Continuous distributions and density functions
Введено| English | Значение |
|---|---|
| probability density/ˌprɒbəˈbɪlɪti ˈdensɪti/ | плотность вероятности |
Is a height the same as a probability?
- A waiting time can take any value in an interval. Its probability comes from area, not the graph's height at one instant.
- This lesson studies probability density 概率密度: A nonnegative function whose integral over an interval gives its probability.
The method
- A density f must be nonnegative and integrate to 1 over its support.
- The cumulative distribution F(x) is the integral up to x.
- For a continuous variable P(X=x)=0 and E(X) is the integral of xf(x).
For a continuous random variable, why is P(X=1)=0 even though the density may be positive at 1?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A density gives probability through integration over an interval. A point interval has zero width, hence zero area, hence zero probability - however tall the density is there.
A worked case
Let f(x)=kx on 0≤x≤2 and zero elsewhere. Normalisation gives integral_0^2 kx dx=2k=1, so k=1/2. Then P(X≤1)=integral_0^1 x/2 dx=1/4 and E(X)=integral_0^2 x²/2 dx=4/3.
Continuous distributions and density functions
A density f must be nonnegative and integrate to 1 over its support
Compare the model with the worked case and explain one change.
Find k when f(x)=kx on [0,2] is a density.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Total density area k[x²/2] from 0 to 2=2k=1, so k=1/2.
Where it goes wrong
- Density can exceed 1 without being invalid; total area must equal 1.
- A cumulative distribution cannot decrease.
- Include the zero-density region outside the support when defining a complete model.
Not true: A continuous variable has P(X=x) equal to the density height f(x). For a continuous variable P(X=x)=0; probabilities are areas under the density curve.
For that density, find P(X≤1).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Probability=integral x/2 from 0 to 1=[x²/4]=1/4.
A continuous variable has P(X=x) equal to the density height f(x).
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
For a continuous variable P(X=x)=0; probabilities are areas under the density curve.
Applying it
- For a uniform distribution, density is reciprocal interval length.
- Find medians or percentiles using the cumulative probability, and use a support-aware integral for moments.
For that density, find E(X). Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
E(X)=integral x²/2 from 0 to 2=[x³/6]=4/3.
Match each density-function requirement to its statement.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
A density may exceed 1 locally; only the total area is fixed at 1. The cumulative function integrates and the expectation weights by x.
In your course
- Pearson Edexcel International A Level Mathematics, unit S2.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Probability density: A nonnegative function whose integral over an interval gives its probability.
- A density f must be nonnegative and integrate to 1 over its support.
- Watch for: density can exceed 1 without being invalid; total area must equal 1.
Write from a uniform mean of 3
- The density is uniform on 0 to 6.
- Find the mean.
A continuous random variable is uniform on the interval from 0 to 6. What is its mean?
The midpoint of 0 and 6 is 3.
What is the mean?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A uniform mean is the midpoint of the interval. Other options — 1/6 (the height of the density); 6 (the upper end).