Mathematics · University
Continuous distributions and densities
A density is integrated over a set; its value at a point is not the point probability.
Prerequisites
Integration and probability.
Study notes
Density versus probability
A density is integrated over a set; its value at a point is not the point probability.
Normalization
A density must be nonnegative and integrate to one.
Expected value
E[g(X)]=∫g(x)f(x)dx when the integral exists.
Cumulative distribution
F(x)=P(X≤x); differentiating an absolutely continuous CDF recovers a density almost everywhere.
Worked examples
Let f(x)=2x for 0≤x≤1 and zero otherwise. Find P(X≤1/2).
Integrate 2x from 0 to 1/2 to obtain 1/4.
0.25
For a continuous random variable with density f(x)=2x on [0,1] and zero elsewhere, find E[X].
Integrate 2x² from 0 to 1.
2/3
Check your understanding
For an absolutely continuous X, what is P(X=a) at a fixed point?
- 1 for every a
- 0
- f(a) always
- Undefined
A uniform variable on [2,8] has what mean?
Explain how a density can exceed 1 without violating probability rules.
For an exponential model with rate 2, find the mean.
Answers and marking guidance
Answers and marking notes
- 0
A single point has zero integral under a density.
- 5
The mean of a uniform interval is its midpoint.
- A density has units inverse to x; probability is area. For example a uniform density of 2 over [0,1/2] integrates to 1.
Credit a valid normalized example and area rather than height.
- 0.5
The mean is 1/rate.
Further reading and review references
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