ECMT1020 Chap.2 Random Variables, Mass and Density
Random Variables, Mass and Density
A random variable attaches a number to each outcome and is described by its distribution. This chapter separates discrete probability mass functions from continuous probability density functions, with close attention to support and normalisation.
It explains why a discrete point can have positive probability while a continuous point has probability zero, and why continuous probabilities are areas over intervals rather than density heights. The support is part of the model, not a decorative annotation. It tells a sum which values to include and tells an integral where the density exists.
Validity checks come before probability calculations: masses must be nonnegative and sum to one, while a density must be nonnegative and integrate to one. Once those conditions hold, event probability is obtained by adding supported masses or measuring area over the requested interval.
That distinction also explains why changing endpoint notation matters for a discrete distribution but usually not for an ordinary continuous one.
What this chapter covers
- 01
Random variables and outcomes
- 02
Discrete supports
- 03
Probability mass functions
- 04
Continuous supports
- 05
Probability density functions
- 06
Normalisation and interval probability
Worked example · free
Validate a simple density
- +1A density must integrate to one over its support.
- +1The integral from zero to two of kx is 2k.
- +1Setting 2k=1 gives k=0.5.
- +1P(X≤1) is the integral from zero to one of 0.5x.
- +1The resulting probability is 0.25.
Key terms
- Random variable
- A numerical function defined on random outcomes.
- Support
- The set of values a random variable can take.
- Mass function
- A function assigning probability to each discrete value.
- Density function
- A nonnegative function whose interval areas are probabilities.
- Normalisation
- The requirement that total probability equals one.
- Point probability
- Probability assigned to one exact value. It may be positive for a discrete variable, but it is zero for an ordinary continuous variable because a single point has no interval width under the density.
- Interval probability
- Probability that a variable lies between specified bounds. Discrete versions add supported masses inside the interval, while continuous versions integrate density over the bounds and therefore depend on area rather than one curve height.
- Density height
- The ordinate of a probability density at one value. It measures local probability concentration per unit and becomes probability only after accumulation across an interval.
Random Variables, Mass and Density FAQ
Can a density exceed one?
Yes. A density height is not itself a probability. The integral over the entire support must equal one, and interval areas must remain between zero and one.
Why is a continuous point probability zero?
A single point has zero width, so its area under an ordinary density is zero. Continuous probability is assigned to intervals rather than isolated values.
How do I reject an invalid distribution?
Check nonnegativity over the stated support and check that all masses sum, or the density integrates, to one. Failure of either condition invalidates the proposal.
Exam move
Draw or list the support before manipulating a distribution. For discrete variables, build a value and mass table and sum the final column. For continuous variables, mark integration bounds and check the total area. After every answer, test whether the reported probability lies from zero to one and whether all omitted regions have been handled. Use a two-column practice routine.
In the left column, classify the variable and write its support. In the right column, state whether the object supplied is a mass, density or cumulative function and list the validity conditions it must satisfy. Do this before solving for any unknown constant. A missing normalisation step often contaminates every later number, so verify the total first even when the question does not explicitly ask.
For a discrete law, enumerate each supported value and preserve exact fractions until the final line. For a continuous law, sketch the support, shade the requested interval and write the integral with bounds before finding an antiderivative. Treat a density height as a rate of probability accumulation, never as the probability of the point itself.
Practise endpoint variants in pairs, such as X less than b and X less than or equal to b. Explain why a mass at b changes the discrete answer and why a single point has zero probability in the continuous case. Add deliberately invalid candidates to revision: a negative mass, a density with total area above one, and a function defined without a complete support.
Diagnose the violated condition in words instead of merely rejecting the function. When checking arithmetic, use three screens. The probability must lie from zero to one, probabilities of disjoint pieces must add, and expanding an event must not reduce its probability. If a result fails one screen, return to support and bounds rather than patching the last numerical line. Finally, translate every answer back to the event.
Saying “0.32” is incomplete; saying “the model assigns probability 0.32 to values in the stated interval” confirms that the calculation and event remained aligned.
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