How can I solve this integral equation with the inverse Laplace Transform?
This question is related to Solving an integral equation with inverse Laplace transform.
Let $alpha,beta,mu>0$ with $alpha/beta>mu$ and $Xsimoperatorname{Gamma}(alpha,beta)$. I am seeking an unbiased estimator $g(X)$ such that
$$
operatorname Eg(X)=left(frac{alpha}{beta}-muright)^{-1}.
$$
This problem can be setup as the following integral equation.
$$
frac{beta^{alpha}}{Gamma(alpha)}int_0^infty g(x)x^{alpha-1}e^{-beta x},mathrm dx=left(frac{alpha}{beta}-muright)^{-1}
$$
Notice that this can be represented by either the Mellin or Laplace transforms as
$$
mathcal M{e^{-beta x}g(x)}(alpha)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}
$$
or
$$
mathcal L{x^{alpha-1}g(x)}(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}.
$$
Here is what I have found:
The inverse Mellin transform of the r.h.s does yield an unbiased estimator for $left(alpha/beta-muright)^{-1}$ when $alpha/beta>mu$ while the inverse Laplace transform also yields and unbiased estimator but only when $alpha/beta<mu$.
My hunch as to what is causing this:
Both inverse transforms are defined as a complex integral where the path of integration is to the right of all poles in the frequency domain function. In the case of the Mellin transform version of the problem we see that $F(alpha)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$ has poles at $alpha=0,-1,-2,dots$ and $alpha=betamu$. The path of integration for the inverse transform is taken along the vertical line running from $alpha-mathrm iinfty$ to $alpha+mathrm iinfty$ where $alpha>betamuimplies alpha/beta>mu$. On the other hand, consider the case for the Laplace transform version of the problem where $F(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$. The function has poles at $beta=0$ and $beta=alpha/mu$ so the path of integration for the inverse transform is taken to be the line running from $beta-mathrm iinfty$ to $beta+mathrm iinfty$ where $beta>alpha/muimplies alpha/beta<mu$.
My question:
I need my unbiased estimator to be independent of $beta$ which requires me to use the Laplace version of the problem, that is, solve $mathcal L{x^{alpha-1}g(x)}(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$ for $g(x)$ where $alpha/beta>mu$. Is there a way to do this using the inverse Laplace transform? Is it possible to use a path of integration for the inverse transform that resides in the domain where $alpha/beta>mu$?
laplace-transform contour-integration integral-transforms estimation-theory gamma-distribution
asked Dec 27 '18 at 18:21
Aaron Hendrickson
579411
add a comment |
This question is related to Solving an integral equation with inverse Laplace transform.
Let $alpha,beta,mu>0$ with $alpha/beta>mu$ and $Xsimoperatorname{Gamma}(alpha,beta)$. I am seeking an unbiased estimator $g(X)$ such that
$$
operatorname Eg(X)=left(frac{alpha}{beta}-muright)^{-1}.
$$
This problem can be setup as the following integral equation.
$$
frac{beta^{alpha}}{Gamma(alpha)}int_0^infty g(x)x^{alpha-1}e^{-beta x},mathrm dx=left(frac{alpha}{beta}-muright)^{-1}
$$
Notice that this can be represented by either the Mellin or Laplace transforms as
$$
mathcal M{e^{-beta x}g(x)}(alpha)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}
$$
or
$$
mathcal L{x^{alpha-1}g(x)}(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}.
$$
Here is what I have found:
The inverse Mellin transform of the r.h.s does yield an unbiased estimator for $left(alpha/beta-muright)^{-1}$ when $alpha/beta>mu$ while the inverse Laplace transform also yields and unbiased estimator but only when $alpha/beta<mu$.
My hunch as to what is causing this:
Both inverse transforms are defined as a complex integral where the path of integration is to the right of all poles in the frequency domain function. In the case of the Mellin transform version of the problem we see that $F(alpha)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$ has poles at $alpha=0,-1,-2,dots$ and $alpha=betamu$. The path of integration for the inverse transform is taken along the vertical line running from $alpha-mathrm iinfty$ to $alpha+mathrm iinfty$ where $alpha>betamuimplies alpha/beta>mu$. On the other hand, consider the case for the Laplace transform version of the problem where $F(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$. The function has poles at $beta=0$ and $beta=alpha/mu$ so the path of integration for the inverse transform is taken to be the line running from $beta-mathrm iinfty$ to $beta+mathrm iinfty$ where $beta>alpha/muimplies alpha/beta<mu$.
My question:
I need my unbiased estimator to be independent of $beta$ which requires me to use the Laplace version of the problem, that is, solve $mathcal L{x^{alpha-1}g(x)}(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$ for $g(x)$ where $alpha/beta>mu$. Is there a way to do this using the inverse Laplace transform? Is it possible to use a path of integration for the inverse transform that resides in the domain where $alpha/beta>mu$?
laplace-transform contour-integration integral-transforms estimation-theory gamma-distribution
asked Dec 27 '18 at 18:21
Aaron Hendrickson
579411
add a comment |
This question is related to Solving an integral equation with inverse Laplace transform.
