Using change of variables to transform density functions












0














I'm was working on some exercises on statistical inference and came across a question I could not solve. After a while I decided to take a look at the solution to hopefully understand the problem better.



Situation: Let $X_{1} dots X_{n}$ be independent and identically distributed random variables, each with the $N(mu,sigma^2)$ distribution. We have that $mu inmathbb{R}$ is a known parameter and $sigma^2$ is unknown. We consider $T=frac{sum^{n}_{i=i}(X_{i}-bar{X})^{2}}{n-1}$ and can use the fact that $frac{(n-1)}{sigma^2} Tsim Gamma(frac{n-1}{2},frac{1}{2})$.



My problem: In the solution is the following line:




Since $T=cY$, where $c=frac{sigma^2}{n-1}$ and $Ysim
Gamma(frac{n-1}{2},frac{1}{2})$, using a change of variables yields
that $f_{T}(x)=c^{-1}f_{Y}(x/c)$.




How does this change of variables work? How is it computed and why is it even necessary? Is it because the distribution of $T$ itself isn't explicitly given, but rather the distribution of $cT$? Any help is appreciated.






statistics normal-distribution statistical-inference change-of-variable gamma-distribution







share|cite|improve this question





























    0














    I'm was working on some exercises on statistical inference and came across a question I could not solve. After a while I decided to take a look at the solution to hopefully understand the problem better.



    Situation: Let $X_{1} dots X_{n}$ be independent and identically distributed random variables, each with the $N(mu,sigma^2)$ distribution. We have that $mu inmathbb{R}$ is a known parameter and $sigma^2$ is unknown. We consider $T=frac{sum^{n}_{i=i}(X_{i}-bar{X})^{2}}{n-1}$ and can use the fact that $frac{(n-1)}{sigma^2} Tsim Gamma(frac{n-1}{2},frac{1}{2})$.



    My problem: In the solution is the following line:




    Since $T=cY$, where $c=frac{sigma^2}{n-1}$ and $Ysim
    Gamma(frac{n-1}{2},frac{1}{2})$    , using a change of variables yields
    that $f_{T}(x)=c^{-1}f_{Y}(x/c)$.




    How does this change of variables work? How is it computed and why is it even necessary? Is it because the distribution of $T$ itself isn't explicitly given, but rather the distribution of $cT$? Any help is appreciated.






    statistics normal-distribution statistical-inference change-of-variable gamma-distribution







    share|cite|improve this question



























      0












      0








      0







      I'm was working on some exercises on statistical inference and came across a question I could not solve. After a while I decided to take a look at the solution to hopefully understand the problem better.



      Situation: Let $X_{1} dots X_{n}$ be independent and identically distributed random variables, each with the $N(mu,sigma^2)$ distribution. We have that $mu inmathbb{R}$ is a known parameter and $sigma^2$ is unknown. We consider $T=frac{sum^{n}_{i=i}(X_{i}-bar{X})^{2}}{n-1}$ and can use the fact that $frac{(n-1)}{sigma^2} Tsim Gamma(frac{n-1}{2},frac{1}{2})$.



      My problem: In the solution is the following line:




      Since $T=cY$, where $c=frac{sigma^2}{n-1}$ and $Ysim
      Gamma(frac{n-1}{2},frac{1}{2})$      , using a change of variables yields
      that $f_{T}(x)=c^{-1}f_{Y}(x/c)$.




      How does this change of variables work? How is it computed and why is it even necessary? Is it because the distribution of $T$ itself isn't explicitly given, but rather the distribution of $cT$? Any help is appreciated.






      statistics normal-distribution statistical-inference change-of-variable gamma-distribution







      share|cite|improve this question















      I'm was working on some exercises on statistical inference and came across a question I could not solve. After a while I decided to take a look at the solution to hopefully understand the problem better.



