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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.

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The CDF of a discrete random variable $X$ is continuous everywhere except at those discrete points $x_i$ for which $P\{X=x_i\} > 0$. At these points of discontinuity, the limit from the right exists as does the limit from the left, but these two limits have different values.

Let $F_X(a^{-}) = \lim_{x\uparrow a} F_X(x)$ denote the limiting value of $F_X(x)$ as $x$ approaches $a$ from the left (or from below) and let $F_X(a^+)= \lim_{x\uparrow a} F_X(x)$ denote the limit of $F_X(x)$ as $x$ approaches $a$ from the right (or from above).

  • If $F_X(x)$ is continuous at $x=a$, then $F_X(a^-)=F_X(a^+)$ and the value of $F_X(a)$ is this limit. It does not matter whether we call it the limit from the right or the limit from the left, but for consistency with other cases, it is convenient to insist that $F_X(a)$ is the limiting value of $F_X(x)$ as $x$ approaches $a$ from the right: $F_X(a) = F_X(a^+)$

  • As you correctly state, the value of $F_X(x)$ at $x=a$ is defined to be $F_X(a^+)$. If $F_X(x)$ is discontinuous at $x=a$, then since $F_X(x)$ is a nondecreasing function, we have that $F_X(a^+) > F_X(a^-)$ and the difference $F_X(a^+) - F_X(a^-)$ is the value of $P\{X=a\}$. That is, $$\text{For each real number}~ a, P\{X=a\} = F_X(a^+) - F_X(a^-). \tag{1}$$ Note that this result is applicable to continuous random variables as well and leads to the conclusion that $P\{X=a\} = 0$ for every real number $a$: a statement that confuses many beginners in probability theory.

If you like this notation and decide that you will use it, then note that your four probabilities differ only in whether $P\{X=a\}$ and $P\{X=b\}$ are included in the probability of interest. For example, your

(i) $P\{a \leq X \leq b\} = P\{a < X \leq b\} + P\{X = a\} = [F_X(b)-F_X(a)] + P\{X=a\}$ can be expressed via $(1)$ as $$P\{a \leq X \leq b\} = F_X(b) - F_X(a^-) = F_X(b^+)-F_X(a^-)$$ and similarly for the other cases.

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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 :)

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