Continuous random variables

Further Maths · Further Probability & Statistics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A continuous random variable X can take any value in a range, so the probability of one exact value is zero. Probabilities are areas under the probability density function (PDF), f(x). The total area is 1 and f(x) is never negative. The PDF may have different formulae on different parts of the range.

The cumulative distribution function (CDF), F(x), is the probability that X is at most x. It is the area under f to the left of x, so integrating f gives F and differentiating F gives f.

To find the distribution of a function of X, such as Y = X³, work with the CDF: write P(Y ≤ y) as a probability about X, then differentiate to get the PDF of Y.

Rules to remember

f(x) ≥ 0 for all x, and the integral of f(x) over the whole range = 1 (use this to find an unknown constant k)
P(a < X < b) = ∫ f(x) dx from a to b = F(b) − F(a)
F(x) = P(X ≤ x) = ∫ f(t) dt from the lower end of the range up to x; f(x) = F′(x). F rises from 0 to 1 and never decreases
Median m: F(m) = 0.5. Lower quartile: F = 0.25. The p% percentile: F = p/100
E(X) = ∫ x f(x) dx, and in general E(g(X)) = ∫ g(x) f(x) dx; Var(X) = E(X²) − [E(X)]²
Piecewise PDF: integrate each piece over its own interval, and when building F add the area from all earlier pieces
For Y = g(X) with g increasing: F_Y(y) = P(X ≤ g⁻¹(y)) = F_X(g⁻¹(y)); differentiate for the PDF and state the new range of y

Common mistake

To find the PDF of Y = g(X), students just replace x by y in f(x). Probabilities transform through the CDF, so find F_Y(y) first and then differentiate. Also change the range: if 0 ≤ x ≤ 2 and Y = X³, then 0 ≤ y ≤ 8.

Worked example

X has probability density function f(x) = 4x³ for 0 ≤ x ≤ 1, and f(x) = 0 otherwise. Y = X². Find the probability density function of Y.

CDF of X: F(x) = ∫ 4t³ dt from 0 to x = x⁴, for 0 ≤ x ≤ 1.
CDF of Y: P(Y ≤ y) = P(X² ≤ y) = P(X ≤ √y) = (√y)⁴ = y².
Differentiate: the PDF of Y is 2y. Since 0 ≤ x ≤ 1, the range is 0 ≤ y ≤ 1.
Check: ∫ 2y dy from 0 to 1 = 1.

Answer: PDF of Y is 2y for 0 ≤ y ≤ 1, and 0 otherwise

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