Probability Density Functions
A continuous random variable X takes values in an interval (or union of intervals). Unlike discrete variables, P(X = x) = 0 for any specific value. Instead, probability is defined over intervals through integration of the probability density function f(x).
📐 PDF — Two Fundamental Requirements
Non-negativity
f(x) ≥ 0 for all x. A PDF can never be negative.
Total area = 1
∫₋∞^∞ f(x) dx = 1. The total probability over all outcomes is 1.
Probability
P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx. Area under the PDF between a and b.
Point probability
P(X = a) = 0 for any specific value a. So P(a ≤ X ≤ b) = P(a < X < b) = P(a < X ≤ b) etc.
Cumulative Distribution Function (CDF)
Definition
F(x) = P(X ≤ x) = ∫₋∞ˣ f(t) dt
F(x) is the area under the PDF to the left of x. It is always non-decreasing, with F(−∞) = 0 and F(+∞) = 1.
PDF from CDF
f(x) = F'(x) = dF/dx
Differentiate the CDF to recover the PDF.
Interval probability from CDF
P(a ≤ X ≤ b) = F(b) − F(a)
Use CDF differences to avoid integration where possible.
Properties of the CDF
| Property | Statement |
|---|---|
| Range | 0 ≤ F(x) ≤ 1 for all x |
| Monotone | F is non-decreasing: x₁ < x₂ ⟹ F(x₁) ≤ F(x₂) |
| Limits | F(x) → 0 as x → −∞ and F(x) → 1 as x → +∞ |
| Continuity | F is continuous for a CRV (unlike discrete RVs) |
| At boundaries | F(a) = 0 and F(b) = 1 at the endpoints of the support |
💡 Cambridge Exam Pattern — PDF/CDF Questions
- A common question gives f(x) = kx² (or similar) over [a,b] and asks you to find k using the total area = 1 condition.
- Then asks you to find the CDF F(x) by integrating — always include the lower limit and write F(x) piecewise (0 for x < a, formula for a ≤ x ≤ b, 1 for x > b).
- Then asks for a specific probability — use the CDF rather than integrating again.
- Always state the support of the distribution (the range where f(x) > 0).
E
Finding k, CDF, and Probability
A CRV X has PDF f(x) = kx(2 − x) for 0 ≤ x ≤ 2, and f(x) = 0 otherwise.
(i) Find k. (ii) Find F(x). (iii) Find P(0.5 ≤ X ≤ 1.5).
(i) Find k. (ii) Find F(x). (iii) Find P(0.5 ≤ X ≤ 1.5).
Part (i)
∫₀² kx(2−x) dx = 1
k∫₀²(2x − x²) dx = k[x² − x³/3]₀² = k(4 − 8/3) = k · 4/3 = 1
k∫₀²(2x − x²) dx = k[x² − x³/3]₀² = k(4 − 8/3) = k · 4/3 = 1
k = 3/4
Part (ii)
For 0 ≤ x ≤ 2:
F(x) = ∫₀ˣ (3/4)t(2−t) dt = (3/4)[t² − t³/3]₀ˣ = (3/4)(x² − x³/3)
F(x) = ∫₀ˣ (3/4)t(2−t) dt = (3/4)[t² − t³/3]₀ˣ = (3/4)(x² − x³/3)
F(x) = 0 (x < 0); (3/4)(x² − x³/3) (0 ≤ x ≤ 2); 1 (x > 2)
Part (iii)
P(0.5 ≤ X ≤ 1.5) = F(1.5) − F(0.5)
F(1.5) = (3/4)(2.25 − 1.125) = (3/4)(1.125) = 0.84375
F(0.5) = (3/4)(0.25 − 0.125/3) = (3/4)(0.2083) = 0.15625
F(1.5) = (3/4)(2.25 − 1.125) = (3/4)(1.125) = 0.84375
F(0.5) = (3/4)(0.25 − 0.125/3) = (3/4)(0.2083) = 0.15625
P = 0.84375 − 0.15625 = 0.6875 = 11/16
Median, Quartiles, and Percentiles
Median m
F(m) = 0.5 ⟺ ∫₋∞ᵐ f(x) dx = 0.5
Solve F(m) = 0.5 for m. For the lower quartile Q₁: F(Q₁) = 0.25. For the upper quartile Q₃: F(Q₃) = 0.75.
