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?
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 $\Phi$