Continuous Random Variables

This lesson introduces continuous random variables — variables that can take any value in an interval (or the whole real line). Unlike discrete random variables, probabilities are not assigned to individual values. Instead, probabilities are calculated using probability density functions (PDFs) and integration.

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From Discrete to Continuous

For a discrete random variable, $P(X = x)$P(X=x) can be positive. For a continuous random variable:

$P(X = x) = 0 \quad \text{for any single value } x$P(X=x)=0for any single value x

This is because a single point has zero width, so the area under the density curve at a single point is zero. Probabilities are defined over intervals:

$P(a \leq X \leq b) = \int_a^b f(x) \, dx$P(a≤X≤b)=∫ab​f(x)dx

where $f(x)$f(x) is the probability density function (PDF).

Exam Tip: Because $P(X = a) = 0$P(X=a)=0 for a continuous random variable, $P(a \leq X \leq b) = P(a < X < b) = P(a \leq X < b) = P(a < X \leq b)$P(a≤X≤b)=P(a<X<b)=P(a≤X<b)=P(a<X≤b). The inclusion or exclusion of endpoints does not matter for continuous distributions.

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Properties of a PDF

A function $f(x)$f(x) is a valid PDF if and only if:

Property	Requirement
Non-negativity	$f(x) \geq 0$f(x)≥0 for all $x$x
Total area is 1	$\int_{-\infty}^{\infty} f(x) \, dx = 1$∫−∞∞​f(x)dx=1

Note that $f(x)$f(x) can be greater than 1 at some points — it is a density, not a probability. It is the area under the curve that represents probability.

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Worked Example: Verifying a PDF and Finding Probabilities

Let $f(x) = \frac{3}{8}x^2$f(x)=83​x2 for $0 \leq x \leq 2$0≤x≤2, and $f(x) = 0$f(x)=0 otherwise.

Step 1: Check non-negativity.

For $0 \leq x \leq 2$0≤x≤2, $x^2 \geq 0$x2≥0 so $f(x) \geq 0$f(x)≥0. Outside this interval, $f(x) = 0$f(x)=0. Condition satisfied.

Step 2: Check total area.

$\int_0^2 \frac{3}{8}x^2 \, dx = \frac{3}{8} \left[\frac{x^3}{3}\right]_0^2 = \frac{3}{8} \times \frac{8}{3} = 1 \quad \checkmark$∫02​83​x2dx=83​[3x3​]02​=83​×38​=1✓

Step 3: Find $P(1 \leq X \leq 2)$P(1≤X≤2).

$P(1 \leq X \leq 2) = \int_1^2 \frac{3}{8}x^2 \, dx = \frac{3}{8}\left[\frac{x^3}{3}\right]_1^2 = \frac{3}{8}\left(\frac{8}{3} - \frac{1}{3}\right) = \frac{3}{8} \times \frac{7}{3} = \frac{7}{8}$P(1≤X≤2)=∫12​83​x2dx=83​[3x3​]12​=83​(38​−31​)=83​×37​=87​

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Finding an Unknown Constant

A common exam question asks you to find a constant $k$k such that $f(x)$f(x) is a valid PDF.

Example: $f(x) = kx(2 - x)$f(x)=kx(2−x) for $0 \leq x \leq 2$0≤x≤2.

$\int_0^2 kx(2 - x) \, dx = 1$∫02​kx(2−x)dx=1

$k \int_0^2 (2x - x^2) \, dx = k\left[x^2 - \frac{x^3}{3}\right]_0^2 = k\left(4 - \frac{8}{3}\right) = k \times \frac{4}{3} = 1$k∫02​(2x−x2)dx=k[x2−3x3​]02​=k(4−38​)=k×34​=1

$k = \frac{3}{4}$k=43​

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Piecewise PDFs

Some PDFs are defined differently on different intervals.

Example: