Relationship between normal sample variance (mean known or unknown) and Chi-squared












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$begingroup$


I know that $(n-1)s^2/sigma^2$ is Chi-squared with $n-1$ degrees freedom. I'm currently working on a question where the population mean $mu$ is known, i.e. $s^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$.



My textbook states: "Clearly, $frac{sum_{i=1}^n(X_i-mu)^2}{sigma^2}$ is a Chi-squared random variable with $n$ degrees of freedom". I know that in this problem, it holds that $frac{sum_{i=1}^n(X_i-mu)^2}{sigma^2}$=$frac{ns^2}{sigma^2}$.



To me, this makes it seem like $frac{qs^2}{sigma^2}$ is chi squared with $q$ degrees of freedom, for any integer $q$, but I'm relatively positive that this is untrue. Can someone help me understand this discrepancy?






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    0












    $begingroup$


    I know that $(n-1)s^2/sigma^2$ is Chi-squared with $n-1$ degrees freedom. I'm currently working on a question where the population mean $mu$ is known, i.e. $s^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$.



    My textbook states: "Clearly, $frac{sum_{i=1}^n(X_i-mu)^2}{sigma^2}$ is a Chi-squared random variable with $n$ degrees of freedom". I know that in this problem, it holds that $frac{sum_{i=1}^n(X_i-mu)^2}{sigma^2}$=$frac{ns^2}{sigma^2}$.



    To me, this makes it seem like $frac{qs^2}{sigma^2}$ is chi squared with $q$ degrees of freedom, for any integer $q$, but I'm relatively positive that this is untrue. Can someone help me understand this discrepancy?






    probability random-variables







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I know that $(n-1)s^2/sigma^2$ is Chi-squared with $n-1$ degrees freedom. I'm currently working on a question where the population mean $mu$ is known, i.e. $s^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$.



      My textbook states: "Clearly, $frac{sum_{i=1}^n(X_i-mu)^2}{sigma^2}$ is a Chi-squared random variable with $n$ degrees of freedom". I know that in this problem, it holds that $frac{sum_{i=1}^n(X_i-mu)^2}{sigma^2}$=$frac{ns^2}{sigma^2}$.



      To me, this makes it seem like $frac{qs^2}{sigma^2}$ is chi squared with $q$ degrees of freedom, for any integer $q$, but I'm relatively positive that this is untrue. Can someone help me understand this discrepancy?






      probability random-variables







      share|cite|improve this question











      $endgroup$




      I know that $(n-1)s^2/sigma^2$ is Chi-squared with $n-1$ degrees freedom. I'm currently working on a question where the population mean $mu$ is known, i.e. $s^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$.



      My textbook states: "Clearly, $frac{sum_{i=1}^n(X_i-mu)^2}{sigma^2}$ is a Chi-squared random variable with $n$ degrees of freedom". I know that in this problem, it holds that $frac{sum_{i=1}^n(X_i-mu)^2}{sigma^2}$=$frac{ns^2}{sigma^2}$.



      To me, this makes it seem like $frac{qs^2}{sigma^2}$ is chi squared with $q$ degrees of freedom, for any integer $q$, but I'm relatively positive that this is untrue. Can someone help me understand this discrepancy?






      probability random-variables




      probability random-variables






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      edited Dec 9 '18 at 13:51









      GNUSupporter 8964民主女神 地下教會

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      asked Dec 9 '18 at 13:49









      DavidDavid

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          $begingroup$

          So let's be more explicit here. Let me "abuse" the notation a bit and do the following:



          $$s_n^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$$



          Now, divide/ multiply by the variance $sigma^2$ as
          $$s_n^2=n^{-1}sigma^2sum_{i=1}^nfrac{(X_i-mu)^2}{sigma^2}$$
          The part $frac{(X_i-mu)^2}{sigma^2}$ is Chi-square with $n$ d.o.f. (degrees of freedom) this is correct. Let's denote this Chi-square as $chi_n^2$ with mean $n$ and variance $2n$. We have
          $$s_n^2=n^{-1}sigma^2chi_n^2$$
          where $n^{-1}sigma^2$ is a constant. To characterize it more we have
          begin{align}
          E s_n^2 &= n^{-1}sigma^2Echi_n^2 = n^{-1}sigma^2 (n) = sigma^2\
          text{var } s_n^2 &= n^{-1}sigma^2text{var } chi_n^2 = n^{-1}sigma^2 (2n) = 2sigma^2
          end{align}



