Cumulative distribution functions | Year 12

01 / LEARN

Understand the idea.

Accumulate from the left

The CDF $F (x) = P (X \leq x)$ accumulates density area from the left. For bounded support, it is $0$ before the support, increases where positive density adds area, and is $1$ after the support.

A density-based CDF is continuous, never decreases and stays within $[0, 1]$ . Flat sections correspond to zero density. Write all support branches: an interior formula may produce impossible probabilities when extended beyond the interval where it was derived.

\begin{aligned} F (x) & = P (X \leq x) \\ P (a < X \leq b) & = F (b) - F (a) \end{aligned}

Reverse the accumulation

Where differentiable, a CDF recovers its density through $f (x) = F^{'} (x)$ . Conversely, integrating density from the support's lower boundary determines the correct cumulative constant. Interval probabilities follow from $P (a < X \leq b) = F (b) - F (a)$ .

A quantile $q$ satisfies $F (q) = p$ . In particular, a median divides probability into halves, so solve $F (m) = \frac{1}{2}$ and retain a solution in the support. The median need not be the midpoint of the support's length.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

CDF and median

For density $2 x$ on [0,1], find the CDF and median.

Worked solution

1
Accumulate density inside the support.
$F (x) = \int_{0}^{x} 2 t d t = {[t^{2}]}_{0}^{x} = x^{2} (0 \leq x \leq 1)$
Using a dummy variable keeps the integration variable distinct from the moving boundary.
2
Include the outer branches.
$F (x) = {\begin{cases} 0, & x < 0 \\ x^{2}, & 0 \leq x \leq 1 \\ 1, & x > 1 \end{cases}$
No area lies below zero, and all area has accumulated by one.
3
Set the cumulative probability to one half.
$F (m) = \frac{1}{2} \Rightarrow m^{2} = \frac{1}{2}$
The median splits total probability into equal halves.
4
Choose the root in the support.
$m = \frac{1}{\sqrt{2}} \approx 0.7071$
The negative square root is outside the support and cannot be the median.

Answer

Interior $F (x) = x^{2}$ ; median $1 / \sqrt{2}$ .

EXAMPLE 02

Shifted uniform support

X is uniform on [2,6]. Find F(3) and P(3<X<5).

Worked solution

1
Calculate the uniform density and interior CDF.
$\begin{aligned} f (x) & = \frac{1}{6 - 2} = \frac{1}{4} \\ F (x) & = \int_{2}^{x} \frac{1}{4} d t = \frac{x - 2}{4} (2 \leq x \leq 6) \end{aligned}$
Uniform probability is proportional to interval length.
2
Evaluate the required cumulative value.
$F (3) = \frac{3 - 2}{4} = \frac{1}{4}$
The interval from two to three occupies one quarter of the support.
3
Subtract cumulative areas for the middle interval.
$\begin{aligned} P (3 < X < 5) & = F (5) - F (3) \\ = \frac{5 - 2}{4} - \frac{1}{4} = \frac{1}{2} \end{aligned}$
Subtracting the area through three leaves the area between three and five.

Answer

$F (3) = 1 / 4$ ; interval probability $1 / 2$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

If $F (x) = x^{3}$ on [0,1], find the density there.

Show answer & reasoning

$3 x^{2}$ .

Differentiate the interior formula: $f (x) = F^{'} (x) = 3 x^{2}$ for $0 < x < 1$ . Endpoint density values do not affect probabilities.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

For CDF $x^{2}$ on $[0, 1]$ , find $P (0.4 < X < 0.8)$ .
2 marks
Give the CDF of density $3 x^{2}$ on $[0, 1]$ over the whole real line.
3 marks
Find the 90th percentile for CDF $x^{2}$ on $[0, 1]$ .
2 marks
Can a CDF decrease from 0.6 to 0.5 as $x$ increases? Explain.
2 marks
$X$ is uniform on $[4, 10]$ . Find $F (2)$ and $F (12)$ .
2 marks
For CDF $1 - {(1 - x)}^{2}$ on $[0, 1]$ , find the density and median.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For CDF $x^{2}$ on $[0, 1]$ , find $P (0.4 < X < 0.8)$ .
2 marks
${0.8}^{2} - {0.4}^{2} = 0.48$ .
0.48.
Give the CDF of density $3 x^{2}$ on $[0, 1]$ over the whole real line.
3 marks
Integrate $3 t^{2}$ from 0 to $x$ inside the support; use outer cumulative limits.
0 below 0; $x^{3}$ on $[0, 1]$ ; 1 above 1.
Find the 90th percentile for CDF $x^{2}$ on $[0, 1]$ .
2 marks
Solve $q^{2} = 0.9$ in the support.
$\sqrt{0.9} \approx 0.9487$ .
Can a CDF decrease from 0.6 to 0.5 as $x$ increases? Explain.
2 marks
An increasing upper bound cannot remove already accumulated probability.
No.
$X$ is uniform on $[4, 10]$ . Find $F (2)$ and $F (12)$ .
2 marks
2 is below the support; 12 is above it.
0 and 1.
For CDF $1 - {(1 - x)}^{2}$ on $[0, 1]$ , find the density and median.
3 marks
Differentiate the CDF.
Solve ${(1 - m)}^{2} = \frac{1}{2}$ and retain the root in $[0, 1]$ .
$f (x) = 2 (1 - x)$ ; median $1 - \frac{1}{\sqrt{2}}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: continuous random variables, normal distributions and sampling distributions of sample means. Original lessons and practice.

Official curriculum source ↗

In this lesson

Understand the idea.

Accumulate from the left

Reverse the accumulation

See the method in action.

CDF and median

Accumulate density inside the support.

Include the outer branches.

Set the cumulative probability to one half.

Choose the root in the support.

Shifted uniform support

Calculate the uniform density and interior CDF.

Evaluate the required cumulative value.

Subtract cumulative areas for the middle interval.

Try it yourself.

Check your understanding

Keep practising