Proof of relation between Normal and Chi-square












1












$begingroup$


Let $Xsim N(0,1)$, i want to determine the distribution of $Y=X^2$.



By definition, the density of $X$ is $f_X(x)=frac{1}{sigmasqrt{2pi}}e^{-frac{1}{2sigma^2}(x-mu)^2},forall xin mathbb{R}$. I also know that i can't apply the law of transformation of random variables because $Y$ is not a monotonic function. So, i write:



$F_Y (y)=mathbb{P}(Yleq y)=mathbb{P}(X^2leq y)=mathbb{P}(Xleq pm sqrt{y})=mathbb{P}(-sqrt{y}leq X leqsqrt{y})$



Then, i observe that if $Xsim N(0,1)$ can be applied the following equivalences:



1) $mathbb{P}(-sqrt{y}leq Xleq sqrt{y})=mathbb{P}(-sqrt{y}leq X leq0)+mathbb{P}(0leq Xleq sqrt{y})$



2) $mathbb{P}(-sqrt{y}leq X leq0)=mathbb{P}(0leq Xleq sqrt{y})$



So, i can write that



$mathbb{P}(-sqrt{y}leq Xleq sqrt{y})=2mathbb{P}(0leq Xleq sqrt{y})=2[mathbb{P}(Xleq sqrt{y})-mathbb{P}(Xleq 0)]=2[F_X(sqrt{y})-F_X(0)]=2[Phi (sqrt{y})-Phi (0)]=2Phi (sqrt{y})-2Phi (0)=2Phi (sqrt{y})-2(0,5)=2Phi (sqrt{y})-1$



On this way, i have to find the density of $Y$:



$f_Y(y)=frac{d}{dy}F_Y(y)=frac{d}{dy}(2Phi (sqrt{y})-1)$



but now i'm stuck. My book is not quite clear on this point so much so that it has to go straight to the solution, that is $f_Y(y)=frac{1}{sqrt{2pi y}}e^{-frac{1}{2}y}Rightarrow Ysim chi ^2_1 $.



The demonstration is correct? How can calculate that derivative? Why $chi ^2$ has $1$ degree of freedom?



Thanks for any help!






probability linear-transformations normal-distribution density-function







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $Xsim N(0,1)$, i want to determine the distribution of $Y=X^2$.



    By definition, the density of $X$ is $f_X(x)=frac{1}{sigmasqrt{2pi}}e^{-frac{1}{2sigma^2}(x-mu)^2},forall xin mathbb{R}$. I also know that i can't apply the law of transformation of random variables because $Y$ is not a monotonic function. So, i write:



    $F_Y (y)=mathbb{P}(Yleq y)=mathbb{P}(X^2leq y)=mathbb{P}(Xleq pm sqrt{y})=mathbb{P}(-sqrt{y}leq X leqsqrt{y})$



    Then, i observe that if $Xsim N(0,1)$ can be applied the following equivalences:



    1) $mathbb{P}(-sqrt{y}leq Xleq sqrt{y})=mathbb{P}(-sqrt{y}leq X leq0)+mathbb{P}(0leq Xleq sqrt{y})$



    2) $mathbb{P}(-sqrt{y}leq X leq0)=mathbb{P}(0leq Xleq sqrt{y})$



    So, i can write that



    $mathbb{P}(-sqrt{y}leq Xleq sqrt{y})=2mathbb{P}(0leq Xleq sqrt{y})=2[mathbb{P}(Xleq sqrt{y})-mathbb{P}(Xleq 0)]=2[F_X(sqrt{y})-F_X(0)]=2[Phi (sqrt{y})-Phi (0)]=2Phi (sqrt{y})-2Phi (0)=2Phi (sqrt{y})-2(0,5)=2Phi (sqrt{y})-1$



    On this way, i have to find the density of $Y$:



    $f_Y(y)=frac{d}{dy}F_Y(y)=frac{d}{dy}(2Phi (sqrt{y})-1)$



    but now i'm stuck. My book is not quite clear on this point so much so that it has to go straight to the solution, that is $f_Y(y)=frac{1}{sqrt{2pi y}}e^{-frac{1}{2}y}Rightarrow Ysim chi ^2_1 $.



