Suppose X is a continuous random variable with probability density function that is strictly positive (i.e., 0) at every real number. Suppose
𝐹
𝑋
(
𝑥
)
is the cdf of X evaluated at real number x. Then, as x increases this cdf is

Question

Suppose X is a continuous random variable with probability density function that is strictly positive (i.e., > 0) at every real number. Suppose 
𝐹
𝑋
(
𝑥
)
 is the cdf of X evaluated at real number x. Then, as x increases this cdf is

BlackTom AI · Accepted Answer

Because f_X(x) > 0 for every real x, the cdf F_X(x) = ∫_{-∞}^x f_X(t) dt has derivative f_X(x) > 0 and therefore strictly increases with x.

1. Relevant concepts
- Probability density function (pdf) f_X(x) integrates to 1 and for a continuous variable defines the cumulative distribution function (cdf) F_X(x) = ∫_{-∞}^x f_X(t) dt.
- If f_X(x) ≥ 0 then F_X is non-decreasing; if f_X(x) > 0 everywhere then F_X has positive derivative everywhere.
- The Fundamental Theorem of Calculus: d/dx ∫_{-∞}^x f(t) dt = f(x) when f is continuous (or integrable with appropriate regularity).

2. Step-by-step reasoning
- Take any two real numbers x1 < x2.
- Compute the increment: F_X(x2) − F_X(x1) = ∫_{-∞}^{x2} f_X(t) dt − ∫_{-∞}^{x1} f_X(t) dt = ∫_{x1}^{x2} f_X(t) dt.
- Since f_X(t) > 0 for every t in (x1, x2), the integrand is strictly positive on an interval of positive length.

3. Intermediate calculation justification
- The integral of a strictly positive integrand over an interval of positive length is strictly positive, so ∫_{x1}^{x2} f_X(t) dt > 0.
- Therefore F_X(x2) − F_X(x1) > 0 whenever x2 > x1.

4. Why the correct answer is correct
- The strict positivity of f_X everywhere implies F_X is differentiable with derivative f_X(x) > 0 at every point; a function with positive derivative everywhere is strictly increasing. Equivalently, the cdf’s increments over any nonzero interval are strictly positive.

5. Why other possible choices are incorrect
- “Non-decreasing” is true but weaker: it allows flat segments (where increment = 0). Those cannot occur here because f_X > 0 forbids zero increments.
- “Constant,” “non-increasing,” or “decreasing” contradict the positive increments shown above.

6. Conclusion
- Hence the cdf F_X(x) is strictly increasing, which matches the provided answer: strictly increasing.

Suppose X is a continuous random variable with probability density function that is strictly positive (i.e., > 0) at every real number. Suppose 𝐹 𝑋 ( 𝑥 ) is the cdf of X evaluated at real number x. Then, as x increases this cdf is 单项选择题

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