# A random variable with a step function CDF is discrete?

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 $$\{\lambda_1,...,\lambda_J\}$$?

## 1 Answer

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 $$\lambda$$.