What are the properties of a cumulative distribution function?

F is non-decreasing (probability only accumulates), continuous for a continuous random variable, equals 0 below the support and 1 above it, and F(x) is always between 0 and 1. A valid cdf therefore rises from 0 up to 1 across the support.

How do you build the cumulative distribution function from a density?

Integrate the density from the lower end of the support up to x: F(x) = integral from the lower end to x of f(t) dt. Below the support F(x) = 0 and above it F(x) = 1, so F is usually written as a piecewise function with three pieces.

How do you recover the density from the cumulative distribution function?

Differentiate: f(x) = F'(x). The density is the derivative of the cumulative distribution function, since integrating the density gives F. Do not leave the answer as F, and do not integrate F.

How do you find a probability from the cumulative distribution function?

Use P(a a) = 1 - F(a) by the complement. Because a single point has probability 0, the < and <= signs make no difference.

How do you find the median and quartiles from the cumulative distribution function?

Solve F(x) = 0.5 for the median, F(x) = 0.25 for the lower quartile and F(x) = 0.75 for the upper quartile, keeping the solution inside the support. Solve F(x) = 0.5, not f(x) = 0.5, because half the probability lies below the median.

How do you recover the density from the cumulative distribution function?

Differentiate: f(x) = F'(x). Do not leave the answer as F and do not integrate.

How do you find a probability from the cumulative distribution function?

P(a a) = 1 - F(a). The < and <= signs give the same probability.

How do you find the median from the cumulative distribution function?

Solve F(x) = 0.5 (not f(x) = 0.5), keeping the answer inside the support. Use 0.25 and 0.75 for the lower and upper quartiles.

Year 12 Methods (Unit 3 & 4) Continuous random variables

Cumulative distribution functions

Q: What is the cumulative distribution function of a continuous random variable?

F(x) = P(X <= x), the probability that X is at most x, equal to the integral of the density from the lower end of the support up to x.

Q: What are the properties of a cumulative distribution function?

F is non-decreasing and continuous, equals 0 below the support and 1 above it, and always lies between 0 and 1. A valid cdf rises from 0 to 1 across the support.

Q: How do you recover the density from the cumulative distribution function?

Differentiate: f(x) = F'(x). Do not leave the answer as F and do not integrate.

Q: How do you find a probability from the cumulative distribution function?

P(a a) = 1 - F(a). The < and <= signs give the same probability.

Q: How do you find the median from the cumulative distribution function?

Solve F(x) = 0.5 (not f(x) = 0.5), keeping the answer inside the support. Use 0.25 and 0.75 for the lower and upper quartiles.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the cumulative distribution function of a continuous random variable is \(F(x)=P(X\le x)=\int_{\text{lower}}^{x} f(t)\,dt\). It is non-decreasing, equals \(0\) below the support and \(1\) above it. The density is recovered by \(f(x)=F'(x)\); a probability is \(P(amedian and quartiles solve \(F(x)=0.5,\,0.25,\,0.75\).

The cumulative distribution function (cdf) of a continuous random variable \(X\) is \(F(x)=P(X\le x)\): the probability that \(X\) is at most \(x\). It is the running total of probability, so it equals the area under the density up to \(x\), \(F(x)=\int_{\text{lower}}^{x} f(t)\,dt\).

Every cdf has the same shape of behaviour: it is non-decreasing (probability only ever accumulates), continuous for a continuous variable, equals \(0\) below the support and rises to \(1\) above it, and always lies between \(0\) and \(1\). A valid cdf is therefore an increasing ramp from \(0\) up to \(1\).

Because integrating the density gives \(F\), differentiating the cdf recovers the density: \(f(x)=F'(x)\). Probabilities come straight from \(F\): \(P(aa)=1-F(a)\). The median \(m\) splits the probability in half, so it solves \(F(m)=0.5\); the quartiles solve \(F(x)=0.25\) and \(F(x)=0.75\).

Key idea. The cdf accumulates probability: \(F(x)=P(X\le x)=\int_{\text{lower}}^{x} f\,dt\). Integrate the density to get \(F\), differentiate \(F\) to get the density, and solve \(F=0.5\) for the median.

\(F(x)\) is the accumulated area, rising from \(0\) to \(1\)

The median \(m\) solves \(F(m)=0.5\)

Definition of the cumulative distribution function:

\[F(x)=P(X\le x)=\int_{\text{lower}}^{x} f(t)\,dt\]

F (x) = P (X \leq x)

Properties (a valid cdf rises from \(0\) to \(1\)):

\[F \text{ non-decreasing}, \qquad F(\text{below})=0, \qquad F(\text{above})=1\]

Recover the density, and read probabilities, from \(F\):

\[f(x)=F'(x) \qquad P(aa)=1-F(a)\]

f (x) = F^{'} (x)

Median and quartiles. Solve \(F(m)=0.5\) for the median, \(F(x)=0.25\) for \(Q_1\) and \(F(x)=0.75\) for \(Q_3\) — using \(F\), the cumulative function, not the density.

