The Ky-Fan Metric

We shall discuss an exam question of my Probability-3 course and see a proof as to why the given distance function is a metric over the space of random variables and a few related properties.

Published on: 14 June, 2026 | Author: Pragyan Pranay

probabilityanalysisky-fanmetric-spaces

Probability-3 course was a nightmare for me and I even had to appear for backpaper for this course. This question originally came in the end-semester paper for B.Stat 2nd Year in session 2025-26. Yeah, this one was exceptionally hard to be done in exam conditions, although with the right tools, this isn’t very difficult, but producing the details with this much perfection is still quite hard. Our professor won’t leave us if we wrote even a single half-baked argument 🫠

🔍 Problem Statement
💡 Solution
- Part A
- Part B
- Part C
- Part D
Footnotes

🔍 Problem Statement

Problem (The Ky-Fan Metric) :

Let $\mathcal{T}$ be the set of all real-valued random variables defined on a given probability space $(\Omega, \mathcal{F} ,\mathbb{P})$ (where two random variables are said to be equivalent if they are equal with probability $1$ ). Define

d(X,Y) \coloneqq \inf\{\epsilon \ge 0 : \mathbb{P}(|X-Y| > \epsilon) \le \epsilon \}, \quad X,Y \in \mathcal{T}

Show that the infimum in the definition of $d$ is attained.
Show that $d$ is a metric on $\mathcal{T}$ .
Let $X \in \mathcal{T}$ and $\{X_n\}_{n \ge 1}$ be a sequence with $X_n \in \mathcal{T}$ for $n \ge 1$ . Show that
$X_n \overset{p}{\to} X \iff d(X_n, X) \to 0$
as $n \to \infty$ .
Show that $\mathcal{T}$ is complete with respect to metric $d$ .

💡Solution

We will solve the problem part by part. Also, for the entire solution, define

D_{X,Y}\coloneqq \{\epsilon \ge 0 : \mathbb{P}(|X-Y| > \epsilon) \le \epsilon \}.

Part A

$D_{X,Y}$ is non-empty since $1 \in D_{X,Y}$ and $D_{X,Y}$ is bounded below by $0$ by definition. Thus, the infimum exists. Let $\gamma = \inf D_{X,Y}$ , which means there exists a real sequence $a_n \searrow \gamma$ where $a_n \in D_{X,Y}$ . Thus,

\mathbb{P}(|X-Y| > a_n) \le a_n \quad \text{for all } n \ge 1.

Write $\mathbb{P}(|X-Y| > a_n) = 1 - \mathbb{P}(|X-Y| \le a_n)$ and let $Z = |X-Y|$ . Then we get

1 - a_n \le \mathbb{P}(|X-Y| \le a_n) = F_Z(a_n)

Taking $n \to \infty$ , then by the right continuity of $F_Z$ , we get

1 - \gamma \le F_Z(\gamma) = \mathbb{P}(|X-Y| \le \gamma) = 1 - \mathbb{P}(|X-Y| > \gamma).

Rearranging, we get

\mathbb{P}(|X-Y| > \gamma) \le \gamma \implies \gamma \in D_{X,Y}

which completes the proof to the first part. $\blacksquare$

Part B

This one is trickier, lets first recall the definition of a metric space for our context, where the parent set is $\mathcal{T}$ and distance function is $d$ .

Definition : (Metric Spaces)

A pair $(\mathcal{T}, d)$ is a metric space (where $d\colon \mathcal{T} \times \mathcal{T} \to \mathbb{R}$ ) if $d$ satisifies:¹

$d(X,Y) = 0 \iff X = Y\,\,\text{a.s}$ ;
$d(X,Y) = d(Y,X)$ for all $X,Y \in \mathcal{T}$ ; [Symmetry]
$d(X,Z) \le d(X,Y) + d(Y,Z)$ for all $X,Y,Z \in \mathcal{T}$ . [Triangle Inequality]

We will start with the first point. Assume that $d(X,Y) = 0$ . By part a), we know that the infimum is attained, thus we have

\mathbb{P}\left( |X-Y| > 0 \right) \le 0.

