Cumulative distribution functions: Exercises — AI-ML deep dive

Before starting, keep three facts from the parent note in your pocket. A cumulative distribution function == $F_{X} (x) = P (X \leq x)$ == is the running total of probability. It never goes down. And the probability of landing in an interval is $F_{X} (b) - F_{X} (a)$ . Everything below is built from those.

Level 1 — Recognition

These test whether you can tell CDFs apart from other objects, and read basic values off them.

Exercise 1.1

Which of these could be a valid CDF? For each, say yes/no and why.

(a) $F (x) = x^{2}$ for all $x \in R$
(b) $F (x) = \frac{1}{2} + \frac{1}{π} arctan (x)$ for all $x$
(c) A function that equals $0.7$ at $x = 2$ and $0.4$ at $x = 5$

Recall Solution 1.1

A valid CDF must satisfy three things (parent note): non-decreasing, limits $0$ at $- \infty$ and $1$ at $+ \infty$ , and range inside $[0, 1]$ .

(a) No. $x^{2} \to + \infty$ , so it blows past $1$ and never settles at a limit of $1$ . Also $x^{2}$ decreases on $x < 0$ . Two rules broken.

(b) Yes. As $x \to - \infty$ , $arctan (x) \to - \frac{π}{2}$ , giving $\frac{1}{2} + \frac{1}{π} (- \frac{π}{2}) = 0$ . As $x \to + \infty$ , $arctan (x) \to + \frac{π}{2}$ , giving $1$ . And $arctan$ is increasing, so $F$ is non-decreasing. All three rules pass — this is the standard Cauchy distribution CDF.

(c) No. It goes down from $0.7$ (at $x = 2$ ) to $0.4$ (at $x = 5$ ). A CDF can never decrease, because you cannot "un-accumulate" probability.

Exercise 1.2

A discrete CDF jumps from $0.3$ to $0.55$ at the point $x = 4$ . What is $P (X = 4)$ ?

Recall Solution 1.2

For discrete variables, $P (X = x)$ = the height of the jump at $x$ : $P (X = x) = F_{X} (x) - F_{X} (x^{-})$ , where $x^{-}$ is the value just to the left. $P (X = 4) = 0.55 - 0.3 = 0.25$

Level 2 — Application

Now we plug into the machinery: interval probabilities and evaluating CDFs.

Exercise 2.1

$X \sim Uniform (0, 1)$ has CDF $F_{X} (x) = x$ on $[0, 1]$ . Find $P (0.2 < X \leq 0.75)$ .

Recall Solution 2.1

Use the interval formula $P (a < X \leq b) = F_{X} (b) - F_{X} (a)$ . $P (0.2 < X \leq 0.75) = F_{X} (0.75) - F_{X} (0.2) = 0.75 - 0.2 = 0.55$ Makes sense: the interval has width $0.55$ , and a uniform density of height $1$ puts area = width.

Exercise 2.2

$X \sim Exp (λ)$ with $F_{X} (x) = 1 - e^{- λ x}$ for $x \geq 0$ , and $λ = 2$ . Find $P (X > 1)$ .

Recall Solution 2.2

Use the complement rule $P (X > x) = 1 - F_{X} (x)$ . $P (X > 1) = 1 - F_{X} (1) = 1 - (1 - e^{- 2 \cdot 1}) = e^{- 2} \approx 0.1353$ This is the "survival" probability — the chance the event has not happened by time $1$ .

Exercise 2.3

For the fair die from the parent note, $F_{X} (x)$ is a staircase with $F_{X} (x) = k /6$ for $k \leq x < k + 1$ . Find $P (2 < X \leq 5)$ .

Recall Solution 2.3

$F_{X} (5) = 5/6$ and $F_{X} (2) = 2/6$ . $P (2 < X \leq 5) = \frac{5}{6} - \frac{2}{6} = \frac{3}{6} = 0.5$ Check: the outcomes strictly above $2$ and up to $5$ are ${3, 4, 5}$ — three faces, each probability $1/6$ , total $0.5$ . ✓

Level 3 — Analysis

Take the definitions apart: go between PDF and CDF, handle piecewise pieces.

