Visual walkthrough — Continuous random variables — PDF, CDF, percentiles

---

Step 1 — What a "density curve" even is

WHAT. We draw a curve $y=f(x)$. Here $x$ is the value our random thing can take (say, a randomly chosen fraction that is twice as likely to land near $1$ as near $0$), and the height $f(x)$ tells you how thickly probability is piled up right there.

WHY a curve and not a list of probabilities. A continuous thing (an exact length, an exact time) has infinitely many possible values. If each got a positive probability, the total would blow past $1$. So no single value can carry probability — instead we spread probability like jam over bread, and describe the thickness of the jam at each spot. That thickness is the density, written $f(x)$.

PICTURE. The curve $f(x)=2x$ rises from $0$ at the left to $2$ at the right. Notice the height reaches $2$ — bigger than 1 — and that is completely fine, because height is density, not probability.

---

Step 2 — Why AREA is probability (the mass-density argument, drawn)

WHAT. Take a thin vertical strip of the curve between $x$ and $x+\Delta x$. Its area is approximately height $\times$ width $=f(x)\Delta x$. We claim that little area is the probability of landing in that strip.

WHY. Recall the rod analogy from the parent note: mass of a chunk $=$ density $\times$ length. Same algebra here. Density $f(x)$ has units "probability per unit $x$"; multiply by a width $\Delta x$ (units of $x$) and the units cancel to leave pure probability. So:

$$\underset{\text{chance of landing in strip}}{\underbrace{P(x\le X\le x+\Delta x)}}\approx \underset{\text{thickness}}{\underbrace{f(x)}}\cdot \underset{\text{strip width}}{\underbrace{\Delta x}}.$$

WHY the integral, and not just multiplication. One thin strip is an approximation because $f$ changes across the strip. To get an interval $[a,b]$ exactly, slice it into infinitely many infinitely-thin strips and add their areas — that adding-up-of-infinitely-thin-slices is exactly what the integral ${\int }_a^{b}f(x)dx$ means. That is why we reach for calculus and not ordinary multiplication.

$$P(a\le X\le b)={\int }_a^{b}f(x)dx(\text{sum of all strip areas from }a\text{ to }b).$$

PICTURE. One orange strip highlighted; below it, the strip is chopped finer and finer until it becomes smooth shaded area.

---

Step 3 — The total area must equal 1 (and what "zero at a point" looks like)

WHAT. Something must happen, so all the probability added together equals $1$. In area language:

$${\int }_{-\infty }^{\infty }f(x)dx=1.$$

Check our example: ${\int }_0^{1}2xdx=[x^2{]}_0^1=1-0=1$. ✓ The term $[x^2{]}_0^1$ is the antiderivative evaluated at the right edge minus the left edge.

WHY a single point has zero probability. A "point" is a strip of width $\Delta x=0$. Area $=f(c)\cdot 0=0$. So $P(X=c)={\int }_c^{c}f=0$. This is the drawn reason the $<$ vs $\le$ distinction never matters for continuous variables.

PICTURE. The whole region under $f(x)=2x$ on $[0,1]$ shaded (a triangle of area $\frac{1}{2}\cdot 1\cdot 2=1$ — you can even skip calculus and use the triangle-area formula here!), with a single red vertical line of "width zero" carrying no area.

---

Step 4 — Sweeping the area to build the CDF

WHAT. Now stop asking about fixed intervals and instead ask a running question: "How much probability lies to the left of $x$?" Call that running total $F(x)$:

$$F(x)=P(X\le x)={\int }_{-\infty }^{x}f(t)dt.$$

• $F(x)$ ::: the area accumulated from the far left up to $x$ — the Cumulative Distribution Function.
• $t$ ::: a dummy sliding variable that runs from $-\infty$ up to the endpoint $x$. We rename the inside variable so it isn't confused with the moving right edge $x$.

WHY. A running total turns every interval question into subtraction (Step 6) and, crucially, is the quantity we invert to find percentiles (Step 7). It's the master accumulator.

For our example, on $0\le x\le 1$:

$$F(x)={\int }_0^{x}2tdt=[t^2{]}_0^x=x^2.$$

The lower limit is $0$ (not $-\infty$) because $f=0$ to the left of $0$ — no area accumulates there.

PICTURE. Left panel: the $2x$ curve with a teal moving edge at position $x$ and the area behind it shaded. Right panel: the value of that shaded area plotted as a new curve — it climbs from $0$ up to $1$, and that climbing curve is $F(x)=x^2$.

---

Step 5 — Why $F$ can only rise, start at 0, and finish at 1

WHAT. Three shape-facts about every CDF, read straight from Step 4's picture:

1. $F(-\infty )=0$ — nothing accumulated yet.
2. $F(+\infty )=1$ — everything accumulated (Step 3's total).
3. $F$ is non-decreasing — as the teal edge slides right it can only sweep in more area (because $f\ge 0$, area never gets subtracted).

WHY it also has no jumps (for a continuous RV). A jump in $F$ would mean a chunk of probability appears with zero width — that's exactly $P(X=c)>0$, which Step 3 forbade. So the graph of $F$ is an unbroken climb.

PICTURE. The $F(x)$ curve traced left-to-right: flat at $0$, a smooth rise, flat at $1$ — with arrows showing "only goes up."

