I stumbled upon this term in McFadden - Analysis of qualitative choice behavior (page 111).

It is said that

"A random Variable $X$ is translation complete if for a function h of bounded absolute variation with $h(\pm\infty)=0$, the condition $E(h(X+a))=0$ for all real $a$ implies $h\equiv 0$ (except possibly on a set of measure zero)."

Further it is said that most common distributions have this porperty; in particular, the Gumble distribution is translation complete. Another qoute says that

"A distribution whose characteristic function is nonzero for real arguments is translation complete [apply Feller, 1966, An Introduction to Probability Theory and Its Applications, Vol. 2, Wiley, New York, p. 479]..."

if I check there I cannot find any reference.

My question are:

If I google translation complete in connection with random variable I get no feasible results. Is translation completeness connected to sufficient statistics: What are complete sufficient statistics?

Though I understand the mathematical meaning of translation completeness I dont get the intention of this definition.

Why does translation completeness imply $G(X - logK) ) = G(X)^K$ if $G$ is a cdf and $X$ a random variable?


I came across the same statement in the same paper. This is the only reference to translation completeness I could find as well...

To partially answer your last question, it appears that translation completeness implies that the bracketed statement in the following integral must be zero, given that the whole integral equals 0.

To restate the entire result.

For a (translation complete) iid r.v. $X$ with CDF $G$, where there are $K$ realizations of $X$.

Let $P_1=\frac{e^a}{e^a + K e^b} = \int^{+ \infty}_{- \infty} G(X + a-b)^K dG(X)$.

Define $c=b+\log K$, and let

$P_2=\frac{e^a}{e^a + e^c}= \int^{+ \infty}_{- \infty} G(X + a-c) dG(X)$.

Th construction of $c$ implies $P_1=P_2$, implying

$\int^{+ \infty}_{- \infty} [G(X + a -b -\log K) - G(X + a-b)^K] dG(X)=0$,

McFadden's statement is:

This can be true for all values of $a \in (-\infty,+\infty)$ only if the term in brackets is zero, since $G$ is translation complete, implying

$G(a - \log K )=G(a)^K$.

  • 1
    $\begingroup$ Exactly what is the "bracketed statement in the integral"? Clearly this result is not general and must follow from strong assumptions about $G$ and $K$ that have not been stated in the question or here. Because that makes this entire thread incomprehensible to people without access to the reference, we would appreciate if you (or the OP) could fill in these gaps. $\endgroup$ – whuber Oct 6 '14 at 19:35
  • 1
    $\begingroup$ @whuber Added the integral. I came across this because i was interested in what conditions are placed on G as well. $\endgroup$ – Ryan Webb Oct 6 '14 at 20:36
  • 1
    $\begingroup$ Also, K is a positive constant. $\endgroup$ – Ryan Webb Oct 6 '14 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.