How to work with a cumulative distribution function

Build \(F\) from the density. Integrate from the lower end of the support up to \(x\): \(F(x)=\int_{\text{lower}}^{x} f(t)\,dt\). Below the support \(F=0\); above it \(F=1\). Write \(F\) as a piecewise function of three pieces.
Recover the density. Differentiate: \(f(x)=F'(x)\) on the support.
Read a probability. \(P(X\le b)=F(b)\), \(P(aa)=1-F(a)\). Endpoints do not matter, so \(\le\) and \(<\) agree.
Find the median or quartiles. Solve \(F(x)=0.5\) (median), \(F(x)=0.25\) (\(Q_1\)) or \(F(x)=0.75\) (\(Q_3\)), keeping the solution inside the support.

Check. A correct \(F\) satisfies \(F(\text{lower})=0\) and \(F(\text{upper})=1\); if the top value is not \(1\), the integration (or the density) is wrong.

Example 1 — Build \(F\) from a density

\(f(x)=\dfrac{x}{8}\) for \(0\le x\le 4\) (and \(0\) otherwise). Find \(F(x)\) on the support and check \(F(4)\).

Solution

Integrate from \(0\) up to \(x\).

\(F(x)\)	\(=\)	\(\displaystyle\int_0^x \dfrac{t}{8}\,dt=\Big[\dfrac{t^2}{16}\Big]_0^x=\dfrac{x^2}{16}\)
\(F(4)\)	\(=\)	\(\dfrac{16}{16}=1\ \checkmark\)

So \(F(x)=\dfrac{x^2}{16}\) on \([0,4]\).

F (x) = \frac{x^{2}}{16}

Example 2 — Recover the density \(f=F'\)

The cdf is \(F(x)=\dfrac{x^3}{64}\) on \([0,4]\). Find the density \(f(x)\).

Solution

Differentiate the cumulative function.

\(f(x)\)	\(=\)	\(F'(x)=\dfrac{d}{dx}\Big(\dfrac{x^3}{64}\Big)\)
	\(=\)	\(\dfrac{3x^2}{64}\)

So \(f(x)=\dfrac{3x^2}{64}\) on \([0,4]\).

f = \frac{3 x^{2}}{64}

Example 3 — A probability from \(F\)

\(F(x)=\dfrac{x^2}{25}\) on \([0,5]\). Find \(P(1

Solution

Upper value minus lower value.

\(P(1	\(=\)	\(F(3)-F(1)\)
	\(=\)	\(\dfrac{9}{25}-\dfrac{1}{25}=\dfrac{8}{25}\)

So \(P(1

P = \frac{8}{25}

Example 4 — The median from \(F(m)=0.5\)

\(F(x)=\dfrac{x^2}{4}\) on \([0,2]\). Find the median \(m\).

Solution

Half the probability lies below the median.

\(F(m)\)	\(=\)	\(\dfrac{m^2}{4}=\dfrac{1}{2}\)
\(m^2\)	\(=\)	\(2\)
\(m\)	\(=\)	\(\sqrt{2}\approx 1.41\)

So the median is \(m=\sqrt{2}\approx 1.41\).

m = \sqrt{2}

Common pitfalls

Do not confuse \(f\) and \(F\). The density is \(f(x)=F'(x)\); to go from cdf to density you differentiate, and to go from density to cdf you integrate. Leaving \(F\) unchanged, or integrating when you should differentiate, is the most common slip.

\(F\) must reach \(1\). A cdf rises from \(0\) up to \(1\) across the support. If your \(F\) does not give \(1\) at the top of the support, the working is wrong — check the integration and the constant.

The median solves \(F=0.5\), not \(f=0.5\). Half the probability lies below the median, and probability is measured by \(F\). Solving \(f(x)=0.5\) uses the density height and gives the wrong answer.

Subtract in the right order. \(P(a

Frequently asked questions

What is the cumulative distribution function?

\(F(x)=P(X\le x)\), the probability that \(X\) is at most \(x\); for a continuous variable it equals \(\int_{\text{lower}}^{x} f(t)\,dt\).

What are the properties of a cdf?

\(F\) is non-decreasing and continuous, equals \(0\) below the support and \(1\) above it, and always lies between \(0\) and \(1\).

How do you build F from a density?

Integrate the density from the lower end of the support up to \(x\). Below the support \(F=0\) and above it \(F=1\), so \(F\) is a piecewise function.

How do you get the density back from F?

Differentiate: \(f(x)=F'(x)\) on the support.

How do you find P(a < X ≤ b) from F?

\(P(aa)=1-F(a)\) by the complement.

How do you find the median from F?

Solve \(F(m)=0.5\), keeping \(m\) inside the support; use \(0.25\) and \(0.75\) for the quartiles.

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