Linear combination of Chi-squared distrubuted variables with ascending degrees of freedom












2












$begingroup$


If we have i.i.d. random variables$ quad X_1,dots , X_n, text{where} X_k sim mathcal{N} (mu_k,sigma_k^2),$ $quad$ then $$ Y =sum_{k=1}^n a_k X_n sim mathcal{N} (sum_{k=1}^n a_k mu_k,sum_{k=1}^n a_k^2sigma_k^2). $$



But what we can do with similar (or at least simplified) linear combination of $ (X_k)_{k=1}^n , X_k sim chi^2(k) ? $



Example:



Let $X,Y sim mathcal{N}(0,1)$ be independent random variables. What is the distribution of $ Z = XY $ ?



$$ XY = frac{1}{2}(X^2 + 2XY + Y^2) - frac{1}{2}(X^2 + Y^2) =left ( frac{X+Y}{sqrt{2}} right ) ^2 - frac{1}{2}(X^2 + Y^2) $$



Of course we have that $$ frac{X+Y}{sqrt{2}} = frac{frac{X+Y}{2} - 0}{1} sqrt{2} sim mathcal{N}(0,1) $$



Denoting $ Z_1 = left ( frac{X+Y}{sqrt{2}} right )^2 sim chi^2(1) , $ $ Z_2 = X^2 + Y^2 sim chi^2(2) , $ we have



$$ XY = Z_1 + frac{1}{2} Z_2 = sum_{k=1}^2 frac{Z_k}{k}, $$
where $ Z_k sim chi^2(k) .$



What more we can do with this kind of approach? I don't have any ideas, nor I can find anything useful at this matter.






probability-theory probability-distributions random-variables







share|cite|improve this question









$endgroup$












  • $begingroup$
    Should I maybe take product of characteristic functions and try to compute inverse Fourier transformat?
    $endgroup$
    – Kusavil
    Feb 16 '15 at 12:15








  • 1




    $begingroup$
    I don't see any profit in this perspective for $XY$. Especially because $Z_1$ and $Z_2$ are not independent.
    $endgroup$
    – drhab
    Feb 16 '15 at 12:38












  • $begingroup$
    Right! Thank you, so it's a wrong example. But have you by any chance heard if there is any nice formula for $chi^2$ distributed variables (let's say independent ones) with different degrees of freedom? Or at least for variables with the same degree of freedom?
    $endgroup$
    – Kusavil
    Feb 16 '15 at 13:26






  • 1




    $begingroup$
    If $chi_k^2$ and $chi_m^2$ are independent then $chi_k^2+chi_m^2$ and $chi_{k+m}^2$ have the same distribution.
    $endgroup$
    – drhab
    Feb 16 '15 at 14:21


















2












$begingroup$


If we have i.i.d. random variables$ quad X_1,dots , X_n, text{where} X_k sim mathcal{N} (mu_k,sigma_k^2),$ $quad$ then $$ Y =sum_{k=1}^n a_k X_n sim mathcal{N} (sum_{k=1}^n a_k mu_k,sum_{k=1}^n a_k^2sigma_k^2). $$



But what we can do with similar (or at least simplified) linear combination of $ (X_k)_{k=1}^n , X_k sim chi^2(k) ? $



Example:



Let $X,Y sim mathcal{N}(0,1)$ be independent random variables. What is the distribution of $ Z = XY $ ?



$$ XY = frac{1}{2}(X^2 + 2XY + Y^2) - frac{1}{2}(X^2 + Y^2) =left ( frac{X+Y}{sqrt{2}} right ) ^2 - frac{1}{2}(X^2 + Y^2) $$



Of course we have that $$ frac{X+Y}{sqrt{2}} = frac{frac{X+Y}{2} - 0}{1} sqrt{2} sim mathcal{N}(0,1) $$



Denoting $ Z_1 = left ( frac{X+Y}{sqrt{2}} right )^2 sim chi^2(1) , $ $ Z_2 = X^2 + Y^2 sim chi^2(2) , $ we have



$$ XY = Z_1 + frac{1}{2} Z_2 = sum_{k=1}^2 frac{Z_k}{k}, $$
where $ Z_k sim chi^2(k) .$



What more we can do with this kind of approach? I don't have any ideas, nor I can find anything useful at this matter.






probability-theory probability-distributions random-variables







share|cite|improve this question









$endgroup$












  • $begingroup$
    Should I maybe take product of characteristic functions and try to compute inverse Fourier transformat?
    $endgroup$
    – Kusavil
    Feb 16 '15 at 12:15








  • 1




    $begingroup$
    I don't see any profit in this perspective for $XY$. Especially because $Z_1$ and $Z_2$ are not independent.
    $endgroup$
    – drhab
    Feb 16 '15 at 12:38












  • $begingroup$
    Right! Thank you, so it's a wrong example. But have you by any chance heard if there is any nice formula for $chi^2$ distributed variables (let's say independent ones) with different degrees of freedom? Or at least for variables with the same degree of freedom?
    $endgroup$
    – Kusavil
    Feb 16 '15 at 13:26