      Situation: Let $X_{1} dots X_{n}$ be independent and identically distributed random variables, each with the $N(mu,sigma^2)$ distribution. We have that $mu inmathbb{R}$ is a known parameter and $sigma^2$ is unknown. We consider $T=frac{sum^{n}_{i=i}(X_{i}-bar{X})^{2}}{n-1}$ and can use the fact that $frac{(n-1)}{sigma^2} Tsim Gamma(frac{n-1}{2},frac{1}{2})$.



      My problem: In the solution is the following line:




      Since $T=cY$, where $c=frac{sigma^2}{n-1}$ and $Ysim
      Gamma(frac{n-1}{2},frac{1}{2})$      , using a change of variables yields
      that $f_{T}(x)=c^{-1}f_{Y}(x/c)$.




      How does this change of variables work? How is it computed and why is it even necessary? Is it because the distribution of $T$ itself isn't explicitly given, but rather the distribution of $cT$? Any help is appreciated.






      statistics normal-distribution statistical-inference change-of-variable gamma-distribution




      statistics normal-distribution statistical-inference change-of-variable gamma-distribution






      share|cite|improve this question















      share|cite|improve this question













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      edited 2 days ago

























      asked 2 days ago









      S. Crim

      13211




      13211






















          1 Answer
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          You can see this as a simple change of variable in an integral:



          $$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$



          define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$



          This yields



          $$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$



          And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.






          share|cite|improve this answer





















          • The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
            – S. Crim
            2 days ago










          • I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
            – Digitalis
            2 days ago










          • Fair enough, thanks for the help anyway!
            – S. Crim
            2 days ago











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          You can see this as a simple change of variable in an integral:



          $$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$



          define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$



          This yields



          $$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$



          And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.






          share|cite|improve this answer





















          • The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
            – S. Crim
            2 days ago










          • I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
            – Digitalis
            2 days ago










          • Fair enough, thanks for the help anyway!
            – S. Crim
            2 days ago
















          1














          You can see this as a simple change of variable in an integral:



          $$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$



          define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$



          This yields



          $$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$



          And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.






          share|cite|improve this answer





















          • The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
            – S. Crim
            2 days ago










          • I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
            – Digitalis
            2 days ago










          • Fair enough, thanks for the help anyway!
            – S. Crim
            2 days ago














          1












          1








          1






          You can see this as a simple change of variable in an integral:



          $$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$



          define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$



          This yields



          $$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$



          And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.






          share|cite|improve this answer












          You can see this as a simple change of variable in an integral:



          $$ P(T < t) = P(cY < t) = P(Y < frac{t}{c}) =int_{-infty}^{frac{t}{c}} f^Y(x)dx $$



          define $ u = cx Rightarrow " du = c dx" iff dx "=" frac{1}{c}du.$



          This yields



          $$ P(T< t) = int_{-infty}^{frac{t}{c}} f^Y(x)dx = int_{-infty}^t frac{1}{c} f^Y(u/c) du .$$



          And $c^{-1}f^Y(xc^{-1})$ is a probability density function for $T$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Digitalis

          509216




          509216












          • The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
            – S. Crim
            2 days ago










          • I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
            – Digitalis
            2 days ago










          • Fair enough, thanks for the help anyway!
            – S. Crim
            2 days ago


















          • The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
            – S. Crim
            2 days ago










          • I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
            – Digitalis
            2 days ago










          • Fair enough, thanks for the help anyway!
            – S. Crim
            2 days ago
















          The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
          – S. Crim
          2 days ago




          The originial exercise was about computing the Fisher information based on $T$, i.e. $I_{T}(sigma^2)$. Is this step necessary because we only know the distribution of $cT$ and not $T$? Or is there another reason for having to change the variables?
          – S. Crim
          2 days ago












          I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
          – Digitalis
          2 days ago




          I'm afraid I'm not able to help you with that, I'm not very gifted when it comes to computing Fischer information. I'm also not familiar with the concept of calculating the fisher information based on a given statistic $T$. :(
          – Digitalis
          2 days ago












          Fair enough, thanks for the help anyway!
          – S. Crim
          2 days ago




          Fair enough, thanks for the help anyway!
          – S. Crim
          2 days ago


















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