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Asymptotic behavior of this estimator?
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Let $X$,$Y$ two real valued random variables with distributions absolutely continuous with regard to the Lebesgue measure. Denote their distribution functions $F$ and $G$, and $g=G'$ the probability density function of $Y$.
Using the inverse transform theorem, we have:
beginalign*
Y &= G^-1circ F (X)\
& =: f(X).
endalign*
in distribution. Now, one observes $(X_n)_ngeq1$ i.i.d. $X$ and $(Y_n)_ngeq1$ i.i.d. Y. Suppose moreover that $(X_n)_ngeq1$ is independant of $(Y_n)_ngeq1$. Define $hatf_n(x)$ an estimator of $f(x) = G^-1circ F(x)$ for some $xin mathbbR$ as:
beginalign*
hatf_n(x) = G_n^-1circ F_n(x)
endalign*
where $F_n(x)=n^-1sum_i=1^n boldsymbol1_X_ileq x$ is the empirical c.d.f. of the $n$-sample $(X_1, dots, X_n)$ and $G_n^-1$ is the generalized inverse of the c.d.f. of the $n$-sample $(Y_1,dots,Y_n)$, i.e. $G_n^-1(alpha) = infx in mathbbR, G_n(x) geq alpha $ where $G_n(x)=n^-1sum_i=1^n boldsymbol1_Y_ileq x$.
I am interested in the asymptotic behavior of $hatf_n(x)$. One can write:
beginalign*
sqrtn(hatf_n(x) - f(x)) &= sqrtn(G_n^-1circ F_n(x) - G^-1circ F(x))\
&= sqrtn(G_n^-1circ F_n(x) - G^-1circ F_n(x))+sqrtn(G^-1circ F_n(x) - G^-1circ F(x))\
&= A_n + B_n quad textsay.
endalign*
Proving stochastic equicontinuity of the process $Z_n(p), p in [0,1]_n geq 1$ where $Z_n(p) = sqrtn(G_n^-1(p) - G^-1(p))$ allows one to prove that $A_n$ converges in distribution to $mathcalN(0,sigma^2)$ where $sigma^2 = fracF(x)(1-F(x))gcirc f(x)^2$. Moreover the Delta method can be used to prove that $B_n$ converges in distribution also to $mathcalN(0,sigma^2)$.
Now, note that that $A_n$ and $B_n$ are not independent so we can not conclude that $A_n + B_n$ converges to $mathcalN(0,2sigma^2)$. Nevertheless, numerical simulations are consistent with this.
Any idea on how to prove it would be greatly appreciated! Thanks!
probability statistics estimation
asked Aug 3 at 13:45
Alfred F.
385
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up vote
1
down vote
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Let $X$,$Y$ two real valued random variables with distributions absolutely continuous with regard to the Lebesgue measure. Denote their distribution functions $F$ and $G$, and $g=G'$ the probability density function of $Y$.
Using the inverse transform theorem, we have:
beginalign*
Y &= G^-1circ F (X)\
& =: f(X).
endalign*
in distribution. Now, one observes $(X_n)_ngeq1$ i.i.d. $X$ and $(Y_n)_ngeq1$ i.i.d. Y. Suppose moreover that $(X_n)_ngeq1$ is independant of $(Y_n)_ngeq1$. Define $hatf_n(x)$ an estimator of $f(x) = G^-1circ F(x)$ for some $xin mathbbR$ as:
beginalign*
hatf_n(x) = G_n^-1circ F_n(x)
endalign*
where $F_n(x)=n^-1sum_i=1^n boldsymbol1_X_ileq x$ is the empirical c.d.f. of the $n$-sample $(X_1, dots, X_n)$ and $G_n^-1$ is the generalized inverse of the c.d.f. of the $n$-sample $(Y_1,dots,Y_n)$, i.e. $G_n^-1(alpha) = infx in mathbbR, G_n(x) geq alpha $ where $G_n(x)=n^-1sum_i=1^n boldsymbol1_Y_ileq x$.
I am interested in the asymptotic behavior of $hatf_n(x)$. One can write:
beginalign*
sqrtn(hatf_n(x) - f(x)) &= sqrtn(G_n^-1circ F_n(x) - G^-1circ F(x))\
&= sqrtn(G_n^-1circ F_n(x) - G^-1circ F_n(x))+sqrtn(G^-1circ F_n(x) - G^-1circ F(x))\
&= A_n + B_n quad textsay.
endalign*
Proving stochastic equicontinuity of the process $Z_n(p), p in [0,1]_n geq 1$ where $Z_n(p) = sqrtn(G_n^-1(p) - G^-1(p))$ allows one to prove that $A_n$ converges in distribution to $mathcalN(0,sigma^2)$ where $sigma^2 = fracF(x)(1-F(x))gcirc f(x)^2$. Moreover the Delta method can be used to prove that $B_n$ converges in distribution also to $mathcalN(0,sigma^2)$.
Now, note that that $A_n$ and $B_n$ are not independent so we can not conclude that $A_n + B_n$ converges to $mathcalN(0,2sigma^2)$. Nevertheless, numerical simulations are consistent with this.
Any idea on how to prove it would be greatly appreciated! Thanks!
probability statistics estimation
asked Aug 3 at 13:45
Alfred F.
