Probability Density Functions

This lesson deepens your understanding of probability density functions (PDFs) for continuous random variables, covering how to find the mean, variance, mode, and median from a PDF, and how to apply these concepts in exam-style problems.

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Mean (Expected Value) of a Continuous Distribution

The expected value of a continuous random variable $X$X with PDF $f(x)$f(x) is:

$E(X) = \mu = \int_{-\infty}^{\infty} x \cdot f(x) \, dx$E(X)=μ=∫−∞∞​x⋅f(x)dx

More generally, for any function $g(X)$g(X):

$E(g(X)) = \int_{-\infty}^{\infty} g(x) \cdot f(x) \, dx$E(g(X))=∫−∞∞​g(x)⋅f(x)dx

Worked Example

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

$E(X) = \int_0^2 x \cdot \frac{3}{8}x^2 \, dx = \frac{3}{8}\int_0^2 x^3 \, dx = \frac{3}{8}\left[\frac{x^4}{4}\right]_0^2 = \frac{3}{8} \times 4 = \frac{3}{2} = 1.5$E(X)=∫02​x⋅83​x2dx=83​∫02​x3dx=83​[4x4​]02​=83​×4=23​=1.5

$E(X^2) = \int_0^2 x^2 \cdot \frac{3}{8}x^2 \, dx = \frac{3}{8}\int_0^2 x^4 \, dx = \frac{3}{8}\left[\frac{x^5}{5}\right]_0^2 = \frac{3}{8} \times \frac{32}{5} = \frac{12}{5} = 2.4$E(X2)=∫02​x2⋅83​x2dx=83​∫02​x4dx=83​[5x5​]02​=83​×532​=512​=2.4

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Variance of a Continuous Distribution

$\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2

Using the values from above:

$\text{Var}(X) = 2.4 - 1.5^2 = 2.4 - 2.25 = 0.15$Var(X)=2.4−1.52=2.4−2.25=0.15

$\text{SD}(X) = \sqrt{0.15} = 0.387$SD(X)=0.15​=0.387

Exam Tip: Always compute $E(X)$E(X) and $E(X^2)$E(X2) separately, then use $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2. Show both integrals in full.

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Mode of a Continuous Distribution

The mode is the value of $x$x at which $f(x)$f(x) is maximised.

To find the mode:

1. Differentiate $f(x)$f(x) with respect to $x$x.
2. Set $f'(x) = 0$f′(x)=0 and solve.
3. Verify it is a maximum (check $f''(x) < 0$f′′(x)<0 or by inspection).

Example: $f(x) = \frac{3}{4}x(2 - x)$f(x)=43​x(2−x) for $0 \leq x \leq 2$0≤x≤2.

$f(x) = \frac{3}{4}(2x - x^2)$f(x)=43​(2x−x2) $f'(x) = \frac{3}{4}(2 - 2x) = 0 \implies x = 1$f′(x)=43​(2−2x)=0⟹x=1 $f''(x) = \frac{3}{4}(-2) = -\frac{3}{2} < 0 \quad \text{(maximum)}$f′′(x)=43​(−2)=−23​<0(maximum)

The mode is $x = 1$x=1.

Note: For some PDFs (e.g., $f(x) = \frac{3}{8}x^2$f(x)=83​x2 on $[0, 2]$[0,2]), the function is monotonically increasing, so the mode is at the right endpoint $x = 2$x=2. Always check the endpoints of the domain.

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Median of a Continuous Distribution

The median $m$m satisfies:

$P(X \leq m) = \int_{-\infty}^m f(x) \, dx = 0.5$P(X≤m)=∫−∞m​f(x)dx=0.5

Worked Example

Find the median of $f(x) = \frac{3}{8}x^2$f(x)=83​x2 for $0 \leq x \leq 2$0≤x≤2.

$\int_0^m \frac{3}{8}x^2 \, dx = 0.5$∫0m​83​x2dx=0.5

$\frac{3}{8} \times \frac{m^3}{3} = 0.5$83​×3m3​=0.5

$\frac{m^3}{8} = 0.5 \implies m^3 = 4 \implies m = \sqrt[3]{4} \approx 1.587$8m3​=0.5⟹m3=4⟹m=34​≈1.587