Expectation & Variance
📐 Core Formulae — Expectation and Variance
Mean E(X)
E(X) = μ = ∫ x f(x) dx (over the support of X)
E(X²)
E(X²) = ∫ x² f(x) dx
Variance Var(X)
Var(X) = σ² = E(X²) − [E(X)]²
Also written Var(X) = E[(X−μ)²] = ∫(x−μ)² f(x) dx — but the computational form E(X²)−μ² is always faster.
Also written Var(X) = E[(X−μ)²] = ∫(x−μ)² f(x) dx — but the computational form E(X²)−μ² is always faster.
E[g(X)]
E[g(X)] = ∫ g(x) f(x) dx — replace x with g(x) in the integrand.
Linear Transformations
Expectation — linear
E(aX + b) = aE(X) + b
Constants shift/scale the mean proportionally.
Variance — linear
Var(aX + b) = a²Var(X)
Adding b does not change variance. Scaling by a multiplies variance by a².
Sums of Independent Random Variables
E(X + Y)
E(X + Y) = E(X) + E(Y)
Always true — no independence required.
Var(X + Y) — independent X, Y
Var(X + Y) = Var(X) + Var(Y)
Only true when X and Y are independent.
⛔ Common Error — Var(X − Y)
When X and Y are independent: Var(X − Y) = Var(X) + Var(Y) — variances always add, never subtract. This catches many students. For Var(2X): that is Var(aX) with a=2, giving 4Var(X). This is NOT the same as Var(X+X) from two independent copies.
E
Mean and Variance from PDF
X has PDF f(x) = (3/4)x(2−x) for 0 ≤ x ≤ 2 (from previous example). Find E(X), E(X²), and Var(X).
E(X)
E(X) = ∫₀² x · (3/4)x(2−x) dx = (3/4)∫₀²(2x²−x³) dx
= (3/4)[2x³/3 − x⁴/4]₀² = (3/4)(16/3 − 4) = (3/4)(4/3)
= (3/4)[2x³/3 − x⁴/4]₀² = (3/4)(16/3 − 4) = (3/4)(4/3)
E(X) = 1
E(X²)
E(X²) = (3/4)∫₀²(2x³−x⁴) dx = (3/4)[x⁴/2 − x⁵/5]₀²
= (3/4)(8 − 32/5) = (3/4)(8/5)
= (3/4)(8 − 32/5) = (3/4)(8/5)
E(X²) = 6/5
Var(X)
Var(X) = E(X²) − [E(X)]² = 6/5 − 1 = 1/5
Var(X) = 0.2 sd(X) = 1/√5 ≈ 0.447
Mode and Median vs Mean
| Measure | Definition | How to Find |
|---|---|---|
| Mean μ | ∫ x f(x) dx | Integrate x·f(x) over support |
| Median m | F(m) = 0.5 | Solve CDF = ½ for m |
| Mode | x where f(x) is maximum | Differentiate f(x), set f'(x)=0 |
| Quartile Q₁ | F(Q₁) = 0.25 | Solve CDF = ¼ |
| Quartile Q₃ | F(Q₃) = 0.75 | Solve CDF = ¾ |
Key Continuous Distributions
Cambridge 9231 P4 requires detailed knowledge of four continuous distributions. You must be able to state the PDF, CDF, mean, and variance for each — and recognise which applies from context.
