Question

Let \(X\) be a discrete valued random variable with cumulative distribution function \(F(x)\).

Which of the following statements is/are correct?

• A. \(F(x)\) is always a positive function.
• B. \(F(x)\) is a non-decreasing function.
• C. \(F(x)\) has jump discontinuity.
• D. \(F(x)\) is a left continuous function.

Hint:

Check the CDF definition \(F(x)=P(X\leq x)\) directly against each claim: it is non-negative and non-decreasing always, and being right continuous with jumps is specific to discrete random variables.

Answer 1

The Correct Option is B, C

Step 1: Recall the definition of a CDF.
For any random variable \(X\), the cumulative distribution function is \(F(x) = P(X \leq x)\). Since \(X\) is discrete, \(F(x)\) is built by adding up the probabilities of every value \(X\) can take that is less than or equal to \(x\).

Step 2: Check option (A).
\(F(x)\) is a probability, so it always satisfies \(0 \leq F(x) \leq 1\), it can equal 0 (for any \(x\) smaller than the minimum value \(X\) can take) and it can equal 1 (for any \(x\) at or above the maximum value \(X\) can take). Since \(F(x) = 0\) is possible, \(F(x)\) is not "always positive" in the strict sense, it is only ever non-negative. Option (A) is incorrect.

Step 3: Check option (B).
For \(x_1 < x_2\), the event \(\{X \leq x_1\}\) is a subset of \(\{X \leq x_2\}\), so \(P(X \leq x_1) \leq P(X \leq x_2)\), meaning \(F(x_1) \leq F(x_2)\). This holds for every CDF, discrete or continuous, so \(F(x)\) is always non-decreasing. Option (B) is correct.

Step 4: Check option (C).
Because \(X\) is discrete, it only takes a countable set of specific values, each with positive probability. As \(x\) crosses one of these values, \(F(x)\) instantly jumps up by the probability mass at that point, rather than rising smoothly. This gives \(F(x)\) a staircase shape with jump discontinuities exactly at the values \(X\) can take. Option (C) is correct.

Step 5: Check option (D).
By convention and by construction, a CDF is right continuous: \(\lim_{h \to 0^+} F(x+h) = F(x)\), because \(P(X \leq x)\) already includes the point \(x\) itself, so approaching from the right does not add anything new. Approaching from the left, however, \(\lim_{h \to 0^+} F(x-h)\) does not equal \(F(x)\) at a jump point, it equals \(F(x)\) minus the probability mass at \(x\). So \(F(x)\) is right continuous, not left continuous. Option (D) is incorrect.

Final Answer:
Options (B) and (C) are correct. \[ \boxed{\text{(B), (C)}} \]