Guides / Probability

Continuous random variables: probability is area under the density, and the CDF does the integrating

A continuous random variable can take any value in a range: a waiting time, a measurement, a weight in a neural network. No single value has positive probability, so the distribution is described by a density f(x), and probabilities are areas under it. The mistakes come from reading the height of the density as a probability, from mixing up a rate with a mean, and from dividing by the variance where the standard deviation belongs.

Short answer

A density has f(x) ≥ 0 and total area ∫ f(x) dx = 1. Then P(a < X < b) = ∫_a^b f(x) dx = F(b) − F(a), where the CDF F(x) = P(X ≤ x) and F′ = f. The mean and variance are integrals: E[X] = ∫ x f(x) dx and Var(X) = E[X²] − (E[X])². The three to know: uniform on [a, b] with mean (a + b)/2 and variance (b − a)²/12; exponential with rate λ, P(X > x) = e^−λx and mean 1/λ; normal N(μ, σ²), standardized by z = (x − μ)/σ.

Why it works

Think of a histogram of many results with narrower and narrower bars. If each bar's area is the fraction of results that land in it, the tops of the bars settle onto a curve, and that curve is the density. A probability is then an area under the curve, and the probability of hitting one exact value is the area of a line, which is 0. That's why P(X < b) and P(X ≤ b) are the same for a continuous variable.

The height f(x) is probability per unit of x: P(x < X < x + dx) ≈ f(x) dx. It can be bigger than 1 when the values are packed into a short interval. The CDF collects the area to the left of x, so it climbs from 0 to 1, and differentiating it gives the density back, which is just the fundamental theorem of calculus.

Example 1: normalize a density, then find a probability

f(x) = cx² for 0 ≤ x ≤ 3, and 0 elsewhere. Find c, the CDF and P(1 < X < 2).

∫₀³ cx² dx = 9c = 1 ⇒ c = 19The total area must be 1: c · 27/3 = 9c.
F(x) = ∫₀^x t²9 dt = x³27 for 0 ≤ x ≤ 3F is 0 to the left of 0 and 1 to the right of 3. Check: F(3) = 1.
P(1 < X < 2) = F(2) − F(1) = 8 − 127 = 727 ≈ 0.259The CDF has done the integrating, so a probability is a subtraction.

Example 2: the mean and variance by integration

Find E[X], Var(X) and the standard deviation for the density in Example 1.

E[X] = ∫₀³ x · x²9 dx = 19 · 814 = 94 = 2.25The sum Σ x P(X = x) becomes an integral of x times the density.
E[X²] = ∫₀³ x² · x²9 dx = 19 · 2435 = 5.4Square x, keep the same density.
Var(X) = 5.4 − 2.25² = 2780 = 0.3375, SD ≈ 0.581The same shortcut as for a discrete variable. The mean sits right of the middle of [0, 3] because the density grows toward 3.

Example 3: an exponential waiting time

Support tickets arrive at a rate of 0.5 per hour, so the wait X for the next one is exponential with λ = 0.5. Find the mean wait, P(X > 3), the median, and P(X > 5 | X > 2).

E[X] = 1λ = 2 hoursA rate of 0.5 per hour means one every 2 hours on average.
P(X > 3) = e^{−0.5 · 3} = e^−1.5 ≈ 0.2231The exponential's survival function is P(X > x) = e^−λx, so F(x) = 1 − e^−λx.
e^−0.5m = ½ ⇒ m = ln 20.5 ≈ 1.386 hoursThe median is where F reaches ½. It's less than the mean: a few long waits pull the mean up.
P(X > 5 | X > 2) = e^−2.5e⁻¹ = e^−1.5 ≈ 0.2231The same as P(X > 3). The exponential is memoryless: having waited 2 hours, the remaining wait has the same distribution as a fresh one.

Example 4: a normal probability by standardizing

Scores are normal with mean 100 and standard deviation 15. Find P(X > 130) and P(85 < X < 115).

z = 130 − 10015 = 2Subtract the mean and divide by the standard deviation: z counts standard deviations from the mean, and Z = (X − μ)/σ is standard normal.
P(X > 130) = 1 − Φ(2) ≈ 1 − 0.9772 = 0.0228Φ is the standard normal CDF, from a table or a calculator.
P(85 < X < 115) = Φ(1) − Φ(−1) ≈ 0.682785 and 115 are one standard deviation either side of the mean: the 68% of the 68–95–99.7 rule.

Common mistakes, and the exact line they happen on

1. Reading the density as a probability

X is uniform on [0, 0.5].

f(x) = 10.5 = 2 on [0, 0.5]Right so far: the height times the width 0.5 must be 1.
P(X = 0.25) = f(0.25) = 2This is the mistake. No probability can be 2. The density is probability per unit length, so it only becomes a probability when multiplied by a width: P(X = 0.25) = 0, and P(0.2 < X < 0.3) = 2 · 0.1 = 0.2.

2. Using the mean where the rate belongs

The tickets from Example 3, with a mean wait of 2 hours.

P(X > x) = e^−λxRight so far.
λ = 2, so P(X > 3) = e⁻⁶ ≈ 0.0025This is the mistake. λ is the rate, 0.5 per hour, and the mean is 1/λ. Using the mean as the rate makes long waits look almost impossible. The right answer is e^−1.5 ≈ 0.2231. Check the units: λx has to be a plain number, so λ is per hour when x is in hours.

3. Dividing by the variance when standardizing

The scores from Example 4, written as N(100, 225).

μ = 100, σ² = 225Right so far: N(μ, σ²) lists the variance.
z = 130 − 100225 ≈ 0.133, P(X > 130) ≈ 0.447This is the mistake. z is measured in standard deviations, so divide by σ = √225 = 15. Then z = 2 and P(X > 130) ≈ 0.0228. A score two standard deviations up should be rare, and 45% isn't.

Practice

f(x) = c(1 − x²) on [−1, 1]. Find c and P(X > ½).
Show answer
∫₋₁¹ (1 − x²) dx = 43, so c = 34, and P(X > ½) = 34∫_½¹ (1 − x²) dx = 532 ≈ 0.156.
X is uniform on [2, 8]. Find its mean, its variance and P(X < 3.5).
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Mean 5, variance 6²/12 = 3, and P(X < 3.5) = 1.5/6 = 0.25.
A wait is exponential with mean 10 minutes. What is the probability it is under 5 minutes?
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The rate is λ = 0.1 per minute, so P(X < 5) = 1 − e^−0.5 ≈ 0.3935.
X is normal with mean 50 and standard deviation 4. Find P(X < 44).
Show answer
z = (44 − 50)/4 = −1.5, and Φ(−1.5) ≈ 0.0668.
X has CDF F(x) = 1 − 1/x² for x ≥ 1. Find its density, its median and its mean.
Show answer
f(x) = F′(x) = 2/x³. The median solves 1/x² = ½, so it is √2 ≈ 1.414. The mean is ∫₁^∞ x · 2/x³ dx = 2.