Uniform Distribution
X ~ U(a, b)
f(x) = 1/(b−a) for a ≤ x ≤ b
F(x) = (x−a)/(b−a)
f(x) = 0 otherwise
F(x) = (x−a)/(b−a)
f(x) = 0 otherwise
Mean: (a+b)/2 Var: (b−a)²/12
Exponential Distribution
X ~ Exp(λ)
f(x) = λe^(−λx) for x ≥ 0
F(x) = 1 − e^(−λx)
f(x) = 0 for x < 0
F(x) = 1 − e^(−λx)
f(x) = 0 for x < 0
Mean: 1/λ Var: 1/λ²
Normal Distribution
X ~ N(μ, σ²)
f(x) = (1/σ√(2π)) e^(−(x−μ)²/2σ²)
No closed-form CDF
Standardise: Z = (X−μ)/σ ~ N(0,1)
No closed-form CDF
Standardise: Z = (X−μ)/σ ~ N(0,1)
Mean: μ Var: σ²
Beta Distribution
X ~ Beta(α, β)
f(x) = x^(α−1)(1−x)^(β−1)/B(α,β)
Support: 0 ≤ x ≤ 1
B(α,β) = Γ(α)Γ(β)/Γ(α+β)
Support: 0 ≤ x ≤ 1
B(α,β) = Γ(α)Γ(β)/Γ(α+β)
Mean: α/(α+β) Var: αβ/[(α+β)²(α+β+1)]
The Exponential Distribution in Depth
The exponential distribution models the waiting time between events in a Poisson process with rate λ. It has the key memoryless property:
Memoryless property
P(X > s + t | X > s) = P(X > t)
Given that you have already waited s units of time, the distribution of remaining wait time is still Exp(λ). Only the exponential (and geometric, in discrete case) has this property.
📌 Exponential–Poisson Link
If events occur as a Poisson process at rate λ per unit time, then the waiting time between consecutive events follows Exp(λ). This link is often tested in P4 — you may need to switch between Poisson (discrete, counting events) and Exponential (continuous, modelling time).
The Normal Distribution — Standardisation
Standardisation — Z-score
If X ~ N(μ, σ²), then Z = (X − μ)/σ ~ N(0, 1)
Use the standard normal tables (Φ) to find probabilities. Cambridge provides Table 1 (Φ values) and Table 2 (critical values) in the MF19 formula booklet.
Symmetry of N(0,1)
P(Z < −z) = 1 − P(Z < z) = 1 − Φ(z)
Interval probability
P(a < X < b) = Φ((b−μ)/σ) − Φ((a−μ)/σ)
Linear Combinations of Normal Variables
If X ~ N(μ₁, σ₁²) and Y ~ N(μ₂, σ₂²) are independent:
aX + bY
aX + bY ~ N(aμ₁ + bμ₂, a²σ₁² + b²σ₂²)
The sum/difference of independent normal variables is also normal. This is a P4 staple — used in inference and hypothesis testing.
Functions of a Random Variable
If Y = g(X) where X is a CRV, we need the distribution of Y. There are two standard methods in Cambridge 9231.
Method 1 — CDF Method (Always Works)
Steps — CDF Method for Y = g(X)
Step 1
Find the CDF of Y: F_Y(y) = P(Y ≤ y) = P(g(X) ≤ y).
Rearrange the inequality to express in terms of X.
Rearrange the inequality to express in terms of X.
Step 2
Express as an integral of f_X(x) and evaluate using the known CDF of X.
Step 3
Differentiate F_Y(y) to get the PDF f_Y(y) = F_Y'(y). State the support of Y.
E
CDF Method — Y = X²
X ~ U(0, 2). Find the PDF of Y = X².
CDF of X
X ~ U(0,2): f_X(x) = 1/2 for 0 ≤ x ≤ 2, CDF F_X(x) = x/2.
Support of Y
X ∈ [0,2] ⟹ Y = X² ∈ [0,4].