          so addressing your question $frac{qs_n^2}{sigma^2}$ is Chi-square with $n$ d.o.f. with mean and variances as
          begin{align}
          E frac{qs_n^2}{sigma^2} &= frac{q}{sigma^2}Es_n^2 = q\
          text{var } frac{qs_n^2}{sigma^2} &=frac{q}{sigma^2}text{var } chi_n^2 = frac{q}{sigma^2}(2sigma^2) =2q
          end{align}
          In case you meant $frac{qs_q^2}{sigma^2}$, nothing changes except the d.o.f, i.e. we will have $q$ instead of $n$ d.o.f. That is why I insisted on $s_n$ instead of just $s$.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            So let's be more explicit here. Let me "abuse" the notation a bit and do the following:



            $$s_n^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$$



            Now, divide/ multiply by the variance $sigma^2$ as
            $$s_n^2=n^{-1}sigma^2sum_{i=1}^nfrac{(X_i-mu)^2}{sigma^2}$$
            The part $frac{(X_i-mu)^2}{sigma^2}$ is Chi-square with $n$ d.o.f. (degrees of freedom) this is correct. Let's denote this Chi-square as $chi_n^2$ with mean $n$ and variance $2n$. We have
            $$s_n^2=n^{-1}sigma^2chi_n^2$$
            where $n^{-1}sigma^2$ is a constant. To characterize it more we have
            begin{align}
            E s_n^2 &= n^{-1}sigma^2Echi_n^2 = n^{-1}sigma^2 (n) = sigma^2\
            text{var } s_n^2 &= n^{-1}sigma^2text{var } chi_n^2 = n^{-1}sigma^2 (2n) = 2sigma^2
            end{align}



            so addressing your question $frac{qs_n^2}{sigma^2}$ is Chi-square with $n$ d.o.f. with mean and variances as
            begin{align}
            E frac{qs_n^2}{sigma^2} &= frac{q}{sigma^2}Es_n^2 = q\
            text{var } frac{qs_n^2}{sigma^2} &=frac{q}{sigma^2}text{var } chi_n^2 = frac{q}{sigma^2}(2sigma^2) =2q
            end{align}
            In case you meant $frac{qs_q^2}{sigma^2}$, nothing changes except the d.o.f, i.e. we will have $q$ instead of $n$ d.o.f. That is why I insisted on $s_n$ instead of just $s$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              So let's be more explicit here. Let me "abuse" the notation a bit and do the following:



              $$s_n^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$$



              Now, divide/ multiply by the variance $sigma^2$ as
              $$s_n^2=n^{-1}sigma^2sum_{i=1}^nfrac{(X_i-mu)^2}{sigma^2}$$
              The part $frac{(X_i-mu)^2}{sigma^2}$ is Chi-square with $n$ d.o.f. (degrees of freedom) this is correct. Let's denote this Chi-square as $chi_n^2$ with mean $n$ and variance $2n$. We have
              $$s_n^2=n^{-1}sigma^2chi_n^2$$
              where $n^{-1}sigma^2$ is a constant. To characterize it more we have
              begin{align}
              E s_n^2 &= n^{-1}sigma^2Echi_n^2 = n^{-1}sigma^2 (n) = sigma^2\
              text{var } s_n^2 &= n^{-1}sigma^2text{var } chi_n^2 = n^{-1}sigma^2 (2n) = 2sigma^2
              end{align}



