Rieman sum for probability estimation $P(x_1^D leq X_1 leq x_1^U, X_2 = x_2)$ with copula density
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I am looking into a way to estimate my probability of presence of a variable using Riemann sum to estimate density, let us consider two random variables $X_1$ and $X_2$ the probability is given by,
$P(X_1 leq x_1 , X_2 leq x_2) = int_{-infty}^{x_1} int_{-infty}^{x_1}f_{X_1 , X_2} (x , y) dx dy \$
$=P(U_1 leq u_1, U_2 leq u_2)$
$=C(u_1,u_2)$
$=int_{-infty}^{u_1} int_{-infty}^{u_1}c_{u_1 , u_2} (u , v) du dv$
meaning that ,
$P(x_1^D leq X_1 leq x_1^U, X_2 = x_2) = int_{u_1^D}^{u_1^U} c_{u_1 , u_2} (u , u_2) du$
That can be approximated by,
$simeq sum_{u=1}^{n}c(u_i,v_2) Delta u_i$
Is the latest equality true or do I miss something?
Plus is it extendable to the multivariate case?
$sum_{u=1}^{n}c(u_i,u_2,...,u_n) Delta u_i$
Is there a better way to estimate multivariate probability?
Many thanks in advance
EDIT: found my mistake, indeed the equivalence between the copula density and the multivariate density function is not that straightforward, indeed,
$c(u_1,u_2)=frac{partial ^d}{partial u_1 partial u_2} C(u_1,u_2)$
Where $partial ^d C(u_1,u_2)$ will obviously be equal to $f(x_1,x_2)$ but as $u_1$ is equivalent to $F_{X_1}(x_1)$, $partial u_1 partial u_2= f_1(x_1)f_2(x_2)$
So the Riemann approximation should be,
$sum_{u=1}^{n}c(u_i,v_2). f_1(F^{-1}(u_i)).f_2(F^{-1}(v_2)) Delta u_i$
probability
asked Dec 3 '18 at 19:54
HammerPowerHammerPower
62
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migrated from quant.stackexchange.com Dec 4 '18 at 7:24
This question came from our site for finance professionals and academics.
add a comment |
$begingroup$
I am looking into a way to estimate my probability of presence of a variable using Riemann sum to estimate density, let us consider two random variables $X_1$ and $X_2$ the probability is given by,
$P(X_1 leq x_1 , X_2 leq x_2) = int_{-infty}^{x_1} int_{-infty}^{x_1}f_{X_1 , X_2} (x , y) dx dy \$
$=P(U_1 leq u_1, U_2 leq u_2)$
$=C(u_1,u_2)$
$=int_{-infty}^{u_1} int_{-infty}^{u_1}c_{u_1 , u_2} (u , v) du dv$
meaning that ,
$P(x_1^D leq X_1 leq x_1^U, X_2 = x_2) = int_{u_1^D}^{u_1^U} c_{u_1 , u_2} (u , u_2) du$
That can be approximated by,
$simeq sum_{u=1}^{n}c(u_i,v_2) Delta u_i$
Is the latest equality true or do I miss something?
Plus is it extendable to the multivariate case?
$sum_{u=1}^{n}c(u_i,u_2,...,u_n) Delta u_i$
Is there a better way to estimate multivariate probability?
Many thanks in advance
EDIT: found my mistake, indeed the equivalence between the copula density and the multivariate density function is not that straightforward, indeed,
$c(u_1,u_2)=frac{partial ^d}{partial u_1 partial u_2} C(u_1,u_2)$
Where $partial ^d C(u_1,u_2)$ will obviously be equal to $f(x_1,x_2)$ but as $u_1$ is equivalent to $F_{X_1}(x_1)$, $partial u_1 partial u_2= f_1(x_1)f_2(x_2)$
So the Riemann approximation should be,
$sum_{u=1}^{n}c(u_i,v_2). f_1(F^{-1}(u_i)).f_2(F^{-1}(v_2)) Delta u_i$
probability
asked Dec 3 '18 at 19:54
HammerPowerHammerPower
62
$endgroup$
migrated from quant.stackexchange.com Dec 4 '18 at 7:24
This question came from our site for finance professionals and academics.
$begingroup$
There are tons of people here with the math knowledge to help you with this, but frankly this will probably fair better on the Mathematics or MathOverflow sites. I’m not going to issue any close / migration votes until other folks here share what they think on it, though
$endgroup$
– Theodore Weld
Dec 4 '18 at 6:30
add a comment |
$begingroup$
I am looking into a way to estimate my probability of presence of a variable using Riemann sum to estimate density, let us consider two random variables $X_1$ and $X_2$ the probability is given by,
$P(X_1 leq x_1 , X_2 leq x_2) = int_{-infty}^{x_1} int_{-infty}^{x_1}f_{X_1 , X_2} (x , y) dx dy \$
$=P(U_1 leq u_1, U_2 leq u_2)$
$=C(u_1,u_2)$
$=int_{-infty}^{u_1} int_{-infty}^{u_1}c_{u_1 , u_2} (u , v) du dv$
meaning that ,
$P(x_1^D leq X_1 leq x_1^U, X_2 = x_2) = int_{u_1^D}^{u_1^U} c_{u_1 , u_2} (u , u_2) du$
That can be approximated by,
$simeq sum_{u=1}^{n}c(u_i,v_2) Delta u_i$
Is the latest equality true or do I miss something?