Let $alpha,beta,mu>0$ with $alpha/beta>mu$ and $Xsimoperatorname{Gamma}(alpha,beta)$. I am seeking an unbiased estimator $g(X)$ such that
$$
operatorname Eg(X)=left(frac{alpha}{beta}-muright)^{-1}.
$$
This problem can be setup as the following integral equation.
$$
frac{beta^{alpha}}{Gamma(alpha)}int_0^infty g(x)x^{alpha-1}e^{-beta x},mathrm dx=left(frac{alpha}{beta}-muright)^{-1}
$$
Notice that this can be represented by either the Mellin or Laplace transforms as
$$
mathcal M{e^{-beta x}g(x)}(alpha)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}
$$
or
$$
mathcal L{x^{alpha-1}g(x)}(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}.
$$
Here is what I have found:
The inverse Mellin transform of the r.h.s does yield an unbiased estimator for $left(alpha/beta-muright)^{-1}$ when $alpha/beta>mu$ while the inverse Laplace transform also yields and unbiased estimator but only when $alpha/beta<mu$.
My hunch as to what is causing this:
Both inverse transforms are defined as a complex integral where the path of integration is to the right of all poles in the frequency domain function. In the case of the Mellin transform version of the problem we see that $F(alpha)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$ has poles at $alpha=0,-1,-2,dots$ and $alpha=betamu$. The path of integration for the inverse transform is taken along the vertical line running from $alpha-mathrm iinfty$ to $alpha+mathrm iinfty$ where $alpha>betamuimplies alpha/beta>mu$. On the other hand, consider the case for the Laplace transform version of the problem where $F(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$. The function has poles at $beta=0$ and $beta=alpha/mu$ so the path of integration for the inverse transform is taken to be the line running from $beta-mathrm iinfty$ to $beta+mathrm iinfty$ where $beta>alpha/muimplies alpha/beta<mu$.
My question:
I need my unbiased estimator to be independent of $beta$ which requires me to use the Laplace version of the problem, that is, solve $mathcal L{x^{alpha-1}g(x)}(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$ for $g(x)$ where $alpha/beta>mu$. Is there a way to do this using the inverse Laplace transform? Is it possible to use a path of integration for the inverse transform that resides in the domain where $alpha/beta>mu$?
laplace-transform contour-integration integral-transforms estimation-theory gamma-distribution
asked Dec 27 '18 at 18:21
Aaron Hendrickson
579411
This question is related to Solving an integral equation with inverse Laplace transform.
Let $alpha,beta,mu>0$ with $alpha/beta>mu$ and $Xsimoperatorname{Gamma}(alpha,beta)$. I am seeking an unbiased estimator $g(X)$ such that
$$
operatorname Eg(X)=left(frac{alpha}{beta}-muright)^{-1}.
$$
This problem can be setup as the following integral equation.
$$
frac{beta^{alpha}}{Gamma(alpha)}int_0^infty g(x)x^{alpha-1}e^{-beta x},mathrm dx=left(frac{alpha}{beta}-muright)^{-1}
$$
Notice that this can be represented by either the Mellin or Laplace transforms as
$$
mathcal M{e^{-beta x}g(x)}(alpha)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}
$$
or
$$
mathcal L{x^{alpha-1}g(x)}(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}.
$$
Here is what I have found:
The inverse Mellin transform of the r.h.s does yield an unbiased estimator for $left(alpha/beta-muright)^{-1}$ when $alpha/beta>mu$ while the inverse Laplace transform also yields and unbiased estimator but only when $alpha/beta<mu$.
My hunch as to what is causing this:
Both inverse transforms are defined as a complex integral where the path of integration is to the right of all poles in the frequency domain function. In the case of the Mellin transform version of the problem we see that $F(alpha)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$ has poles at $alpha=0,-1,-2,dots$ and $alpha=betamu$. The path of integration for the inverse transform is taken along the vertical line running from $alpha-mathrm iinfty$ to $alpha+mathrm iinfty$ where $alpha>betamuimplies alpha/beta>mu$. On the other hand, consider the case for the Laplace transform version of the problem where $F(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$. The function has poles at $beta=0$ and $beta=alpha/mu$ so the path of integration for the inverse transform is taken to be the line running from $beta-mathrm iinfty$ to $beta+mathrm iinfty$ where $beta>alpha/muimplies alpha/beta<mu$.
My question:
I need my unbiased estimator to be independent of $beta$ which requires me to use the Laplace version of the problem, that is, solve $mathcal L{x^{alpha-1}g(x)}(beta)=Gamma(alpha)beta^{-alpha}(alpha/beta-mu)^{-1}$ for $g(x)$ where $alpha/beta>mu$. Is there a way to do this using the inverse Laplace transform? Is it possible to use a path of integration for the inverse transform that resides in the domain where $alpha/beta>mu$?
laplace-transform contour-integration integral-transforms estimation-theory gamma-distribution
laplace-transform contour-integration integral-transforms estimation-theory gamma-distribution
asked Dec 27 '18 at 18:21
Aaron Hendrickson
579411
asked Dec 27 '18 at 18:21
Aaron Hendrickson
579411
edited Dec 27 '18 at 18:33
asked Dec 27 '18 at 18:21
Aaron Hendrickson
579411
asked Dec 27 '18 at 18:21
Aaron Hendrickson
579411
asked Dec 27 '18 at 18:21
Aaron Hendrickson
579411
579411
add a comment |
add a comment |
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