A random variable assigns a number to each outcome. Suppose X takes 0,2,5 with probabilities 1/5,1/2,3/10. Before plotting a histogram, ask for P(X≤x) as x moves along the number line. This cumulative probability records every mass already passed.
Predict
Predict F(2), F(2−), and the jump at 2. The event X≤2 includes the mass at 2; approaching from the left does not. That endpoint distinction is essential for atoms and disappears only when there is no mass at the endpoint.
Experiment
The CDF F(x)=P(X≤x) is zero below 0, equals 1/5 from 0 up to but not including 2, equals 7/10 from 2 up to 5, and equals one from 5 onward. Draw a step graph with closed dots at the right-continuous values.
Point
Left limit
CDF value
Jump
0
0
1/5
1/5
2
1/5
7/10
1/2
5
7/10
1
3/10
Recover every original probability from its jump. Then calculate P(0<X≤5)=F(5)−F(0)=4/5, noting which endpoint was excluded.
Understand
A CDF is nondecreasing, right continuous, and tends to zero and one at the negative and positive ends of the real line. These properties characterize distribution functions. It can have flat stretches and jumps; differentiability and strict increase are not required. Its left limit F(a−) exists because of monotonicity.
Interval formulas depend on endpoint inclusion. P(a<X≤b)=F(b)−F(a), while P(a≤X≤b)=F(b)−F(a−). The point mass at a is F(a)−F(a−). Writing these events first is safer than memorizing a subtraction without its inequalities.
For an absolutely continuous law, a density integrates to the CDF and individual points have probability zero. For a discrete law, the CDF accumulates masses. Mixed laws combine both types. A process that is exactly zero with some probability and otherwise has a continuous positive value cannot be represented fully by an ordinary density alone.
For example, let X be zero with probability 1/3 and otherwise uniform on (0,3). Then F(x)=0 below zero, F(x)=1/3+2x/9 on 0≤x<3, and F(x)=1 at or above 3. The jump is 1/3 and the continuous density is 2/9 on (0,3), whose integral supplies the remaining 2/3. Differentiating the CDF away from zero loses the atom unless it is recorded separately.
A random variable’s distribution does not reveal every feature of the original sample space or its relationship with other variables. Two different constructions can share the same CDF. To answer joint or conditional questions, additional information about their combined law is needed.
The term random variable includes a measurability requirement: inverse images of suitable numeric sets must be events. In this calculus-based unit, the explicit finite and density constructions meet that requirement; a deeper measure course develops it systematically.
Apply
Write a complete piecewise CDF for the practice two-point law, including all endpoint inequalities. Then inspect the mixed example and explain why its derivative integrates to less than one. The missing mass is not an integration error; it is the jump that differentiation omitted.
Continue
Next study discrete families generated by different trial mechanisms. Define what the random variable counts before choosing a formula.
Apply · explain · revise
Practice and reasoning
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X takes −1,0,2 with probabilities 1/4,1/2,1/4. Find F(0).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For that X, find P(0<X≤2).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
At a point a, P(X=a) equals:
Every CDF must be:
Write the full CDF for P(X=1)=1/3 and P(X=4)=2/3, and recover both masses from jumps.
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X is zero with probability 1/2 and otherwise uniform on (0,2). Derive its CDF and explain why an ordinary density alone cannot describe the whole law.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
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