    The demonstration is correct? How can calculate that derivative? Why $chi ^2$ has $1$ degree of freedom?



    Thanks for any help!






    probability linear-transformations normal-distribution density-function







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $Xsim N(0,1)$, i want to determine the distribution of $Y=X^2$.



      By definition, the density of $X$ is $f_X(x)=frac{1}{sigmasqrt{2pi}}e^{-frac{1}{2sigma^2}(x-mu)^2},forall xin mathbb{R}$. I also know that i can't apply the law of transformation of random variables because $Y$ is not a monotonic function. So, i write:



      $F_Y (y)=mathbb{P}(Yleq y)=mathbb{P}(X^2leq y)=mathbb{P}(Xleq pm sqrt{y})=mathbb{P}(-sqrt{y}leq X leqsqrt{y})$



      Then, i observe that if $Xsim N(0,1)$ can be applied the following equivalences:



      1) $mathbb{P}(-sqrt{y}leq Xleq sqrt{y})=mathbb{P}(-sqrt{y}leq X leq0)+mathbb{P}(0leq Xleq sqrt{y})$



      2) $mathbb{P}(-sqrt{y}leq X leq0)=mathbb{P}(0leq Xleq sqrt{y})$



      So, i can write that



      $mathbb{P}(-sqrt{y}leq Xleq sqrt{y})=2mathbb{P}(0leq Xleq sqrt{y})=2[mathbb{P}(Xleq sqrt{y})-mathbb{P}(Xleq 0)]=2[F_X(sqrt{y})-F_X(0)]=2[Phi (sqrt{y})-Phi (0)]=2Phi (sqrt{y})-2Phi (0)=2Phi (sqrt{y})-2(0,5)=2Phi (sqrt{y})-1$



      On this way, i have to find the density of $Y$:



      $f_Y(y)=frac{d}{dy}F_Y(y)=frac{d}{dy}(2Phi (sqrt{y})-1)$



      but now i'm stuck. My book is not quite clear on this point so much so that it has to go straight to the solution, that is $f_Y(y)=frac{1}{sqrt{2pi y}}e^{-frac{1}{2}y}Rightarrow Ysim chi ^2_1 $.



      The demonstration is correct? How can calculate that derivative? Why $chi ^2$ has $1$ degree of freedom?



      Thanks for any help!






      probability linear-transformations normal-distribution density-function







      share|cite|improve this question









      $endgroup$




      Let $Xsim N(0,1)$, i want to determine the distribution of $Y=X^2$.



      By definition, the density of $X$ is $f_X(x)=frac{1}{sigmasqrt{2pi}}e^{-frac{1}{2sigma^2}(x-mu)^2},forall xin mathbb{R}$. I also know that i can't apply the law of transformation of random variables because $Y$ is not a monotonic function. So, i write:



      $F_Y (y)=mathbb{P}(Yleq y)=mathbb{P}(X^2leq y)=mathbb{P}(Xleq pm sqrt{y})=mathbb{P}(-sqrt{y}leq X leqsqrt{y})$



      Then, i observe that if $Xsim N(0,1)$ can be applied the following equivalences:



      1) $mathbb{P}(-sqrt{y}leq Xleq sqrt{y})=mathbb{P}(-sqrt{y}leq X leq0)+mathbb{P}(0leq Xleq sqrt{y})$



      2) $mathbb{P}(-sqrt{y}leq X leq0)=mathbb{P}(0leq Xleq sqrt{y})$



      So, i can write that



      $mathbb{P}(-sqrt{y}leq Xleq sqrt{y})=2mathbb{P}(0leq Xleq sqrt{y})=2[mathbb{P}(Xleq sqrt{y})-mathbb{P}(Xleq 0)]=2[F_X(sqrt{y})-F_X(0)]=2[Phi (sqrt{y})-Phi (0)]=2Phi (sqrt{y})-2Phi (0)=2Phi (sqrt{y})-2(0,5)=2Phi (sqrt{y})-1$



