A random variable with a step function CDF is discrete?

Question

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

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

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

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

• $\lambda\equiv (\lambda_1,...,\lambda_J)$

• $\mu\equiv (\mu_1,...,\mu_J)$

Take a random variable $Y$ with CDF given by $\Delta(\cdot; \lambda,\mu)$. Is it correct to say that $Y$ should be necessarily a discrete random variables with support $\{\mu_1,...,\mu_J\}$ and with probabilities masses equal to $\{\lambda_1,...,\lambda_J\}$?

Answer 1

Let's write each of the possibilities (let $a^-$ be number arbitrarily close to $a$, but smaller than $a$):

$P(Y=\mu_1)=P(Y\leq\mu_1)-P(Y<\mu_1)=\Delta(\mu_1)-\Delta(\mu_1^-)=\lambda_1$ $P(Y=\mu_2)=P(Y\leq\mu_2)-P(Y<\mu_2)=\Delta(\mu_2)-\Delta(\mu_2^-)=\lambda_1+\lambda_2-\lambda_1=\lambda_2$

continuing this way ...

$P(Y=\mu_k)=P(Y\leq\mu_J)-P(Y<\mu_J)=\Delta(\mu_J)-\Delta(\mu_J^-)=\sum_{i=1}^{K}\lambda_i-\sum_{i=1}^{K-1}\lambda_i=\lambda_K$

So, yes. You have a discrete RV with support $\mu$, and masses $\lambda$.