Plus is it extendable to the multivariate case?
$sum_{u=1}^{n}c(u_i,u_2,...,u_n) Delta u_i$
Is there a better way to estimate multivariate probability?
Many thanks in advance
EDIT: found my mistake, indeed the equivalence between the copula density and the multivariate density function is not that straightforward, indeed,
$c(u_1,u_2)=frac{partial ^d}{partial u_1 partial u_2} C(u_1,u_2)$
Where $partial ^d C(u_1,u_2)$ will obviously be equal to $f(x_1,x_2)$ but as $u_1$ is equivalent to $F_{X_1}(x_1)$, $partial u_1 partial u_2= f_1(x_1)f_2(x_2)$
So the Riemann approximation should be,
$sum_{u=1}^{n}c(u_i,v_2). f_1(F^{-1}(u_i)).f_2(F^{-1}(v_2)) Delta u_i$
probability
asked Dec 3 '18 at 19:54
HammerPowerHammerPower
62
$endgroup$
I am looking into a way to estimate my probability of presence of a variable using Riemann sum to estimate density, let us consider two random variables $X_1$ and $X_2$ the probability is given by,
$P(X_1 leq x_1 , X_2 leq x_2) = int_{-infty}^{x_1} int_{-infty}^{x_1}f_{X_1 , X_2} (x , y) dx dy \$
$=P(U_1 leq u_1, U_2 leq u_2)$
$=C(u_1,u_2)$
$=int_{-infty}^{u_1} int_{-infty}^{u_1}c_{u_1 , u_2} (u , v) du dv$
meaning that ,
$P(x_1^D leq X_1 leq x_1^U, X_2 = x_2) = int_{u_1^D}^{u_1^U} c_{u_1 , u_2} (u , u_2) du$
That can be approximated by,
$simeq sum_{u=1}^{n}c(u_i,v_2) Delta u_i$
Is the latest equality true or do I miss something?
Plus is it extendable to the multivariate case?
$sum_{u=1}^{n}c(u_i,u_2,...,u_n) Delta u_i$
Is there a better way to estimate multivariate probability?
Many thanks in advance
EDIT: found my mistake, indeed the equivalence between the copula density and the multivariate density function is not that straightforward, indeed,
$c(u_1,u_2)=frac{partial ^d}{partial u_1 partial u_2} C(u_1,u_2)$
Where $partial ^d C(u_1,u_2)$ will obviously be equal to $f(x_1,x_2)$ but as $u_1$ is equivalent to $F_{X_1}(x_1)$, $partial u_1 partial u_2= f_1(x_1)f_2(x_2)$
So the Riemann approximation should be,
$sum_{u=1}^{n}c(u_i,v_2). f_1(F^{-1}(u_i)).f_2(F^{-1}(v_2)) Delta u_i$
probability
probability
asked Dec 3 '18 at 19:54
HammerPowerHammerPower
62
asked Dec 3 '18 at 19:54
HammerPowerHammerPower
62
edited Dec 10 '18 at 21:03
HammerPower
asked Dec 3 '18 at 19:54
HammerPowerHammerPower
62
asked Dec 3 '18 at 19:54
HammerPowerHammerPower
62
asked Dec 3 '18 at 19:54
HammerPowerHammerPower
62
62
migrated from quant.stackexchange.com Dec 4 '18 at 7:24
This question came from our site for finance professionals and academics.
migrated from quant.stackexchange.com Dec 4 '18 at 7:24
This question came from our site for finance professionals and academics.
$begingroup$
There are tons of people here with the math knowledge to help you with this, but frankly this will probably fair better on the Mathematics or MathOverflow sites. I’m not going to issue any close / migration votes until other folks here share what they think on it, though
$endgroup$
– Theodore Weld
Dec 4 '18 at 6:30
add a comment |
$begingroup$
There are tons of people here with the math knowledge to help you with this, but frankly this will probably fair better on the Mathematics or MathOverflow sites. I’m not going to issue any close / migration votes until other folks here share what they think on it, though
$endgroup$
– Theodore Weld
Dec 4 '18 at 6:30
$begingroup$
There are tons of people here with the math knowledge to help you with this, but frankly this will probably fair better on the Mathematics or MathOverflow sites. I’m not going to issue any close / migration votes until other folks here share what they think on it, though
$endgroup$
– Theodore Weld
Dec 4 '18 at 6:30
$begingroup$
There are tons of people here with the math knowledge to help you with this, but frankly this will probably fair better on the Mathematics or MathOverflow sites. I’m not going to issue any close / migration votes until other folks here share what they think on it, though
$endgroup$
– Theodore Weld
Dec 4 '18 at 6:30
add a comment |
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$begingroup$
There are tons of people here with the math knowledge to help you with this, but frankly this will probably fair better on the Mathematics or MathOverflow sites. I’m not going to issue any close / migration votes until other folks here share what they think on it, though
$endgroup$
– Theodore Weld
Dec 4 '18 at 6:30