9231 · 4.1
Continuous random variables — FAQ
Frequently asked questions for 9231 Continuous random variables. Direct answers first, then deeper explanation — then practise with marking.
Why is P(X=a) = 0 for a continuous random variable? It seems like the variable must take *some* value.
While the variable will indeed take a specific value, the probability of it being exactly that one value out of an infinite number of possibilities is infinitesimally small, and thus is defined as zero. Probability for a CRV is the area under the PDF curve over an interval. The area corresponding to a single point (an interval of width zero) is zero. We can only talk about the probability of X being in a range, e.g., P(a ≤ X ≤ b).
Can the probability density function f(x) have a value greater than 1?
Yes, it can. Unlike a probability, the value of f(x) is a measure of 'probability density', not probability itself. For example, a uniform distribution on the interval [0, 0.5] has a PDF of f(x) = 2 for 0 ≤ x ≤ 0.5. The key constraint is that the area under the f(x) curve, which represents the total probability, must integrate to 1.
When finding the CDF, F(x), why do I need to define it for values outside the main range of the PDF?
The cumulative distribution function, F(x) = P(X ≤ x), must be defined for all real numbers x from -∞ to +∞. For a PDF defined on an interval [a, b], the probability of X being less than 'a' is zero, so F(x) = 0 for x < a. The probability of X being less than or equal to any value greater than 'b' is one, as all possible values are within that range. Therefore, F(x) = 1 for x ≥ b. A complete piecewise definition is required for a full description of the distribution.