Since $\mathbb{P}(\bullet) \ge 0$ , we get

\begin{align*} \mathbb{P}\left(|X-Y| > 0 \right) &= 0 \\ \implies \mathbb{P}\left( |X-Y| \le 0 \right) &= 1. \end{align*}

Since $|r| \ge 0$ for any $r \in \mathbb{R}$ , we get $\mathbb{P}(0 \le |X-Y| \le 0) = 1 \implies P(X-Y = 0) = 1$ . Thus

\mathbb{P}(X = Y) = 1,

which shows $d(X,Y) = 0 \implies X = Y\,\,\text{a.s.}$ Now assume that $X = Y\,\, \text{a.s}$ , then

0 \in D_{X,Y} \iff \mathbb{P}(|X-Y| > 0) \le 0 \iff \mathbb{P}(|X-Y| > 0) = 0 \iff \mathbb{P}(|X-Y| \le 0) = 1.

Finally, this means $0 \in D_{X,Y} \iff \mathbb{P}(X - Y = 0) = 1 \iff X = Y\,\,\text{a.s.}$ Recall that $0 \le t$ for all $t \in D_{X,Y}$ , and now we have shown $0 \in D_{X,Y}$ , which means $d(X,Y) = \inf D_{X,Y} = 0$ as desired.

Proving symmetry is simple. Just note that

D_{X,Y} = \{\epsilon \ge 0 : \mathbb{P}(|X-Y| > \epsilon) \le \epsilon \} = \{\epsilon \ge 0 : \mathbb{P}(|Y-X| > \epsilon) \le \epsilon \} = D_{Y,X}

where the equality of probabilities follows from the simple fact that $|r| = |-r|$ for all $r \in \mathbb{R}$ . This gives $\inf D_{X,Y} = \inf D_{Y,X} \implies d(X,Y) = d(Y,X)$ and done.

We shall now prove the triangle inequality. This is one is quite tricky. For $X,Y,Z \in \mathcal{T}$ , we shall prove

d(X,Y) \le d(X,Z) + d(Z,Y) \iff c \le a + b

where $c = d(X,Y)$ , $a = d(X,Z)$ and $b = d(Z,Y)$ . Define

\begin{align*} A &\coloneqq \{\omega : |X(\omega) - Y(\omega)| > a+b \} \\ B &\coloneqq \{\omega : |X(\omega) - Z(\omega)| > a \} \\ C &\coloneqq \{\omega : |Z(\omega) - Y(\omega)| > b \} \end{align*}

We claim the following.

Claim —

A \subseteq B \cup C

Let $\alpha \in A$ . We wish to show that $\alpha \in B \cup C$ . Assume otherwise, $\alpha \not\in B \cup C$ if and only if $\alpha \in (B \cup C)' = B' \cap C'$ . Since $\alpha \in B' \cap C'$ , we must have

\begin{align*} |X(\alpha) - Z(\alpha)| &\le a \\ |Z(\alpha) - Y(\alpha)| &\le b. \end{align*}

Adding the above equations and using triangle inequality in $\mathbb{R}$ gives

|X(\alpha) - Y(\alpha)| \le a + b.

This is a contradiction since $\alpha \in A$ gives $|X(\alpha) - Y(\alpha)| > a + b$ . Hence, the claim is proved. $\square$

With the above claim and with the union bound $\mathbb{P}(A) \le \mathbb{P}(B \cup C) \le \mathbb{P}(B) + \mathbb{P}(C)$ we get

\mathbb{P}(|X-Y| > a + b) = \mathbb{P}(A) \le \mathbb{P}(|X-Z| > a) + \mathbb{P}(|Z-Y| > b) = a + b

which ultimately shows $a+b \in D_{X,Y} \implies c = d(X,Y) = \inf D_{X,Y} \le a + b$ completing the proof to Part B. $\blacksquare$

Part C

Assume $X_n \overset{p}{\to} X$ . Fix $\varepsilon > 0$ , we now need to show that there exists $N \in \mathbb{N}$ such that $d(X_n, X) = |d(X_n, X) - 0| < \varepsilon$ for all $n > N$ . Since $X_n \overset{p}{\to} X$ , for all $\eta > 0$ there exists $N_1 \in \mathbb{N}$ such that