385
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$,$Y$ two real valued random variables with distributions absolutely continuous with regard to the Lebesgue measure. Denote their distribution functions $F$ and $G$, and $g=G'$ the probability density function of $Y$.
Using the inverse transform theorem, we have:
beginalign*
Y &= G^-1circ F (X)\
& =: f(X).
endalign*
in distribution. Now, one observes $(X_n)_ngeq1$ i.i.d. $X$ and $(Y_n)_ngeq1$ i.i.d. Y. Suppose moreover that $(X_n)_ngeq1$ is independant of $(Y_n)_ngeq1$. Define $hatf_n(x)$ an estimator of $f(x) = G^-1circ F(x)$ for some $xin mathbbR$ as:
beginalign*
hatf_n(x) = G_n^-1circ F_n(x)
endalign*
where $F_n(x)=n^-1sum_i=1^n boldsymbol1_X_ileq x$ is the empirical c.d.f. of the $n$-sample $(X_1, dots, X_n)$ and $G_n^-1$ is the generalized inverse of the c.d.f. of the $n$-sample $(Y_1,dots,Y_n)$, i.e. $G_n^-1(alpha) = infx in mathbbR, G_n(x) geq alpha $ where $G_n(x)=n^-1sum_i=1^n boldsymbol1_Y_ileq x$.
I am interested in the asymptotic behavior of $hatf_n(x)$. One can write:
beginalign*
sqrtn(hatf_n(x) - f(x)) &= sqrtn(G_n^-1circ F_n(x) - G^-1circ F(x))\
&= sqrtn(G_n^-1circ F_n(x) - G^-1circ F_n(x))+sqrtn(G^-1circ F_n(x) - G^-1circ F(x))\
&= A_n + B_n quad textsay.
endalign*
Proving stochastic equicontinuity of the process $Z_n(p), p in [0,1]_n geq 1$ where $Z_n(p) = sqrtn(G_n^-1(p) - G^-1(p))$ allows one to prove that $A_n$ converges in distribution to $mathcalN(0,sigma^2)$ where $sigma^2 = fracF(x)(1-F(x))gcirc f(x)^2$. Moreover the Delta method can be used to prove that $B_n$ converges in distribution also to $mathcalN(0,sigma^2)$.
Now, note that that $A_n$ and $B_n$ are not independent so we can not conclude that $A_n + B_n$ converges to $mathcalN(0,2sigma^2)$. Nevertheless, numerical simulations are consistent with this.
Any idea on how to prove it would be greatly appreciated! Thanks!
probability statistics estimation
asked Aug 3 at 13:45
Alfred F.
385
Let $X$,$Y$ two real valued random variables with distributions absolutely continuous with regard to the Lebesgue measure. Denote their distribution functions $F$ and $G$, and $g=G'$ the probability density function of $Y$.
Using the inverse transform theorem, we have:
beginalign*
Y &= G^-1circ F (X)\
& =: f(X).
endalign*
in distribution. Now, one observes $(X_n)_ngeq1$ i.i.d. $X$ and $(Y_n)_ngeq1$ i.i.d. Y. Suppose moreover that $(X_n)_ngeq1$ is independant of $(Y_n)_ngeq1$. Define $hatf_n(x)$ an estimator of $f(x) = G^-1circ F(x)$ for some $xin mathbbR$ as:
beginalign*
hatf_n(x) = G_n^-1circ F_n(x)
endalign*
where $F_n(x)=n^-1sum_i=1^n boldsymbol1_X_ileq x$ is the empirical c.d.f. of the $n$-sample $(X_1, dots, X_n)$ and $G_n^-1$ is the generalized inverse of the c.d.f. of the $n$-sample $(Y_1,dots,Y_n)$, i.e. $G_n^-1(alpha) = infx in mathbbR, G_n(x) geq alpha $ where $G_n(x)=n^-1sum_i=1^n boldsymbol1_Y_ileq x$.
I am interested in the asymptotic behavior of $hatf_n(x)$. One can write:
beginalign*
sqrtn(hatf_n(x) - f(x)) &= sqrtn(G_n^-1circ F_n(x) - G^-1circ F(x))\
&= sqrtn(G_n^-1circ F_n(x) - G^-1circ F_n(x))+sqrtn(G^-1circ F_n(x) - G^-1circ F(x))\
&= A_n + B_n quad textsay.
endalign*
Proving stochastic equicontinuity of the process $Z_n(p), p in [0,1]_n geq 1$ where $Z_n(p) = sqrtn(G_n^-1(p) - G^-1(p))$ allows one to prove that $A_n$ converges in distribution to $mathcalN(0,sigma^2)$ where $sigma^2 = fracF(x)(1-F(x))gcirc f(x)^2$. Moreover the Delta method can be used to prove that $B_n$ converges in distribution also to $mathcalN(0,sigma^2)$.
Now, note that that $A_n$ and $B_n$ are not independent so we can not conclude that $A_n + B_n$ converges to $mathcalN(0,2sigma^2)$. Nevertheless, numerical simulations are consistent with this.
Any idea on how to prove it would be greatly appreciated! Thanks!
probability statistics estimation
asked Aug 3 at 13:45
Alfred F.
385
edited Aug 3 at 13:50
asked Aug 3 at 13:45
Alfred F.
385
asked Aug 3 at 13:45
Alfred F.
385
asked Aug 3 at 13:45
Alfred F.
385
385
add a comment |Â
add a comment |Â
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