      On this way, i have to find the density of $Y$:



      $f_Y(y)=frac{d}{dy}F_Y(y)=frac{d}{dy}(2Phi (sqrt{y})-1)$



      but now i'm stuck. My book is not quite clear on this point so much so that it has to go straight to the solution, that is $f_Y(y)=frac{1}{sqrt{2pi y}}e^{-frac{1}{2}y}Rightarrow Ysim chi ^2_1 $.



      The demonstration is correct? How can calculate that derivative? Why $chi ^2$ has $1$ degree of freedom?



      Thanks for any help!






      probability linear-transformations normal-distribution density-function




      probability linear-transformations normal-distribution density-function






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 7 '18 at 17:24









      Marco PittellaMarco Pittella

      1288




      1288






















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          $begingroup$

          Use the chain rule. Let $phi$ be the density of a standard normal. Then we have that
          $$
          Phi'=phi
          $$
          and so
          $$
          frac{d}{dy}(2Phi (sqrt{y})-1)=2phi(sqrt y)times frac{1}{2sqrt y}=frac{1}{sqrt{2pi y}}e^{-y/2}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your answer! Clear!
            $endgroup$
            – Marco Pittella
            Dec 7 '18 at 17:55











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          $begingroup$

          Use the chain rule. Let $phi$ be the density of a standard normal. Then we have that
          $$
          Phi'=phi
          $$
          and so
          $$
          frac{d}{dy}(2Phi (sqrt{y})-1)=2phi(sqrt y)times frac{1}{2sqrt y}=frac{1}{sqrt{2pi y}}e^{-y/2}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your answer! Clear!
            $endgroup$
            – Marco Pittella
            Dec 7 '18 at 17:55
















          2












          $begingroup$

          Use the chain rule. Let $phi$ be the density of a standard normal. Then we have that
          $$
          Phi'=phi
          $$
          and so
          $$
          frac{d}{dy}(2Phi (sqrt{y})-1)=2phi(sqrt y)times frac{1}{2sqrt y}=frac{1}{sqrt{2pi y}}e^{-y/2}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your answer! Clear!
            $endgroup$
            – Marco Pittella
            Dec 7 '18 at 17:55














          2












          2








          2





          $begingroup$

          Use the chain rule. Let $phi$ be the density of a standard normal. Then we have that
          $$
          Phi'=phi
          $$
          and so
          $$
          frac{d}{dy}(2Phi (sqrt{y})-1)=2phi(sqrt y)times frac{1}{2sqrt y}=frac{1}{sqrt{2pi y}}e^{-y/2}
          $$






          share|cite|improve this answer









          $endgroup$



          Use the chain rule. Let $phi$ be the density of a standard normal. Then we have that
          $$
          Phi'=phi
          $$
          and so
          $$
          frac{d}{dy}(2Phi (sqrt{y})-1)=2phi(sqrt y)times frac{1}{2sqrt y}=frac{1}{sqrt{2pi y}}e^{-y/2}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 17:30









          Foobaz JohnFoobaz John

          22.1k41352




          22.1k41352












          • $begingroup$
            Thanks for your answer! Clear!
            $endgroup$
            – Marco Pittella
            Dec 7 '18 at 17:55


















          • $begingroup$
            Thanks for your answer! Clear!
            $endgroup$
            – Marco Pittella
            Dec 7 '18 at 17:55
















          $begingroup$
          Thanks for your answer! Clear!
          $endgroup$
          – Marco Pittella
          Dec 7 '18 at 17:55




          $begingroup$
          Thanks for your answer! Clear!
          $endgroup$
          – Marco Pittella
          Dec 7 '18 at 17:55


















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