Cumulative Distribution Function

The key idea. The cumulative distribution function $F$ answers a single question for every value $x$: how much probability sits at or below $x$? Formally,

$F(x) = P(X \le x) = \int_{-\infty}^{x} f(t)\,dt$

As $x$ slides from left to right, $F$ sweeps up area under the pdf, so it starts at $0$ (no area yet) and climbs to $1$ (all the area) once $x$ passes the top of the range. Because area can never be negative, $F$ never decreases — its graph is the characteristic S-shape below.

Why use a CDF at all? Once you have $F$, most questions become one-liners: probabilities, the median and any percentile are read straight off it with no further integration. The CDF is the "answer machine" you build once and reuse.

Cambridge tip. Always sanity-check a CDF at the ends of the range: a correct $F$ must satisfy $F(\text{lower}) = 0$ and $F(\text{upper}) = 1$. If it does not, your integration or constant is wrong.

Integration and differentiation are inverses, so the CDF and pdf are two faces of the same information:

$F(x) = \int_{-\infty}^{x} f(t)\,dt \qquad \Longleftrightarrow \qquad f(x) = F'(x)$

Going pdf $\to$ CDF (integrate). Add a constant of integration determined by $F(\text{lower}) = 0$, or — more simply — integrate from the bottom of the range up to $x$.

Going CDF $\to$ pdf (differentiate). Differentiate $F$ on the active interval only. Where $F$ is flat (constant $0$ below the range or constant $1$ above it), its slope is zero, so $f = 0$ there. This is why the pdf is always quoted with an "otherwise $0$" tail.

A worked illustration: if $F(x) = \dfrac{x^2}{16}$ on $0 \le x \le 4$, then

$f(x) = F'(x) = \frac{2x}{16} = \frac{x}{8}, \quad 0 \le x \le 4; \qquad f(x) = 0 \text{ otherwise.}$

Cambridge tip. When you differentiate a CDF, state the pdf on its interval and add the "$f(x) = 0$ otherwise" line — examiners reserve a mark for it.

Once $F$ is in hand, three jobs become near-instant.

1. Probabilities — subtract two CDF values, no integration: $P(a \le X \le b) = F(b) - F(a), \qquad P(X > a) = 1 - F(a).$

2. Percentiles and the median — set $F$ equal to the required cumulative probability and solve: $F(m) = 0.5 \;\;(\text{median}), \qquad F(x_p) = p \;\;(\text{the } 100p\text{-th percentile}).$ For example the lower quartile $q_1$ solves $F(q_1) = 0.25$. Solving $F(m) = 0.5$ is almost always quicker than the alternative of integrating the pdf and setting the area to $\tfrac12$.

3. The full piecewise definition. $F$ is defined for all real $x$, so a complete answer has three pieces:

• below the range: $F(x) = 0$;
• within the range: the integrated expression;
• above the range: $F(x) = 1$.

For a pdf $f(x) = \frac{3}{8}x^2$ on $0 \le x \le 2$: $F(x) = 0 \;\;(x<0); \quad F(x) = \frac{x^3}{8} \;\;(0 \le x \le 2); \quad F(x) = 1 \;\;(x>2).$

For a piecewise pdf, never restart the integral from zero on a new piece — carry the accumulated probability from the previous join: $F(x) = F(\text{join}) + \int_{\text{join}}^{x} f$.

Cambridge tip. Quote percentiles to 3 significant figures unless they are exact, and confirm continuity at each join ($F$ must agree from both sides) — examiners award credit for that check.

Suppose $Y = h(X)$ — for example $Y = X^2$ or $Y = \sqrt{X}$ — and you need the distribution of $Y$. The CDF method is the safe, mark-scheme-friendly route. It has four reliable steps.

Step 1 — find the range of $Y$. Push the range of $X$ through $h$. If $X \in [0,1]$ and $Y = X^2$, then $Y \in [0,1]$.

Step 2 — write the CDF of $Y$ as an event about $X$. Start from the definition and unwrap the inequality: $F_Y(y) = P(Y \le y) = P\big(h(X) \le y\big).$ For $Y = X^2$ with $X \ge 0$, $\;X^2 \le y \Leftrightarrow X \le \sqrt{y}$, so $F_Y(y) = P(X \le \sqrt{y}) = F_X(\sqrt{y})$. For $Y = \sqrt{X}$ (increasing), $\;\sqrt{X} \le y \Leftrightarrow X \le y^2$, so $F_Y(y) = F_X(y^2)$.

Step 3 — substitute the known $F_X$. Replace the argument and simplify to get $F_Y(y)$ as an explicit function of $y$.

Step 4 — differentiate for the pdf of $Y$. $\;g(y) = F_Y'(y)$. Take care with the chain rule: $\frac{d}{dy}e^{-y^2} = -2y\,e^{-y^2}$, and $\frac{d}{dy}y^{3/2} = \tfrac{3}{2}y^{1/2}$. Finish with the "otherwise $0$" tail, and ideally verify $\int g\,dy = 1$.

Cambridge tip. If the question asks for the pdf of $Y$, do not stop at $F_Y$ — the final, mark-bearing step is always $g(y) = F_Y'(y)$. And state the range of $X$ first so you take the correct (positive) root.