Suppose a variable has density f(x)=2x on 0<x<1 and zero elsewhere. Its density is larger near one, meaning equal-width intervals near one carry more probability than equal-width intervals near zero. Predict why f(0.9)=1.8 is allowed even though probabilities cannot exceed one.
Predict
A density has units of probability per measurement unit. Its height is not the probability of exactly that input. Predict P(X=0.9) and compare it with the probability of a small interval around 0.9. One point has zero width; an interval accumulates area.
Experiment
Integrate the density: ∫₀¹2x dx=1, so it is normalized. The CDF inside the support is F(x)=x². Therefore P(1/4<X<3/4)=9/16−1/16=1/2.
Draw the line from (0,0) to (1,2), shade the region between the two endpoints and label its area rather than its height. Compare two intervals of width 1/4: [0,1/4] has mass 1/16, while [3/4,1] has mass 7/16. The same width does not imply the same probability unless the density is constant.
Understand
An absolutely continuous probability law has a nonnegative integrable density f with total integral one. Event probabilities are integrals of f over the event set. The CDF is F(x)=∫ from negative infinity to x of f(t)dt. Where the appropriate derivative exists, it recovers the density. Values of a density at individual points can be changed without changing the law.
For such a law, singleton probabilities are zero. Including or excluding a finite number of interval endpoints therefore has no effect on probability. This differs from a discrete or mixed law with atoms. Always determine which type of distribution is being used before discarding endpoint distinctions.
A density may exceed one. Uniform(0,1/2) has height two and width one half, producing total area one. A density may also be unbounded at an endpoint while remaining integrable. Neither feature alone invalidates it. Nonnegativity and total integral are the relevant requirements.
Normalization by a constant cannot repair a function that changes sign on a set of positive length. A proposed f(x)=2x−1 on (0,2) is negative near zero; dividing by its positive integral leaves that negativity. Check support and sign before solving for a normalization constant.
A density statement includes its support. Writing only f(x)=2x without restricting the domain would fail both nonnegativity on negative inputs and integrability over the whole real line. A piecewise CDF likewise needs values below and above the support, not only its interior expression.
For a small width Δx near a point of continuity, probability is approximately f(x)Δx. This local approximation explains the units and graphical meaning, but finite intervals generally require integration. A density peak can be narrow enough to carry little total probability compared with a broad low-density region.
Apply
Check the practice normalization on its full stated support, then derive a CDF with all branches. Construct an example of a tall valid density and a normalized-looking invalid signed function. Explain the difference using integrals and signs rather than a rule that all plotted heights must be below one.
Continue
Next study uniform, exponential and normal families. Their formulas describe different mechanisms and need different assumptions, even when means and standard deviations happen to match.
Apply · explain · revise
Practice and reasoning
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f(x)=cx on 0<x<2 and zero elsewhere. Find c.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For that density, find P(1<X<2).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
A density value may exceed 1:
For an absolutely continuous law, P(X=a) is:
Derive the CDF for f(x)=3x² on (0,1), zero elsewhere, and calculate P(1/2<X≤1).
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Can f(x)=2x−1 on (0,2) become a valid density merely by dividing by its total integral? Explain.
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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