgroup$




Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(theta) = -E(frac{partial^2 log L}{partial theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(theta) = frac{n}{theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$log L = prod log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$log L = sum_{i}(log x_{i} -frac{2}{theta} - frac{x_{i}^2}{2theta^2}) $$ Then I take the partial derivative with respect to $theta$ two times and obtain $$frac{partial^2 log L}{partial theta^2} = frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} $$ So $I(theta) = -E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $theta$?






probability-distributions expectation




probability-distributions expectation






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asked Nov 16 '14 at 13:06









user29163user29163

360313




360313












  • $begingroup$
    how to take an expected value with respect to the $x_i^2$?
    $endgroup$
    – Karl
    Nov 16 '14 at 13:26










  • $begingroup$
    The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
    $endgroup$
    – Sayan
    Nov 16 '14 at 13:44












  • $begingroup$
    I did take them with respect to $x_{i}$'s. I will post the result asap.
    $endgroup$
    – user29163
    Nov 16 '14 at 13:54


















  • $begingroup$
    how to take an expected value with respect to the $x_i^2$?
    $endgroup$
    – Karl
    Nov 16 '14 at 13:26










  • $begingroup$
    The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
    $endgroup$
    – Sayan
    Nov 16 '14 at 13:44












  • $begingroup$
    I did take them with respect to $x_{i}$'s. I will post the result asap.
    $endgroup$
    – user29163
    Nov 16 '14 at 13:54
















$begingroup$
how to take an expected value with respect to the $x_i^2$?
$endgroup$
– Karl
Nov 16 '14 at 13:26




$begingroup$
how to take an expected value with respect to the $x_i^2$?
$endgroup$
– Karl
Nov 16 '14 at 13:26












$begingroup$
The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
$endgroup$
– Sayan
Nov 16 '14 at 13:44






$begingroup$
The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
$endgroup$
– Sayan
Nov 16 '14 at 13:44














$begingroup$
I did take them with respect to $x_{i}$'s. I will post the result asap.
$endgroup$
– user29163
Nov 16 '14 at 13:54




$begingroup$
I did take them with respect to $x_{i}$'s. I will post the result asap.
$endgroup$
– user29163
Nov 16 '14 at 13:54










2 Answers
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Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
$$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$






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    The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.






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      2 Answers
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      2 Answers
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      $begingroup$

      Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
      $$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
      now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$






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        $begingroup$

        Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
        $$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
        now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$






        share|cite|improve this answer









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          $begingroup$

          Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
          $$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
          now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$






          share|cite|improve this answer









          $endgroup$



          Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
          $$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
          now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 16 '14 at 19:12









          user29163user29163

          360313




          360313























              0












              $begingroup$

              The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.






                  share|cite|improve this answer









                  $endgroup$



                  The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 14 '16 at 23:16









                  Filippo PalombaFilippo Palomba

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