  • 1




    $begingroup$
    If $chi_k^2$ and $chi_m^2$ are independent then $chi_k^2+chi_m^2$ and $chi_{k+m}^2$ have the same distribution.
    $endgroup$
    – drhab
    Feb 16 '15 at 14:21
















2












2








2





$begingroup$


If we have i.i.d. random variables$ quad X_1,dots , X_n, text{where} X_k sim mathcal{N} (mu_k,sigma_k^2),$ $quad$ then $$ Y =sum_{k=1}^n a_k X_n sim mathcal{N} (sum_{k=1}^n a_k mu_k,sum_{k=1}^n a_k^2sigma_k^2). $$



But what we can do with similar (or at least simplified) linear combination of $ (X_k)_{k=1}^n , X_k sim chi^2(k) ? $



Example:



Let $X,Y sim mathcal{N}(0,1)$ be independent random variables. What is the distribution of $ Z = XY $ ?



$$ XY = frac{1}{2}(X^2 + 2XY + Y^2) - frac{1}{2}(X^2 + Y^2) =left ( frac{X+Y}{sqrt{2}} right ) ^2 - frac{1}{2}(X^2 + Y^2) $$



Of course we have that $$ frac{X+Y}{sqrt{2}} = frac{frac{X+Y}{2} - 0}{1} sqrt{2} sim mathcal{N}(0,1) $$



Denoting $ Z_1 = left ( frac{X+Y}{sqrt{2}} right )^2 sim chi^2(1) , $ $ Z_2 = X^2 + Y^2 sim chi^2(2) , $ we have



$$ XY = Z_1 + frac{1}{2} Z_2 = sum_{k=1}^2 frac{Z_k}{k}, $$
where $ Z_k sim chi^2(k) .$



What more we can do with this kind of approach? I don't have any ideas, nor I can find anything useful at this matter.






probability-theory probability-distributions random-variables







share|cite|improve this question









$endgroup$




If we have i.i.d. random variables$ quad X_1,dots , X_n, text{where} X_k sim mathcal{N} (mu_k,sigma_k^2),$ $quad$ then $$ Y =sum_{k=1}^n a_k X_n sim mathcal{N} (sum_{k=1}^n a_k mu_k,sum_{k=1}^n a_k^2sigma_k^2). $$



But what we can do with similar (or at least simplified) linear combination of $ (X_k)_{k=1}^n , X_k sim chi^2(k) ? $



Example:



Let $X,Y sim mathcal{N}(0,1)$ be independent random variables. What is the distribution of $ Z = XY $ ?



$$ XY = frac{1}{2}(X^2 + 2XY + Y^2) - frac{1}{2}(X^2 + Y^2) =left ( frac{X+Y}{sqrt{2}} right ) ^2 - frac{1}{2}(X^2 + Y^2) $$



Of course we have that $$ frac{X+Y}{sqrt{2}} = frac{frac{X+Y}{2} - 0}{1} sqrt{2} sim mathcal{N}(0,1) $$



Denoting $ Z_1 = left ( frac{X+Y}{sqrt{2}} right )^2 sim chi^2(1) , $ $ Z_2 = X^2 + Y^2 sim chi^2(2) , $ we have



$$ XY = Z_1 + frac{1}{2} Z_2 = sum_{k=1}^2 frac{Z_k}{k}, $$
where $ Z_k sim chi^2(k) .$



What more we can do with this kind of approach? I don't have any ideas, nor I can find anything useful at this matter.






probability-theory probability-distributions random-variables




probability-theory probability-distributions random-variables






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 16 '15 at 12:07









KusavilKusavil

363212




363212












  • $begingroup$
    Should I maybe take product of characteristic functions and try to compute inverse Fourier transformat?
    $endgroup$
    – Kusavil
    Feb 16 '15 at 12:15








  • 1




    $begingroup$
    I don't see any profit in this perspective for $XY$. Especially because $Z_1$ and $Z_2$ are not independent.
    $endgroup$
    – drhab
    Feb 16 '15 at 12:38












  • $begingroup$
    Right! Thank you, so it's a wrong example. But have you by any chance heard if there is any nice formula for $chi^2$ distributed variables (let's say independent ones) with different degrees of freedom? Or at least for variables with the same degree of freedom?
    $endgroup$
    – Kusavil
    Feb 16 '15 at 13:26






  • 1




    $begingroup$
    If $chi_k^2$ and $chi_m^2$ are independent then $chi_k^2+chi_m^2$ and $chi_{k+m}^2$ have the same distribution.
    $endgroup$
    – drhab
    Feb 16 '15 at 14:21




















  • $begingroup$
    Should I maybe take product of characteristic functions and try to compute inverse Fourier transformat?
    $endgroup$
    – Kusavil
    Feb 16 '15 at 12:15








  • 1




    $begingroup$
    I don't see any profit in this perspective for $XY$. Especially because $Z_1$ and $Z_2$ are not independent.
    $endgroup$
    – drhab
    Feb 16 '15 at 12:38












