# Continuous probability distributions and PDFs

> IB Mathematics AA HL · IB AA HL
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u4-continuous-probability-distributions-and-pdfs/

This sub-topic introduces continuous probability distributions, where random variables take values over intervals rather than discrete points. You will learn to define, verify, and use probability density functions (PDFs) to calculate probabilities.

**Prerequisites:** [Discrete probability distributions](https://www.owlsprep.com/study/ib-math-aa-hl-u3-discrete-probability/); [Definite integration](https://www.owlsprep.com/study/ib-math-aa-hl-u2-definite-integration/)

## Learning objectives

- Distinguish between discrete and continuous random variables
- State and apply the core properties of a valid probability density function (PDF)
- Calculate interval probabilities for continuous random variables from a PDF
- Find the mode of a continuous probability distribution

## Discrete vs Continuous Random Variables

Discrete random variables take countable distinct values (e.g. number of heads in 10 coin flips), while continuous random variables can take any value within one or more intervals (e.g. height of a randomly selected adult, wait time for a bus). Probabilities for discrete distributions are assigned to individual outcomes, but this is not true for continuous distributions.

**Continuous random variable** — A random variable $X$ is continuous if its set of possible values is an uncountable interval, and the probability that $X$ equals any single value is 0.

*Example:* The mass of an apple, measured to any precision, is a continuous random variable.

> **info**
>
> Because $P(X = a) = 0$ for any single $a$, $P(a < X < b) = P(a \leq X \leq b)$ for all continuous distributions. This removes any ambiguity when calculating interval probabilities.

**Worked example:** Explain why $P(X = 2.5) = 0$ if $X$ is the height of a person in meters.

1. The probability of $X$ falling into an interval decreases as the interval width shrinks. For a single point, the interval width approaches 0.
2. $$\lim_{\delta \to 0} P(x - \delta/2 < X < x + \delta/2) = 0$$
3. Therefore, the probability that $X$ equals exactly 2.5 is 0.

## Properties of a Valid Probability Density Function

Instead of assigning probability to single points, we use a function called the probability density function (PDF) to find the probability that a continuous random variable falls in an interval. The probability equals the area under the PDF curve over the interval.

**Probability Density Function (PDF)** — For a continuous random variable $X$ defined over $(-\infty, \infty)$, $f(x)$ is the PDF of $X$ if it satisfies two core conditions: (1) $f(x) \geq 0$ for all $x$, and (2) $\int_{-\infty}^{\infty} f(x) dx = 1$. Any probability $P(a < X < b)$ is given by $\int_a^b f(x) dx$.

*Notation:* f(x)

**Worked example:** Verify if $f(x) = \begin{cases} 2x & 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$ is a valid PDF.

1. First check the non-negativity condition. For $0 \leq x \leq 1$, $2x \geq 0$, and $f(x) = 0$ elsewhere. So $f(x) \geq 0$ for all $x$, first condition satisfied.
2. Next, integrate $f(x)$ over the entire domain to check total area is 1:
3. $$\int_{-\infty}^{\infty} f(x) dx = \int_0^1 2x dx$$
4. $$= \left[ x^2 \right]_0^1 = 1^2 - 0^2 = 1$$
5. Both conditions are satisfied, so this is a valid PDF.

## Calculating Probabilities from a PDF

Once you confirm a PDF is valid, you calculate the probability that $X$ lies between two values by integrating the PDF between those bounds. This corresponds directly to finding the area under the PDF curve over the interval of interest.

**Worked example:** For the valid PDF $f(x) = \begin{cases} 2x & 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$, find $P(0.2 < X < 0.6)$.

1. We integrate the PDF from 0.2 to 0.6, since $f(x) = 0$ outside this interval:
2. $$P(0.2 < X < 0.6) = \int_{0.2}^{0.6} 2x dx$$
3. $$= \left[ x^2 \right]_{0.2}^{0.6} = (0.6)^2 - (0.2)^2$$
4. $$= 0.36 - 0.04 = 0.32$$
5. So the required probability is 0.32.

