3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 $\lambda$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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