Exercise 3.1

A continuous $X$ has PDF $f_{X} (x) = 2 x$ for $0 \leq x \leq 1$ (and $0$ elsewhere). Derive $F_{X} (x)$ for all real $x$ , covering every case.

Recall Solution 3.1

The CDF is the accumulated area: $F_{X} (x) = \int_{- \infty}^{x} f_{X} (t) d t$ . We split the real line into the three regions the density defines.

Case $x < 0$ : no density yet, nothing accumulated. $F_{X} (x) = 0$ .

Case $0 \leq x \leq 1$ : integrate the triangle density from $0$ to $x$ . $F_{X} (x) = \int_{0}^{x} 2 t d t = [t^{2}]_{0}^{x} = x^{2}$

Case $x > 1$ : all the mass is already collected. $F_{X} (x) = \int_{0}^{1} 2 t d t = [t^{2}]_{0}^{1} = 1$

Full answer: $F_{X} (x) = ⎩ ⎨ ⎧ 0 x^{2} 1 x < 0 0 \leq x \leq 1 x > 1$ Sanity check: differentiate the middle piece, $\frac{d}{d x} x^{2} = 2 x = f_{X} (x)$ . ✓ (Fundamental Theorem of Calculus, as the parent note showed.)

Exercise 3.2

Using the CDF from 3.1, find $P (0.25 < X \leq 0.5)$ and verify the total probability is $1$ .

Recall Solution 3.2

$P (0.25 < X \leq 0.5) = F_{X} (0.5) - F_{X} (0.25) = 0. 5^{2} - 0.2 5^{2} = 0.25 - 0.0625 = 0.1875$ Total probability: $F_{X} (1) - F_{X} (0) = 1 - 0 = 1$ . ✓ The whole distribution is accounted for.

Exercise 3.3

Look at figure below. It shows a mixed variable: a continuous ramp with one jump. What is $P (X = 2)$ and what is $P (X < 2)$ ?

Figure — Cumulative distribution functions

Recall Solution 3.3

Read the graph at $x = 2$ . The curve rises smoothly up to $F = 0.4$ (approaching from the left, the value $F_{X} (2^{-}) = 0.4$ ), then jumps to $F = 0.7$ .

$P (X = 2)$ is the jump height: $P (X = 2) = F_{X} (2) - F_{X} (2^{-}) = 0.7 - 0.4 = 0.3$

$P (X < 2)$ is the accumulated probability strictly below $2$ , which is the value just before the jump: $F_{X} (2^{-}) = 0.4$ .

This is why right-continuity matters: $F_{X} (2)$ itself is $0.7$ (the top of the jump), because $\leq$ includes the point.

Level 4 — Synthesis

Combine CDFs with inverses, medians, and sampling — ideas that link to Quantile Functions and Inverse Transform Sampling.

Exercise 4.1

For $X \sim Exp (λ)$ with $λ = 0.5$ , find the value $x_{q}$ such that $75%$ of the distribution is below it (the third quartile). This uses the quantile / inverse CDF.

Recall Solution 4.1

We want $F_{X} (x_{q}) = 0.75$ , i.e. invert the CDF. $1 - e^{- 0.5 x_{q}} = 0.75 ⟹ e^{- 0.5 x_{q}} = 0.25$ $- 0.5 x_{q} = ln (0.25) = - ln 4 ⟹ x_{q} = \frac{l n 4}{0.5} = 2 ln 4 \approx 2.7726$ So $75%$ of events occur before $t \approx 2.77$ .

Exercise 4.2

You have a uniform random number generator producing $U \sim Uniform (0, 1)$ . Using inverse transform sampling, derive the formula that converts $U$ into a sample from $Exp (λ)$ , then compute the sample when $U = 0.6$ , $λ = 1$ .

Recall Solution 4.2

Inverse transform sampling says: if $U$ is uniform, then $X = F_{X}^{- 1} (U)$ has the distribution with CDF $F_{X}$ . So we invert $F_{X} (x) = 1 - e^{- λ x}$ . $U = 1 - e^{- λ x} ⟹ e^{- λ x} = 1 - U ⟹ x = - \frac{1}{λ} ln (1 - U)$ With $U = 0.6$ , $λ = 1$ : $x = - ln (1 - 0.6) = - ln (0.4) \approx 0.9163$ This is exactly how libraries generate exponential samples from a uniform source.