CDF of Y
F_Y(y) = P(Y ≤ y) = P(X² ≤ y) = P(X ≤ √y) = F_X(√y) = √y/2
(valid for 0 ≤ y ≤ 4)
(valid for 0 ≤ y ≤ 4)
PDF of Y
f_Y(y) = d/dy [√y/2] = 1/(4√y)
f_Y(y) = 1/(4√y) for 0 < y ≤ 4, 0 otherwise
Verify
∫₀⁴ 1/(4√y) dy = [√y/2]₀⁴ = 2/2 = 1 ✓
Method 2 — Change of Variable Formula (Monotone g)
For monotone increasing g with inverse x = h(y)
f_Y(y) = f_X(h(y)) · |h'(y)|
The |Jacobian| factor |h'(y)| = |dx/dy| corrects for the change of scale. If g is decreasing, the same formula applies due to the absolute value.
E
Change of Variable — Y = e^X
X ~ U(0,1). Find the PDF of Y = e^X.
Inverse
y = eˣ ⟹ x = h(y) = ln y. Support: y ∈ [e⁰, e¹] = [1, e].
Jacobian
|h'(y)| = |d/dy (ln y)| = 1/y
PDF of Y
f_Y(y) = f_X(ln y) · (1/y) = 1 · (1/y) = 1/y
f_Y(y) = 1/y for 1 ≤ y ≤ e, 0 otherwise
Worked Examples
1
Piecewise PDF — Full Treatment
X has PDF f(x) = ax for 0 ≤ x ≤ 1, f(x) = a(2−x) for 1 < x ≤ 2, and 0 otherwise.
(i) Find a. (ii) Find F(x). (iii) Find the median. (iv) Find E(X) and Var(X).
(i) Find a. (ii) Find F(x). (iii) Find the median. (iv) Find E(X) and Var(X).
Part (i) — k
∫₀¹ ax dx + ∫₁² a(2−x) dx = 1
a[x²/2]₀¹ + a[2x − x²/2]₁² = a(1/2) + a(4−2 − 2+1/2) = a/2 + a/2 = a = 1
a[x²/2]₀¹ + a[2x − x²/2]₁² = a(1/2) + a(4−2 − 2+1/2) = a/2 + a/2 = a = 1
a = 1 (triangular distribution on [0,2])
Part (ii) — F(x)
For 0 ≤ x ≤ 1: F(x) = ∫₀ˣ t dt = x²/2
For 1 < x ≤ 2: F(x) = F(1) + ∫₁ˣ (2−t) dt = 1/2 + [2t−t²/2]₁ˣ = 1/2 + (2x−x²/2) − (2−1/2)
= 1/2 + 2x − x²/2 − 3/2 = 2x − x²/2 − 1
For 1 < x ≤ 2: F(x) = F(1) + ∫₁ˣ (2−t) dt = 1/2 + [2t−t²/2]₁ˣ = 1/2 + (2x−x²/2) − (2−1/2)
= 1/2 + 2x − x²/2 − 3/2 = 2x − x²/2 − 1
F(x) = x²/2 (0≤x≤1); 2x − x²/2 − 1 (1<x≤2); 0 or 1 elsewhere
Part (iii) — Median
By symmetry of the triangular distribution, median = 1. Verify: F(1) = 1/2 ✓
Part (iv) — E(X)
By symmetry about x=1: E(X) = 1.
E(X²) = ∫₀¹ x²·x dx + ∫₁² x²(2−x) dx = [x⁴/4]₀¹ + [2x³/3−x⁴/4]₁²
= 1/4 + (16/3−4) − (2/3−1/4) = 1/4 + 4/3 − 5/12 = 7/6
Var(X) = 7/6 − 1 = 1/6 ≈ 0.167
E(X²) = ∫₀¹ x²·x dx + ∫₁² x²(2−x) dx = [x⁴/4]₀¹ + [2x³/3−x⁴/4]₁²
= 1/4 + (16/3−4) − (2/3−1/4) = 1/4 + 4/3 − 5/12 = 7/6
Var(X) = 7/6 − 1 = 1/6 ≈ 0.167
E(X) = 1, Var(X) = 1/6
2
Exponential Distribution — Poisson Link
Emails arrive at a rate of 5 per hour following a Poisson process. Let T be the waiting time (in hours) until the next email.
