Continuous RVs: PDF, CDF & Uniform

Ask a friend how long they waited for the bus. They might say “about eight minutes.” Not 8.0000 minutes. The bus arriving at exactly one specific instant has, in a sense, zero chance — there are infinitely many instants in any window. So how do you talk about probability for a continuous quantity at all? You stop pinning probability to individual points and start spreading it as area under a curve. The curve is the probability density function, and once you can read areas off it you can answer every continuous-RV question GATE throws at you — and it is the same density picture behind every loss curve and likelihood you will later fit in machine learning.

Density, not probability

The PDF f(x) describes how thickly probability is packed near each value. Two rules define it:

• f(x) ≥ 0 everywhere (density is never negative), and
• the total area under it is 1: the integral of f over all x equals 1.

Probability is then the area over an interval:

P(a ≤ X ≤ b) = ∫ from a to b of f(x) dx

Two consequences trip people up. First, the probability of any single value is zero: the area over a point of zero width is zero, so P(X = a) = 0 for a continuous RV. (That is why P(a ≤ X ≤ b) and P(a < X < b) are equal — the endpoints add nothing.) Second, f(x) is a density, not a probability: it can exceed 1. On the interval [0, 0.5] a valid density could reach a height of 2, because what must stay ≤ 1 is the area, not the height.

CDF: the running total

The cumulative distribution function F(x) = P(X ≤ x) accumulates probability from the left — it is the integral of f up to x:

F(x) = ∫ from −∞ to x of f(t) dt

So F runs from 0 up to 1, never decreasing. Going the other way, the density is the derivative of the CDF: f(x) = F prime(x). PDF and CDF are an integrate/differentiate pair — given either one you can recover the other.

Drag the two handles to bracket a region and watch the shaded area — that area is the probability P(a < X < b), no integration by hand. Flip between Normal, Exponential, Beta, and Uniform to see how the same idea (probability = area) works for every continuous distribution you will meet on the exam.

The continuous uniform

The simplest continuous distribution: X is uniform on [a, b] when its density is flat. To enclose area 1 over a width of b − a, the height must be 1/(b − a):

• f(x) = 1/(b − a) for x in [a, b], and 0 outside.
• mean = (a + b)/2 (the midpoint, by symmetry).
• variance = (b − a)² / 12.

For uniform(0, 10): mean = 5, variance = 100/12 ≈ 8.33.

How GATE asks this

A reliable NAT. Three flavours recur: (1) find the constant that makes a given f(x) a valid PDF (set its total integral to 1 and solve); (2) compute a probability P(a ≤ X ≤ b) by integrating the density; (3) find the median m from the CDF by solving F(m) = 0.5. Continuous uniform questions are pure plug-in once you write f = 1/(b − a).

Worked example

Find c so that f(x) = c·x on [0, 2] (and 0 elsewhere) is a valid PDF. Then compute P(X ≤ 1).

A valid PDF must integrate to 1 over its support:

∫ from 0 to 2 of c·x dx = c · [x²/2] from 0 to 2 = c · (4/2) = 2c

Set 2c = 1  ⟹  c = 0.5

Now integrate the density up to 1 to get the probability:

P(X ≤ 1) = ∫ from 0 to 1 of 0.5·x dx = 0.5 · [x²/2] from 0 to 1
         = 0.5 · (1/2) = 0.25

So c = 0.5 and P(X ≤ 1) = 0.25. (Note f(2) = 0.5·2 = 1 here, and a density can even exceed 1 on a narrower interval — only the total area is pinned to 1.)

Quick check