**Check your understanding**

Check your understanding of core PDF properties

1. What is $P(X = 0.5)$ for the PDF in the worked example above?

   - 0
   - 0.5
   - 1
   - Cannot be determined

   *Why:* For any continuous distribution, the probability of $X$ taking a single exact value is always 0, regardless of the PDF value at that point.

## Mode of a Continuous Distribution

For discrete distributions, the mode is the most likely individual value. For continuous distributions, the mode is the value of $x$ where the PDF $f(x)$ reaches its maximum value. We use calculus to find this maximum.

**Mode of a continuous distribution** — The mode is the value of $x$ at which the PDF $f(x)$ attains its global maximum value. A distribution can be unimodal (one mode), bimodal (two modes), or have no unique mode.

**Worked example:** Find the mode of $f(x) = \begin{cases} 12x^2(1-x) & 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$.

1. Expand and differentiate $f(x)$ to find critical points:
2. $$f(x) = 12x^2 - 12x^3$$
3. $$f'(x) = 24x - 36x^2 = 12x(2 - 3x)$$
4. Set derivative equal to 0 to find critical points:
5. $$12x(2 - 3x) = 0 \implies x = 0 \text{ or } x = \frac{2}{3}$$
6. Use the second derivative test to confirm a maximum:
7. $$f''(x) = 24 - 72x. \quad At \ x = \frac{2}{3}, f''(\frac{2}{3}) = 24 - 48 = -24 < 0$$
8. Endpoints $x=0$ and $x=1$ give $f(x)=0$, which is the minimum. So the global maximum is at $x = \frac{2}{3}$, so the mode is $\frac{2}{3}$.

## Common pitfalls

- **Wrong:** Claiming $P(X = a) = f(a)$
  - Why it fails: For continuous distributions, $f(a)$ is the density at $a$, not a probability. Probability is always area over an interval.
  - Correct: Always state $P(X = a) = 0$ for any continuous random variable, regardless of $f(a)$.
- **Wrong:** Using $f(b) - f(a)$ instead of integrating to find $P(a < X < b)$
  - Why it fails: Many students mix up PDF properties with the fundamental theorem of calculus for CDFs, and incorrectly use the difference of PDF values.
  - Correct: Always calculate $P(a < X < b)$ as the definite integral of $f(x)$ from $a$ to $b$.
- **Wrong:** Only checking that the total integral equals 1 when verifying a PDF, ignoring non-negativity
  - Why it fails: A function with a negative region cannot be a valid PDF, even if its total integral equals 1. Examiners frequently test this.
  - Correct: Always explicitly check both conditions: $f(x) \geq 0$ for all $x$, and $\int_{-\infty}^{\infty} f(x) dx = 1$.
- **Wrong:** Stopping at the first internal critical point when finding the mode, ignoring endpoints
  - Why it fails: The maximum of $f(x)$ can occur at the endpoint of the domain, not just at a turning point inside the interval.
  - Correct: Always evaluate $f(x)$ at all critical points and endpoints to confirm the global maximum for the mode.

## Cheatsheet

| PDF Property | Rule |
| --- | --- |
| Non-negativity | $f(x) \geq 0$ for all $x$ |
| Total area under curve | $\int_{-\infty}^{\infty} f(x) dx = 1$ |
| Interval probability | $P(a < X < b) = \int_a^b f(x) dx$ |
| Single point probability | $P(X = a) = 0$ |
| Mode | Value of $x$ where $f(x)$ is maximized |

## What's next

PDFs are the foundation of all continuous probability topics you will encounter in IB AA HL, and mastery of these core rules is essential for exam success. Next, you will extend this knowledge to cumulative distribution functions (CDFs), which let you calculate cumulative probabilities, percentiles, and quantiles. From there, you will study common named continuous distributions including the uniform, exponential, and normal distribution, all of which are heavily tested in IB exams. You will also learn to calculate expectation and variance for continuous distributions, building on the PDF integration skills you practiced here.

- [Normal Distribution](https://www.owlsprep.com/study/ib-math-aa-hl-u4-normal-distribution/)
- [Bivariate data: correlation and regression](https://www.owlsprep.com/study/ib-math-aa-hl-u4-bivariate-data-correlation-and-regression/)

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