  • $begingroup$
    Right! Thank you, so it's a wrong example. But have you by any chance heard if there is any nice formula for $chi^2$ distributed variables (let's say independent ones) with different degrees of freedom? Or at least for variables with the same degree of freedom?
    $endgroup$
    – Kusavil
    Feb 16 '15 at 13:26






  • 1




    $begingroup$
    If $chi_k^2$ and $chi_m^2$ are independent then $chi_k^2+chi_m^2$ and $chi_{k+m}^2$ have the same distribution.
    $endgroup$
    – drhab
    Feb 16 '15 at 14:21


















$begingroup$
Should I maybe take product of characteristic functions and try to compute inverse Fourier transformat?
$endgroup$
– Kusavil
Feb 16 '15 at 12:15






$begingroup$
Should I maybe take product of characteristic functions and try to compute inverse Fourier transformat?
$endgroup$
– Kusavil
Feb 16 '15 at 12:15






1




1




$begingroup$
I don't see any profit in this perspective for $XY$. Especially because $Z_1$ and $Z_2$ are not independent.
$endgroup$
– drhab
Feb 16 '15 at 12:38






$begingroup$
I don't see any profit in this perspective for $XY$. Especially because $Z_1$ and $Z_2$ are not independent.
$endgroup$
– drhab
Feb 16 '15 at 12:38














$begingroup$
Right! Thank you, so it's a wrong example. But have you by any chance heard if there is any nice formula for $chi^2$ distributed variables (let's say independent ones) with different degrees of freedom? Or at least for variables with the same degree of freedom?
$endgroup$
– Kusavil
Feb 16 '15 at 13:26




$begingroup$
Right! Thank you, so it's a wrong example. But have you by any chance heard if there is any nice formula for $chi^2$ distributed variables (let's say independent ones) with different degrees of freedom? Or at least for variables with the same degree of freedom?
$endgroup$
– Kusavil
Feb 16 '15 at 13:26




1




1




$begingroup$
If $chi_k^2$ and $chi_m^2$ are independent then $chi_k^2+chi_m^2$ and $chi_{k+m}^2$ have the same distribution.
$endgroup$
– drhab
Feb 16 '15 at 14:21






$begingroup$
If $chi_k^2$ and $chi_m^2$ are independent then $chi_k^2+chi_m^2$ and $chi_{k+m}^2$ have the same distribution.
$endgroup$
– drhab
Feb 16 '15 at 14:21












1 Answer
1






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$begingroup$

Your $Z_1$ and $Z_2$ are correlated so the sum is not like a sum of independent chi-square distributed variables.



If you use instead $Z_1 = X-Y$ and $Z_2 = X+Y$ which are two independent variables distributed as $N(0,2)$, then $$XY = frac{1}{4}Z_1^2- frac{1}{4} Z_2^2$$ thus the product XY it is distributed as the difference of two chi-square distributed variables.



This has a variance-gamma distribution as explained in https://math.stackexchange.com/a/85525/466748






share|cite|improve this answer









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    $begingroup$

    Your $Z_1$ and $Z_2$ are correlated so the sum is not like a sum of independent chi-square distributed variables.



    If you use instead $Z_1 = X-Y$ and $Z_2 = X+Y$ which are two independent variables distributed as $N(0,2)$, then $$XY = frac{1}{4}Z_1^2- frac{1}{4} Z_2^2$$ thus the product XY it is distributed as the difference of two chi-square distributed variables.



    This has a variance-gamma distribution as explained in https://math.stackexchange.com/a/85525/466748






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your $Z_1$ and $Z_2$ are correlated so the sum is not like a sum of independent chi-square distributed variables.



      If you use instead $Z_1 = X-Y$ and $Z_2 = X+Y$ which are two independent variables distributed as $N(0,2)$, then $$XY = frac{1}{4}Z_1^2- frac{1}{4} Z_2^2$$ thus the product XY it is distributed as the difference of two chi-square distributed variables.



      This has a variance-gamma distribution as explained in https://math.stackexchange.com/a/85525/466748






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your $Z_1$ and $Z_2$ are correlated so the sum is not like a sum of independent chi-square distributed variables.



        If you use instead $Z_1 = X-Y$ and $Z_2 = X+Y$ which are two independent variables distributed as $N(0,2)$, then $$XY = frac{1}{4}Z_1^2- frac{1}{4} Z_2^2$$ thus the product XY it is distributed as the difference of two chi-square distributed variables.



        This has a variance-gamma distribution as explained in https://math.stackexchange.com/a/85525/466748






        share|cite|improve this answer









        $endgroup$



        Your $Z_1$ and $Z_2$ are correlated so the sum is not like a sum of independent chi-square distributed variables.



        If you use instead $Z_1 = X-Y$ and $Z_2 = X+Y$ which are two independent variables distributed as $N(0,2)$, then $$XY = frac{1}{4}Z_1^2- frac{1}{4} Z_2^2$$ thus the product XY it is distributed as the difference of two chi-square distributed variables.



        This has a variance-gamma distribution as explained in https://math.stackexchange.com/a/85525/466748







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 7 '18 at 8:14









        Martijn WeteringsMartijn Weterings

        17010




        17010






























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