-2
$\begingroup$

Consider $$ \Phi(\frac{\beta_1+\beta_2X_2+\beta_3X_3+...+\beta_KX_k}{\sigma}) $$ where

  • $\Phi: \mathbb{R}\rightarrow [0,1]$ is a known function
  • $\beta_1, ..., \beta_K, \sigma$ are unknown real-valued parameters with $\sigma>0$
  • $X_2, ..., X_K$ are known real numbers

It can be seen that $$ (\star) \hspace{1cm} \Phi(\frac{\beta_1+\beta_2X_2+\beta_3X_3+...+\beta_KX_k}{\sigma})= \Phi(\frac{\tilde{\beta}_1+\tilde{\beta}_2X_2+\tilde{\beta}_3X_3+...+\tilde{\beta}_KX_k}{\tilde{ \sigma}}) $$ where $\tilde{\beta}_k\equiv\alpha \beta_k$ $\forall k=1,..., K$ and $\tilde{\sigma}\equiv \alpha \sigma$ for any $\alpha>0$

Suppose that now I impose the constraint $\beta_1^2+ \beta_2^2+...+\beta_K^2=1$

Question: can we still find an $\alpha>0$ such that $(\star)$ holds together with $\tilde{\beta}_1^2+ \tilde{\beta}_2^2+...+\tilde{\beta}_K^2=1$?

My intuition is that we can't. Can you help me to formally show this? Do we need to use some conditions on $\Phi$, such that continuity or strict monotonicity?

$\endgroup$
6
  • $\begingroup$ I don’t understand your problems, because equality $(\star)$ holds for any choice of $\alpha$, $\sigma$, $\beta$’s and $X$’s, because the argument of the function from the left hand side equals the argument of the function from the left hand side, right? $\endgroup$ Sep 8, 2017 at 20:22
  • $\begingroup$ Equality $(\star)$ holds for any choice of $\alpha, \sigma,\beta$. $\endgroup$
    – TEX
    Sep 9, 2017 at 8:09
  • $\begingroup$ Suppose now I instead suppose the following: $$ \Phi(\frac{\beta_1+\beta_2X_2+\beta_3X_3+...+\beta_KX_k}{\sigma}) $$ where $\Phi: \mathbb{R}\rightarrow [0,1]$ is a known function; $\beta_1, ..., \beta_K, \sigma$ are unknown real-valued parameters with $\sigma>0$ AND $\beta_1^2+...+\beta^2_K=1$; $X_2, ..., X_K$ are known real numbers $\endgroup$
    – TEX
    Sep 9, 2017 at 8:11
  • $\begingroup$ My question is: does $(\star)$ still hold? $\endgroup$
    – TEX
    Sep 9, 2017 at 8:11
  • $\begingroup$ @STF Yes, it still holds. I might still be misunderstanding but those expression are equal for all reals, given your definitions of $\alpha$ and $ \sigma$ (given the obvious $\sigma \neq 0$) really there is no constraint on the $\beta$'s where it would not hold. There are also no values for $\alpha$ where it would not hold. This leads me to believe I still misunderstand your question. What does the function have to do with it? The expressions where the function is evaluated are per definition equal so the function really doesn't matter. $\endgroup$
    – zen
    Sep 11, 2017 at 9:19

1 Answer 1

0
+50
$\begingroup$

The answer to the updated question is that yes such an $\alpha$ exists and is equal to one. This is trivial to see since then $\tilde{\beta}_K=\beta_{K} \forall K$. so if one restriction holds the other holds as well and it is trivial that * holds. There is no ther solution since if $\beta_1^2+ \beta_2^2+...+\beta_K^2=1$ then $\tilde{\beta}_1^2+ \tilde{\beta}_2^2+...+\tilde{\beta}_K^2=(\alpha\beta_1)^2+ (\alpha\beta_2)^2+...+(\alpha\beta_K)^2=\alpha^2\beta_1^2+ \alpha^2\beta_2^2+...+\alpha^2\beta_K^2=\alpha^2(\beta_1^2+ \beta_2^2+...+\beta_K^2)=\alpha^2(1)=\alpha^2$

so if both constraints are true and $\alpha>0$ then this implies $\alpha=1$ This is indepedent of the function $\Phi$

$\endgroup$
5
  • $\begingroup$ This answer is plain wrong. As all "tilted" variables have the factor $\alpha$, it cancels out in the fraction and the property is valid for any $\alpha$, regardless any constraint on the $\beta$. $\endgroup$
    – user65203
    Sep 11, 2017 at 13:24
  • $\begingroup$ @YvesDaoust you are ofcourse right witch I also mentioned in a comment. However OP has edited the question so that there are two constraints. These two constraints imply that $\alpha=1$ is the only answer. You are right that (*) does not imply anything about $\alpha$. $\endgroup$
    – zen
    Sep 11, 2017 at 13:44
  • $\begingroup$ You are right, I get it now. (Though there is a single constraint, not two.) $\endgroup$
    – user65203
    Sep 11, 2017 at 13:47
  • $\begingroup$ @YvesDaoust $ \beta_1^2+ \beta_2^2+...+\beta_K^2=1$ is one and the other is $\tilde{\beta}_1^2+ \tilde{\beta}_2^2+...+\tilde{\beta}_K^2=1$ $\endgroup$
    – zen
    Sep 11, 2017 at 13:48
  • $\begingroup$ Ok, sorry. So formulated, the question is so unexpected. (Obviously $\beta=\tilde\beta$). $\endgroup$
    – user65203
    Sep 11, 2017 at 13:50

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