Properties of a step function

Consider the step function $$\Delta: \mathbb{R}\rightarrow [0,1]$$ $$\Delta(x;\lambda,\mu)\equiv \sum_{j=1}^J \lambda_j\times 1\{\mu_j\leq x\}$$ where

• $$\lambda\equiv (\lambda_1,...,\lambda_J)$$ is a parameter

• $$\mu\equiv (\mu_1,...,\mu_J)$$ is a parameter

• $$J=3$$

• $$\lambda_j\geq 0$$ $$\forall j$$; $$\sum_{j=1}^J \lambda_j=1$$

• $$\mu_j\in \mathbb{R}$$ $$\forall j$$; $$\mu_1<...<\mu_J$$

• $$1\{\cdot\}$$ is an indicator function taking value $$1$$ if the condition inside is satisfied and zero otherwise

Claim: If $$\lambda_1\times \lambda_2\times \lambda_3\neq 0$$ and $$\frac{\mu_2-\mu_1}{\mu_3-\mu_2}\neq 1$$, then $$\Delta(\cdot; \lambda,\mu)$$ cannot be such that $$\Delta(x;\lambda,\mu)=1-\Delta(-x;\lambda,\mu)$$.

Question: Could you help me to show this claim? Even just the intuition would be OK.

Also, is there a way to generalise this claim to any $$J$$?

Some thoughts: one way to show the claim is to prove that

"If $$\Delta(x;\lambda,\mu)=1-\Delta(-x;\lambda,\mu)$$ $$\forall x$$, then $$\lambda_1\times \lambda_2\times \lambda_3= 0$$ or $$\frac{\mu_2-\mu_1}{\mu_3-\mu_2}= 1$$."

I have tried to picture in my mind the cases in which we can have $$\Delta(x;\lambda,\mu)=1-\Delta(-x;\lambda,\mu)$$ $$\forall x$$ when $$J=3$$. Given that when $$J=3$$ we can have 4 steps at most, I ended up with 2 cases only:

1) $$\lambda_j=\lambda_k=0$$, $$\lambda_h=1$$, $$\mu_h=0$$

2) $$\lambda_j=0$$, $$\lambda_k=\lambda_h=\frac{1}{2}$$, $$\mu_k=-\mu_h$$

These two cases definitively satisfy $$\lambda_1\times \lambda_2\times \lambda_3=0$$. I can't picture a case implying $$\frac{\mu_2-\mu_1}{\mu_3-\mu_2}= 1$$.

Hint: Since all $$\lambda_j$$ are nonzero and $$\mu_3-\mu_2\neq \mu_2-\mu_1$$ it should be easy to see that the function $$\Delta(x,\lambda,\mu)=\begin{cases}0,x<\mu_1\\\lambda_1,\mu_1\leq x<\mu_2\\ \lambda_1+\lambda_2,\mu_2\leq x<\mu_3\\ 1=\lambda_1+\lambda_2+\lambda_3,\mu_3\leq x \end{cases}$$ doesn´t fulfill $$\Delta(\delta,\lambda,\mu)=1-\Delta(-\delta,\lambda,\mu)$$ for all $$\delta>0$$.
• Thanks, but I can't go too far with that hint. I have added some thoughts to the question which I hope clarify where I got stuck. Also, I would like a generalisation of the statement for any $J$. – STF Feb 17 at 13:26