On Monday 30 January 2006 14:24, Franzese, Robert wrote:
> This struck us as problematic; given Jensen's inequality, it would seem,
> if we are correct, that this would be accurate,

Dear all,
educate me if I am wrong - why is Jensen's inequality relevant here?
Inlow's suggestion does not involve the function of an integral.
If anything gets transformed (a function is applied to it), it is the solution
to an equation (which involves the unknown upper limit of an integral - see
the TeX stuff below).
But who would do a logit or probit transform to tail probabilities anyway?

Best,

Martin


TeX starts here:
--------------------------------------------------
\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator{\diag}{diag}
\DeclareMathOperator{\dif}{d}
\DeclareMathOperator{\E}{\textrm{E}\,}
\DeclareMathOperator{\VAR}{\textrm{VAR}}
\DeclareMathOperator{\AVAR}{\textrm{AVAR}}
\DeclareMathOperator{\COV}{\textrm{COV}}
\DeclareMathOperator{\Dt}{\textrm{Dt}}
\DeclareMathOperator{\Mt}{\textrm{Mt}}
\DeclareMathOperator{\dt}{\textrm{dt}}
% \DeclareMathOperator{\Pr}{\textrm{Pr}}
\usepackage{algorithmic}

\usepackage{graphicx}
\DeclareMathOperator{\T}{\textsf{T}}
\begin{document}
\title{A quick guess regarding confidence limits
for predictions in logistic regression}
\author{Martin Elff, University of Mannheim}
\maketitle
Finding confidence limits for confidence level $\alpha$
along the lines of Inlow's proposal
means solving first the equations
\[
\int_{-\infty}^{q^*} f_N(t;x^{\T}b,\AVAR(x^{\T}b)) \dif t =
\frac12\alpha\textrm{ and }
\int_{-\infty}^{q^*} f_N(t;x^{\T}b,\AVAR(x^{\T}b)) \dif t = 1-\frac12\alpha
\]
for $q^*$
(where $f_N(x,\mu,\sigma^2)$ denotes the normal density function)
and then transform the solutions
$q^*_{\frac12\alpha}$ and $q^*_{1-\frac12\alpha}$
to
\[
p\left(q^*_{\frac12\alpha}\right)=
\frac{\exp\left(q^*_{\frac12\alpha}\right)}
{1+\exp\left(q^*_{\frac12\alpha}\right)}\textrm{ and }
p\left(q^*_{1-\frac12\alpha}\right)=
\frac{\exp\left(q^*_{1-\frac12\alpha}\right)}
{1+\exp\left(q^*_{1-\frac12\alpha}\right)}
\]
and use these $p\left(q^*_{\frac12\alpha}\right)$ and
$p\left(q^*_{1-\frac12\alpha}\right)$
as confidence limits.

The ``only really correct way'' would instead mean solving
\[
\int_0^{q}
\frac{\exp\left(t\right)}{1+\exp\left(t\right)}
f_N(t;x^{\T}b,\AVAR(x^{\T}b)) \dif t = \frac12\alpha\textrm{ and }
\int_0^{q}
\frac{\exp\left(t\right)}{1+\exp\left(t\right)}
f_N(t;x^{\T}b,\AVAR(x^{\T}b)) \dif t = 1-\frac12\alpha
\]
for $q$
and using the solutions
$q_{\frac12\alpha}$ and $q_{1-\frac12\alpha}$ as confidence limits.


Inlow rejects (I guess, rightly so) the following approach:
Solve the equations
\[
\int_{-\infty}^{q^{**}}
f_N\left(t;
\frac{\exp\left(x^{\T}b\right)}{1+\exp\left(x^{\T}b\right)},
\left(\frac{\exp\left(x^{\T}b\right)}{1+\exp\left(x^{\T}b\right)}\right)^2
\AVAR(x^{\T}b)
\right) \dif t = \frac12\alpha
\]
and
\[
\int_{-\infty}^{q^{**}}
f_N\left(t;
\frac{\exp\left(x^{\T}b\right)}{1+\exp\left(x^{\T}b\right)},
\left(\frac{\exp\left(x^{\T}b\right)}{1+\exp\left(x^{\T}b\right)}\right)^2
\AVAR(x^{\T}b)
\right) \dif t = 1-\frac12\alpha
\]
for $q^{**}$ and use the solutions
$q^{**}_{\frac12\alpha}$ and $q^{**}_{1-\frac12\alpha}$

In neither case is the function of an integral involved.
So I would suppose that it is not Jensen's inequality that is relevant
here but the rules of integration by parts.

Of course, I might completely be mistaken. Please enlighten me if this is
so ...
\end{document}



--
-------------------------------------------------
Dr. Martin Elff
Department of Social Sciences
University of Mannheim
D7, 27, Room 407
68131 Mannheim
Germany

Phone: +49-621-181-2093
Fax: +49-621-181-2099
E-Mail: [log in to unmask]
Homepage:
http://webrum.uni-mannheim.de/sowi/elff
http://www.sowi.uni-mannheim.de/lspwivs/
-------------------------------------------------


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