(i) State the distribution of T. (ii) Find P(T > 0.3). (iii) Given T > 0.1, find P(T > 0.4).
(i) State the distribution of T. (ii) Find P(T > 0.3). (iii) Given T > 0.1, find P(T > 0.4).
Part (i)
T ~ Exp(5) [rate λ=5 per hour, mean=1/5 hour=12 minutes]
Part (ii)
P(T > 0.3) = 1 − F(0.3) = e^(−5×0.3) = e^(−1.5)
= 0.2231
Part (iii) — Memoryless
P(T > 0.4 | T > 0.1) = P(T > 0.4 − 0.1) = P(T > 0.3) = e^(−1.5)
= 0.2231 [same answer — memoryless property]
Practice Questions
Question 1 — PDF, CDF, Median
[7 marks]
A CRV X has PDF f(x) = kx²(3−x) for 0 ≤ x ≤ 3, 0 otherwise.
(i) Show k = 4/27. (ii) Find F(x) for 0 ≤ x ≤ 3. (iii) Find the median, correct to 3 d.p.
(i) Show k = 4/27. (ii) Find F(x) for 0 ≤ x ≤ 3. (iii) Find the median, correct to 3 d.p.
For (i): ∫₀³ kx²(3−x) dx = k[x³ − x⁴/4]₀³ = k(27−81/4) = 27k/4 = 1. For (iii): solve F(m) = 0.5 numerically.
✓ Solution
(i) ∫₀³ kx²(3−x) dx = k∫₀³(3x²−x³)dx = k[x³−x⁴/4]₀³ = k(27−81/4) = 27k/4 = 1 ⟹ k = 4/27 ✓(ii) F(x) = (4/27)∫₀ˣ t²(3−t) dt = (4/27)[t³−t⁴/4]₀ˣ = (4/27)(x³ − x⁴/4)
F(x) = (4x³/27)(1 − x/12) = 4x³/27 − x⁴/27 for 0 ≤ x ≤ 3
(iii) Solve F(m) = 0.5: 4m³/27 − m⁴/108 = 0.5Try m=2: F(2) = 32/27 − 16/108 = 32/27 − 4/27 = 28/27 — too big; wait 28/27 > 1 — recheck.
F(2) = (4×8/27)(1−2/12) = (32/27)(10/12) = 320/324 = 0.988 — too big.
Try m=1.6: F(1.6) = 4(4.096)/27 − (6.5536)/108 = 0.608 − 0.061 = 0.547
Try m=1.5: F(1.5) = 4(3.375)/27 − 5.0625/108 = 0.500 − 0.047 = 0.453
Median ≈ between 1.5 and 1.6; bisect to get
m ≈ 1.554
Question 2 — Expectation and Variance
[5 marks]
X ~ U(2, 8). Without integration, state E(X) and Var(X). Find P(|X − E(X)| < 1).
For U(a,b): E(X) = (a+b)/2, Var(X) = (b−a)²/12. For the probability, note |X−5| < 1 means 4 < X < 6, which is an interval of length 2 out of the total range 6.
✓ Solution
E(X) = (2+8)/2 = 5Var(X) = (8−2)²/12 = 36/12 = 3
P(|X−5| < 1) = P(4 < X < 6) = (6−4)/(8−2) = 2/6 =
1/3
Question 3 — Exponential Distribution
[5 marks]
The lifetime T (years) of a component follows Exp(0.2).
(i) Find the mean and standard deviation of T.
(ii) Find P(T > 8).
(iii) Find the value of t such that P(T < t) = 0.9.
(i) Find the mean and standard deviation of T.
(ii) Find P(T > 8).
(iii) Find the value of t such that P(T < t) = 0.9.
For Exp(λ): mean = 1/λ, var = 1/λ². For (iii): F(t) = 1−e^(−λt) = 0.9 ⟹ e^(−0.2t) = 0.1 ⟹ t = −ln(0.1)/0.2.
