Linear combination of Chi-squared distrubuted variables with ascending degrees of freedom
$begingroup$
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
asked Feb 16 '15 at 12:07
KusavilKusavil
363213
$endgroup$
add a comment |
$begingroup$
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
asked Feb 16 '15 at 12:07
KusavilKusavil
363213
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Feb 16 '15 at 12:07
KusavilKusavil
363213
$endgroup$
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
asked Feb 16 '15 at 12:07
KusavilKusavil
363213
asked Feb 16 '15 at 12:07
KusavilKusavil
363213
asked Feb 16 '15 at 12:07
KusavilKusavil
363213
asked Feb 16 '15 at 12:07
KusavilKusavil
363213
asked Feb 16 '15 at 12:07
KusavilKusavil
363213
363213
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 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
answered Jun 7 '18 at 8:14
Martijn WeteringsMartijn Weterings
17010
$endgroup$
add a comment |
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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
answered Jun 7 '18 at 8:14
Martijn WeteringsMartijn Weterings
17010
$endgroup$
add a comment |
$begingroup$
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
answered Jun 7 '18 at 8:14
Martijn WeteringsMartijn Weterings
17010
$endgroup$
add a comment |
$begingroup$
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
answered Jun 7 '18 at 8:14
Martijn WeteringsMartijn Weterings
17010
$endgroup$
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
answered Jun 7 '18 at 8:14
Martijn WeteringsMartijn Weterings
17010
answered Jun 7 '18 at 8:14
Martijn WeteringsMartijn Weterings
17010
answered Jun 7 '18 at 8:14
Martijn WeteringsMartijn Weterings
17010
answered Jun 7 '18 at 8:14
Martijn WeteringsMartijn Weterings
17010
17010
add a comment |
add a comment |
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$begingroup$
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