---

Step 6 — Undoing it: slope of $F$ gives back $f$, and difference gives intervals

WHAT (differentiate). $F$ was built by piling up $f$. So the rate at which $F$ climbs at a point is exactly how much density is being added there — that's the slope. By the Fundamental Theorem of Calculus:

$$f(x)=\frac{d}{dx}F(x)=F^{\prime }(x).$$

Check: $\frac{d}{dx}{x}^2=2x=f(x)$. ✓ The steeper $F$ is, the taller $f$ is.

WHAT (subtract). For a fixed interval, area between $a$ and $b$ = (area up to $b$) minus (area up to $a$):

$$P(a\le X\le b)={\int }_a^{b}f=F(b)-F(a).$$

Example: $P(0.2\le X\le 0.5)=F(0.5)-F(0.2)=0.5^2-0.2^2=0.25-0.04=\mathrm{0.21.}$

PICTURE. Top: the slope (tangent line) of $F$ at some $x$, with a caption "slope here $=$ height of $f$ there." Bottom: two shaded areas $F(b)$ and $F(a)$ whose difference is the middle strip.

---

Step 7 — Reading percentiles by going sideways on $F$

WHAT. A percentile flips the CDF question around. The CDF asks "given a position $x$, how much area is behind it?" The percentile asks "given an amount of area $p$, where is the position $x_p$?" Formally:

$$F(x_p)=p⟺x_p=F^{-1}(p),$$

where $F^{-1}$ is the inverse CDF (the quantile function).

• $p$ ::: a target fraction of probability, e.g. $0.5$ for the median.
• $x_p$ ::: the value on the $x$-axis where that fraction has accumulated.

WHY it's always solvable here. Because $f>0$ across $[0,1]$, $F$ strictly rises (never flat), so each height $p$ is hit at exactly one $x_p$ — a unique answer.

Median of our example: solve $F(m)=m^2=0.5$, so $m=\sqrt{0.5}\approx 0.707$. Notice the median sits past the midpoint $0.5$ — because the density leans right, more of the "jam" is on the right, so the halfway split shifts right.

PICTURE. The $F$ curve: go UP the $y$-axis to $p=0.5$, travel RIGHT until you hit the curve, then DOWN to read off $x_p=0.707$. Percentiles are a horizontal-then-vertical read; the CDF is a vertical-then-horizontal read.

---

Step 8 — The degenerate cases you must still handle

WHAT. Real problems hand you flat spots, edges, and the outside of the support. Handle them explicitly:

• Outside the support. For our example $F(x)=0$ for $x<0$ and $F(x)=1$ for $x>1$. If someone asks for the "$150\%$ percentile," there is no such $p$ — percentiles only exist for $0<p<1$.
• A flat density (uniform). For the Uniform distribution on $[0,1]$, $f=1$ (constant) and $F(x)=x$ (a straight diagonal). Percentiles read off trivially: $x_p=p$. This is the "no lean" case — median sits exactly at $0.5$.
• Endpoints carry no probability. $P(X\le 0)=F(0)=0$ and $P(X\ge 1)=1-F(1)=0$. Including or excluding the edge changes nothing (Step 3).
• A gap where $f=0$. If a density is zero on some stretch, $F$ is flat there — it briefly stops accumulating. Then $F$ is not strictly increasing on that gap, and a percentile landing exactly on the flat spot is not unique. (Our $2x$ example avoids this since $f>0$ throughout the support.)

PICTURE. Two mini-panels side by side: left = uniform (flat $f$, diagonal $F$); right = a density with a zero-gap producing a flat plateau in $F$.

---

The one-picture summary

Everything on one canvas: the density $f=2x$ (orange) with a shaded strip = probability; the CDF $F=x^2$ (teal) built by sweeping that area; and the median read off by the sideways trick (plum). Three characters, one story: density spreads, cumulative collects, percentile asks where.

Recall Feynman retelling — the whole walkthrough in plain words

Picture spreading 1 kg of jam on bread, but thicker on the right. The thickness at each spot is the PDF $f$ — and thickness can be huge (over "1") as long as the total jam is 1 kg. To find the chance of landing in a stretch of bread, you don't read the thickness; you scoop that strip and weigh it — that weight is the area, an integral. Now walk a knife from the far left edge toward the right and keep a running weight of everything scooped so far: that running total is the CDF $F$. It starts at 0, only climbs (you can't un-scoop jam), and ends at 1 when you've swept it all. If you want the thickness back, look at how fast the running total is climbing — steep climb means thick jam; that's differentiating. Finally, to find the median, don't hunt on the thickness curve — go up the running-total curve to the height $0.5$, slide across to the curve, drop down, and that spot has exactly half the jam on each side. Because the jam leaned right, that halfway spot ($0.707$) sits a little right of the middle.

---

Connections

• Continuous random variables — PDF, CDF, percentiles (parent — this page derives its central link visually)
• Fundamental Theorem of Calculus (Step 6: slope of accumulated area returns the density)
• Uniform distribution (Step 8: the flat-density / diagonal-CDF base case)
• Exponential distribution (another support-from-0 density to sweep the same way)
• Normal distribution (its CDF $\Phi$ is read with the exact sideways trick of Step 7)
• Quantile function and inverse-transform sampling (Step 7's $F^{-1}$ in action)
• Expectation and Variance of continuous RVs (next: weighting these areas by $x$)
• Discrete random variables — PMF (the summing cousin of this integrating story)

Concept Map