[Math] Continuous and Discrete random variable distribution function

probabilityprobability distributionsprobability theoryrandom variables

I have a very basic question in probability. It pertains to the difference between a continuous random variable distribution function and a discrete one. This question has confused me many times.

Suppose the continuous r.v. distribution is given by $F_X(x)$.
Then as $F_X$ is having both right and left continuity, I can say that

(i) $P(a \leq X \leq b)=F_X(b)-F_X(a)$

(ii) $P(a < X \leq b)=F_X(b)-F_X(a)$

(iii) $P(a \leq X < b)=F_X(b)-F_X(a)$

(iv) $P(a < X < b)=F_X(b)-F_X(a)$

Eqn. (i) to (iv) holds simply because $F_X$ is continuous. However how to compute these probabilities in the discrete case has always confused me.

Suppose the discrete r.v. distribution is given by $F_X(x)$.
Then as $F_X$ is ONLY having right continuity, I can say that

(i) $P(a \leq X \leq b)=F_X(b)-F_X(a) + P(X==a)$

(ii) $P(a < X \leq b)=F_X(b)-F_X(a)$

(iii) $P(a \leq X < b)=???$

(iv) $P(a < X < b)=???$

Please don't down vote just because it's easy. If an answer exists do point it out to me.

Thanks in advance.

Best Answer

Since you're working with a discrete random variable, the probability of X being lower than b equals the probability of X being equal to or lower than the number before b.

Imagine we're working with a fair dice:

$P(2\leq X < 5) = P(2\leq X \leq 4)$. Sounds quite fair, doesn't it?

I think you can answer your question yourself now :)