              so addressing your question $frac{qs_n^2}{sigma^2}$ is Chi-square with $n$ d.o.f. with mean and variances as
              begin{align}
              E frac{qs_n^2}{sigma^2} &= frac{q}{sigma^2}Es_n^2 = q\
              text{var } frac{qs_n^2}{sigma^2} &=frac{q}{sigma^2}text{var } chi_n^2 = frac{q}{sigma^2}(2sigma^2) =2q
              end{align}
              In case you meant $frac{qs_q^2}{sigma^2}$, nothing changes except the d.o.f, i.e. we will have $q$ instead of $n$ d.o.f. That is why I insisted on $s_n$ instead of just $s$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                So let's be more explicit here. Let me "abuse" the notation a bit and do the following:



                $$s_n^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$$



                Now, divide/ multiply by the variance $sigma^2$ as
                $$s_n^2=n^{-1}sigma^2sum_{i=1}^nfrac{(X_i-mu)^2}{sigma^2}$$
                The part $frac{(X_i-mu)^2}{sigma^2}$ is Chi-square with $n$ d.o.f. (degrees of freedom) this is correct. Let's denote this Chi-square as $chi_n^2$ with mean $n$ and variance $2n$. We have
                $$s_n^2=n^{-1}sigma^2chi_n^2$$
                where $n^{-1}sigma^2$ is a constant. To characterize it more we have
                begin{align}
                E s_n^2 &= n^{-1}sigma^2Echi_n^2 = n^{-1}sigma^2 (n) = sigma^2\
                text{var } s_n^2 &= n^{-1}sigma^2text{var } chi_n^2 = n^{-1}sigma^2 (2n) = 2sigma^2
                end{align}



                so addressing your question $frac{qs_n^2}{sigma^2}$ is Chi-square with $n$ d.o.f. with mean and variances as
                begin{align}
                E frac{qs_n^2}{sigma^2} &= frac{q}{sigma^2}Es_n^2 = q\
                text{var } frac{qs_n^2}{sigma^2} &=frac{q}{sigma^2}text{var } chi_n^2 = frac{q}{sigma^2}(2sigma^2) =2q
                end{align}
                In case you meant $frac{qs_q^2}{sigma^2}$, nothing changes except the d.o.f, i.e. we will have $q$ instead of $n$ d.o.f. That is why I insisted on $s_n$ instead of just $s$.






                share|cite|improve this answer











                $endgroup$



                So let's be more explicit here. Let me "abuse" the notation a bit and do the following:



                $$s_n^2=n^{-1}sum_{i=1}^n(X_i-mu)^2$$



                Now, divide/ multiply by the variance $sigma^2$ as
                $$s_n^2=n^{-1}sigma^2sum_{i=1}^nfrac{(X_i-mu)^2}{sigma^2}$$
                The part $frac{(X_i-mu)^2}{sigma^2}$ is Chi-square with $n$ d.o.f. (degrees of freedom) this is correct. Let's denote this Chi-square as $chi_n^2$ with mean $n$ and variance $2n$. We have
                $$s_n^2=n^{-1}sigma^2chi_n^2$$
                where $n^{-1}sigma^2$ is a constant. To characterize it more we have
                begin{align}
                E s_n^2 &= n^{-1}sigma^2Echi_n^2 = n^{-1}sigma^2 (n) = sigma^2\
                text{var } s_n^2 &= n^{-1}sigma^2text{var } chi_n^2 = n^{-1}sigma^2 (2n) = 2sigma^2
                end{align}



                so addressing your question $frac{qs_n^2}{sigma^2}$ is Chi-square with $n$ d.o.f. with mean and variances as
                begin{align}
                E frac{qs_n^2}{sigma^2} &= frac{q}{sigma^2}Es_n^2 = q\
                text{var } frac{qs_n^2}{sigma^2} &=frac{q}{sigma^2}text{var } chi_n^2 = frac{q}{sigma^2}(2sigma^2) =2q
                end{align}
                In case you meant $frac{qs_q^2}{sigma^2}$, nothing changes except the d.o.f, i.e. we will have $q$ instead of $n$ d.o.f. That is why I insisted on $s_n$ instead of just $s$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 9 '18 at 14:03

























                answered Dec 9 '18 at 13:57









                Ahmad BazziAhmad Bazzi

                8,0012724




                8,0012724






























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