Exercise 4.3

For the triangular variable of 3.1 ( $F_{X} (x) = x^{2}$ on $[0, 1]$ ), find the median.

Recall Solution 4.3

Set $F_{X} (m) = 0.5$ : $m^{2} = 0.5 ⟹ m = 0.5 = \frac{1}{2} \approx 0.7071$ Note the median sits right of centre ( $> 0.5$ ) because this density $2 x$ leans toward larger values — more mass on the right.

Level 5 — Mastery

The real-work tier: expectation from the CDF, empirical CDFs, and comparing distributions.

Exercise 5.1

Prove that for a non-negative continuous $X$ , $E [X] = \int_{0}^{\infty} [1 - F_{X} (x)] d x$ , then verify it on $Exp (λ)$ with $λ = 2$ .

Recall Solution 5.1

Proof sketch. $1 - F_{X} (x) = P (X > x) = \int_{x}^{\infty} f_{X} (t) d t$ . Integrate over $x$ from $0$ to $\infty$ and swap the order (Fubini): $\int_{0}^{\infty} \int_{x}^{\infty} f_{X} (t) d t d x = \int_{0}^{\infty} f_{X} (t) (\int_{0}^{t} d x) d t = \int_{0}^{\infty} t f_{X} (t) d t = E [X] .$ The inner swap works because for each fixed $t$ , the region ${0 \leq x \leq t}$ contributes length $t$ .

Verify on $Exp (2)$ . Here $1 - F_{X} (x) = e^{- 2 x}$ . $E [X] = \int_{0}^{\infty} e^{- 2 x} d x = [- \frac{1}{2} e^{- 2 x}]_{0}^{\infty} = \frac{1}{2}$ And indeed the mean of $Exp (λ)$ is $1/ λ = 1/2$ . ✓ Links to Expected Value and Variance.

Exercise 5.2

You observe five sorted data points: $1, 3, 3, 6, 9$ . The empirical CDF is the staircase $\hat{F} (x) = \frac{# { x _{i} \leq x }}{n}$ . Compute $\hat{F} (3)$ and $\hat{F} (5)$ .

Recall Solution 5.2

$n = 5$ .

$\hat{F} (3)$ : count how many of ${1, 3, 3, 6, 9}$ are $\leq 3$ . That's $1, 3, 3$ → three points. $\hat{F} (3) = \frac{3}{5} = 0.6$

$\hat{F} (5)$ : points $\leq 5$ are $1, 3, 3$ → still three (the $6$ and $9$ are above). $\hat{F} (5) = \frac{3}{5} = 0.6$ The empirical CDF is flat between data points and jumps by $1/ n$ at each observation (by $2/ n$ where a value repeats, as at $x = 3$ ).

Exercise 5.3

A Kolmogorov–Smirnov comparison uses the largest gap between two CDFs. Suppose at some point the empirical CDF reads $\hat{F} (x) = 0.6$ and the theoretical CDF reads $F_{0} (x) = 0.42$ . If this is the maximum gap, what is the KS statistic $D$ ?

Recall Solution 5.3

The KS statistic is the supremum of the absolute vertical gap: $D = sup_{x} ∣ \hat{F} (x) - F_{0} (x) ∣ = ∣0.6 - 0.42∣ = 0.18$ Larger $D$ means the sample looks less like the theoretical model. This is a workhorse goodness-of-fit test.

Recall One-line self-check

The interval formula is $P (a < X \leq b)$ equals what? ::: $F_{X} (b) - F_{X} (a)$ The inverse-transform sampling formula for a target CDF $F$ is $X =$ ? ::: $F^{- 1} (U)$ with $U \sim Uniform (0, 1)$ $P (X = x)$ for a discrete variable equals which two CDF values? ::: $F_{X} (x) - F_{X} (x^{-})$ , the jump height