# Continuous probability distributions

> CIE A-Level Further Mathematics · 9231
> Source: https://www.owlsprep.com/study/cie-9231-u4-continuous-probability-distributions/

This sub-topic introduces continuous random variables and their core properties. You will learn to work with probability density functions and cumulative distribution functions, and calculate key descriptive statistics for continuous distributions.

**Prerequisites:** [Discrete probability distributions](https://www.owlsprep.com/study/cie-9231-u4-discrete-probability-distributions/); Core integration techniques for A-Level Mathematics

## Learning objectives

- Distinguish between discrete and continuous probability distributions
- Verify that a function is a valid probability density function
- Calculate probabilities, cumulative distribution functions from pdfs
- Compute expectation, variance, median and mode for continuous distributions

## Valid Probability Density Functions

A continuous random variable can take any value within an interval of real numbers, unlike discrete variables which only take distinct, separate values. The probability distribution of a continuous variable is described by a probability density function (pdf).

**Probability Density Function (pdf)** — A non-negative function $f(x)$ that describes the relative likelihood for the continuous random variable $X$ to take a given value. The total area under the curve of $f(x)$ over the full range of $X$ must equal 1.

*Notation:* $f_X(x)$

*Example:* $f(x) = 2x$ for $0 \leq x \leq 1$ is a valid pdf.

**Worked example:** Show that $f(x) = \frac{1}{8}(x+1)$ for $2 \leq x \leq 4$ is a valid pdf.

1. Check the two required conditions for validity: non-negativity and total area = 1.
2. For $2 \leq x \leq 4$, $x+1$ is always positive, so $f(x) > 0$, satisfying the non-negativity condition.
3. Integrate $f(x)$ over the full range of $x$:
4. $$\int_{2}^{4} \frac{1}{8}(x+1) dx = \frac{1}{8} \left[ \frac{x^2}{2} + x \right]_{2}^{4}$$
5. Evaluate the definite integral:
6. $$\frac{1}{8} \left( (8 + 4) - (2 + 2) \right) = \frac{1}{8} (12 - 4) = 1$$
7. Both conditions are satisfied, so $f(x)$ is a valid pdf.

> **Exam tip:** Always check both conditions for validity: non-negativity AND total area = 1. Examiners regularly penalise candidates who only check one condition.

## Calculating Probabilities from pdfs

For continuous distributions, probability equals the area under the pdf curve between the bounds of the interval. A key property of continuous random variables is that the probability of $X$ taking any single specific value is always zero.

$$P(a \leq X \leq b) = \int_{a}^{b} f(x) dx$$

> **info**
>
> Because $P(X=a) = 0$ for any $a$, all interval bounds give the same probability: $P(a \leq X \leq b) = P(a < X < b) = P(a \leq X < b) = P(a < X \leq b)$.

**Worked example:** For the valid pdf $f(x) = \frac{1}{8}(x+1)$ for $2 \leq x \leq 4$, calculate $P(2.5 < X < 3.5)$.

1. Set up the integral of $f(x)$ between the given bounds:
2. $$P(2.5 < X < 3.5) = \int_{2.5}^{3.5} \frac{1}{8}(x+1) dx$$
3. Compute the antiderivative and evaluate:
4. $$= \frac{1}{8} \left[ \frac{x^2}{2} + x \right]_{2.5}^{3.5}$$
5. $$= \frac{1}{8} \left( (\frac{12.25}{2} + 3.5) - (\frac{6.25}{2} + 2.5) \right) = 0.5$$

## Cumulative Distribution Functions (cdfs)

The cumulative distribution function $F(x)$ gives the probability that $X$ is less than or equal to a given value $x$, and is derived by integrating the pdf from the lower bound of $X$ up to $x$.

**Cumulative Distribution Function (cdf)** — $F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) dt$. For a pdf defined over $a \leq x \leq b$, $F(x) = 0$ for $x < a$ and $F(x) = 1$ for $x > b$.

*Notation:* $F_X(x)$

**Worked example:** Find the cumulative distribution function for the pdf $f(x) = \frac{1}{8}(x+1)$, $2 \leq x \leq 4$.

1. For $x < 2$, there is no probability, so $F(x) = 0$.
2. For $2 \leq x \leq 4$, integrate the pdf from 2 to $x$:
3. $$F(x) = \int_{2}^{x} \frac{1}{8}(t+1) dt = \frac{1}{8} \left[ \frac{t^2}{2} + t \right]_{2}^{x}$$
4. Simplify the expression:
5. $$= \frac{1}{8} \left( \frac{x^2}{2} + x - 4 \right) = \frac{x^2 + 2x - 8}{16}$$
6. For $x > 4$, $F(x) = 1$, which checks out at $x=4$: $\frac{16 + 8 - 8}{16} = 1$. The full cdf is:
7. $$F(x) = \begin{cases} 0 & x < 2 \\ \frac{x^2 + 2x - 8}{16} & 2 \leq x \leq 4 \\ 1 & x > 4 \end{cases}$$

> **tip**
>
> Always write the cdf as a piecewise function covering all values of $x$, not just the interval where the pdf is non-zero. This is a common, easy-to-earn marking point.