✓ Solution
(i) Mean = 1/0.2 = 5 years. Var = 1/0.04 = 25. SD = 5 years.(ii) P(T > 8) = e^(−0.2×8) = e^(−1.6) ≈
0.2019
(iii) 1−e^(−0.2t) = 0.9 ⟹ e^(−0.2t) = 0.1 ⟹ t = ln(10)/0.2 =
11.51 years
Question 4 — Function of a CRV
[6 marks]
X has PDF f(x) = 2x for 0 ≤ x ≤ 1. Find the PDF of Y = 1 − X².
Use the CDF method. For 0 ≤ x ≤ 1, Y = 1−X² ∈ [0,1]. F_Y(y) = P(Y ≤ y) = P(1−X² ≤ y) = P(X² ≥ 1−y) = P(X ≥ √(1−y)) = 1−F_X(√(1−y)).
✓ Solution
X ∈ [0,1] ⟹ Y = 1−X² ∈ [0,1] (Y decreases as X increases).F_X(x) = x² (integrate f(x)=2x).
F_Y(y) = P(1−X² ≤ y) = P(X² ≥ 1−y) = P(X ≥ √(1−y)) = 1 − F_X(√(1−y)) = 1 − (1−y) = y
f_Y(y) = d/dy [y] = 1
Y ~ U(0,1)
(Interesting result — the probability integral transform: if X has CDF F, then F(X) ~ U(0,1).)
Interactive Distribution Tool
Select a distribution, adjust parameters, and see the PDF and CDF curves with live probability calculations.
PDF & CDF Explorer
—Mean
—Variance
—P(X ≤ a)
—P(a ≤ X ≤ b)
Formula Sheet — Continuous Random Variables
PDF Requirements
f(x) ≥ 0Non-negativity
∫ f(x) dx = 1Total area = 1
P(a≤X≤b) = ∫ₐᵇ f(x) dx—
P(X=a) = 0Always for CRV
CDF
F(x) = P(X≤x)= ∫₋∞ˣ f(t) dt
f(x) = F'(x)Differentiate CDF
P(a≤X≤b)= F(b) − F(a)
0 ≤ F(x) ≤ 1Monotone non-dec.
Expectation & Variance
E(X) = ∫ x f(x) dx—
Var(X) = E(X²) − [E(X)]²Computational
E(aX+b)= aE(X)+b
Var(aX+b)= a²Var(X)
Var(X−Y) indep.= Var(X)+Var(Y)
Uniform U(a,b)
f(x) = 1/(b−a)a ≤ x ≤ b
F(x) = (x−a)/(b−a)—
Mean(a+b)/2
Variance(b−a)²/12
Exponential Exp(λ)
f(x) = λe^(−λx)x ≥ 0
F(x) = 1−e^(−λx)—
Mean = 1/λVar = 1/λ²
MemorylessP(X>s+t|X>s)=P(X>t)
Functions of X — CDF Method
F_Y(y) = P(g(X)≤y)Rearrange
f_Y(y) = F_Y'(y)Differentiate
Jacobian methodf_Y(y)=f_X(h(y))|h'(y)|
Check support of YAlways state it
📋 Cambridge Exam Strategy — CRV
- Finding k: Integrate f(x) over the stated support, set equal to 1, solve. Never integrate over the whole real line if the support is finite.
- Writing F(x): Always give F(x) piecewise — F(x) = 0 below support, formula in between, F(x) = 1 above support. Cambridge deducts marks for missing the boundary cases.
- Mode: Differentiate f(x) and set f'(x) = 0. Check it is a maximum (second derivative or sign check). The mode is where the PDF is highest, not where F = 0.5.
- Var(X−Y): This equals Var(X) + Var(Y) for independent X,Y. Never Var(X) − Var(Y).
- Functions of X: Always use the CDF method for safety — it always works. The Jacobian method is faster only for monotone transformations.
- Exponential memoryless: P(T > s+t | T > s) = P(T > t). State this explicitly — Cambridge awards a mark for invoking the memoryless property by name.