## Descriptive Statistics for Continuous Distributions

We calculate common descriptive statistics for continuous distributions using integration, following these standard rules:

1. Expectation (mean): $E(X) = \int_{-\infty}^{\infty} x f(x) dx$
2. Variance: $Var(X) = E(X^2) - [E(X)]^2$, where $E(X^2) = \int_{-\infty}^{\infty} x^2 f(x) dx$
3. Median: The value $m$ that satisfies $F(m) = 0.5$
4. Mode: The value of $x$ that maximises $f(x)$, found via differentiation

**Worked example:** Calculate the median of the distribution with pdf $f(x) = \frac{1}{8}(x+1)$, $2 \leq x \leq 4$.

1. Use the cdf derived earlier, and set $F(m) = 0.5$:
2. $$\frac{m^2 + 2m - 8}{16} = 0.5$$
3. Rearrange to form a quadratic equation:
4. $$m^2 + 2m - 16 = 0$$
5. Solve the quadratic, take the root in the interval $[2,4]$:
6. $$m = -1 + \sqrt{17} \approx 3.12$$
7. The median is approximately 3.12 (3 significant figures).

## Common pitfalls

- **Wrong:** Forgetting to check the non-negativity condition when testing for a valid pdf
  - Why it fails: Many students only check that the total integral equals 1, but a function can integrate to 1 and be negative over part of its range
  - Correct: Always confirm that $f(x) \geq 0$ for all $x$ in the range of $X$ before checking the integral condition
- **Wrong:** Calculating $P(X = a)$ as $f(a)$ for continuous distributions
  - Why it fails: Confusion between discrete probability mass functions and continuous probability density functions
  - Correct: Remember that $P(X = a) = 0$ for any continuous random variable, probability is only non-zero over intervals
- **Wrong:** Only writing the expression for cdf over the interval where $f(x)$ is non-zero, omitting the other cases
  - Why it fails: Skipping the cases for $x$ below the lower bound or above the upper bound of $X$
  - Correct: Always specify $F(x) = 0$ for $x < \min(X)$ and $F(x) = 1$ for $x > \max(X)$
- **Wrong:** Calculating variance as $[E(X)]^2 - E(X^2)$
  - Why it fails: Mixing up the order of terms in the variance formula, leading to negative variance
  - Correct: Memorise that $Var(X) = E(X^2) - [E(X)]^2$, which always gives a non-negative result
- **Wrong:** Stopping at the first stationary point when finding the mode, not checking endpoints
  - Why it fails: Assuming the mode must be an interior critical point, when the maximum could be at an endpoint
  - Correct: Check all critical points and endpoints of the range to find the maximum value of $f(x)$

## Cheatsheet

| Quantity | Formula |
| --- | --- |
| Valid pdf conditions | $f(x) \geq 0 \ \forall x, \ \int_{-\infty}^{\infty} f(x) dx = 1$ |
| $P(a \leq X \leq b)$ | $\int_{a}^{b} f(x) dx$ |
| cdf from pdf | $F(x) = \int_{-\infty}^{x} f(t) dt$ |
| Expectation $E(X)$ | $\int_{-\infty}^{\infty} x f(x) dx$ |
| Variance $Var(X)$ | $E(X^2) - [E(X)]^2$ |
| Median $m$ | $F(m) = 0.5$ |
| Mode | Value of $x$ maximising $f(x)$ |

## What's next

Continuous probability distributions form the foundation of all further work in statistics for CIE A-Level Further Maths. You will apply these core integration and manipulation skills to specific named continuous distributions, including uniform, exponential, normal and chi-squared distributions. These skills are also critical for more advanced topics like probability generating functions, continuous hypothesis testing and linear combinations of random variables later in the unit.

- [Statistical Inference](https://www.owlsprep.com/study/cie-9231-u4-statistical-inference/)
- [Generating Functions](https://www.owlsprep.com/study/cie-9231-u4-generating-functions/)
- [Non-parametric tests](https://www.owlsprep.com/study/cie-9231-u4-non-parametric-tests/)

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