\mathbb{P}(|X_n - X| > \varepsilon/2) < \eta \quad \forall n > N_1 \implies \mathbb{P}(|X_n - X| > \varepsilon/2) \le \eta \quad \text{for all }n > N_1

Since $\eta$ is free, put $\eta = \varepsilon/2$ to get

\mathbb{P}(|X_n - X| > \varepsilon/2) \le \frac{\varepsilon}{2} \quad \text{for all } n > N_1 \implies \varepsilon/2 \in D_{X_n, X} \quad \text{for all }n > N_1.

To finish it off, this means $(\inf D_{X_n, X}) \le \varepsilon/2$ for all $n > N_1$ , which gives

d(X_n, X) \le \frac{\varepsilon}{2} < \varepsilon \quad \text{for all }n > N_1

which proves $\displaystyle\lim_{n \to \infty} d(X_n, X) = 0$ as needed.

The other direction is quite easy as well, assume $\displaystyle\lim_{n \to \infty} d(X_n, X) = 0$ and define $a_n \coloneqq d(X_n,X)$ . By definition of $a_n$ , we know that $\mathbb{P}(|X_n-X| > a_n) \le a_n$ for all $n \in \mathbb{N}$ . Fix $\varepsilon, \eta > 0$ . Since $a_n > 0$ and $a_n \to 0$ , we can find $N_1 \in \mathbb{N}$ such that $a_n < \varepsilon$ for $n > N_1$ . Moreover, we can also find $N_2 \in \mathbb{N}$ such that $a_n < \eta$ for $n > N_2$ . For $n > \max(N_1,N_2)$ we have

\begin{align*} \mathbb{P}(|X_n - X| > \varepsilon) \le \mathbb{P}(|X_n - X| > a_n) \le a_n < \eta \\ \implies \mathbb{P}(|X_n - X| > \varepsilon) < \eta \quad \text{for all } n > \max(N_1,N_2) \end{align*}

which proves that $X_n \overset{p}{\to} X$ . This completes the proof to Part C. $\blacksquare$

Part D

This one will use a few tools, specifically the Cauchy Criteria for convergence in probability of random variables. We shall now define what cauchy in probability means and then prove the theorem which will trivialize the problem.

Definition : (Cauchy in Probability)

We say a that a sequence of random variables $\{X_n\}_{n \ge 1}$ is Cauchy in Probability if for all $\varepsilon, \eta > 0$ there exists $N \in \mathbb{N}$ such that

\mathbb{P}(|X_m - X_n| > \varepsilon) < \eta

for all $m,n > N$ .

Theorem : (Cauchy Criteria for Convergence in Probability)

For a sequence of real-valued random variables $\{X_n\}_{n \ge 1}$

X_n \overset{p}{\to} X \iff \{X_n\}_{n \ge 1}\,\,\text{is Cauchy in Probability.}

Assume $X_n \overset{p}{\to} X$ . Fix $\varepsilon, \eta > 0$ . By convergence in probability, we know there exists $N \in \mathbb{N}$ such that

\mathbb{P}(|X_m - X| > \varepsilon/2) < \frac \eta 2

for all $m > N$ . Define

\begin{align*} A &\coloneqq \{\omega : |X_m(\omega) - X_n(\omega)| > \varepsilon \} \\ B &\coloneqq \left\{\omega : |X_m(\omega) - X(\omega)| > \frac\varepsilon 2\right\} \\ C &\coloneqq \left\{\omega : |X_n(\omega) - X(\omega)| > \frac\varepsilon 2 \right\} \end{align*}

Claim —

A \subseteq B \cup C.

We ommit the proof, since its analogous to the claim in Part B. For $m,n > N$ , using the claim in conjunction with the union bound² like in Part B we get

\mathbb{P}(|X_m-X_n| > \varepsilon) \le \mathbb{P}\left(|X_m - X| > \frac\varepsilon 2\right) + \mathbb{P}\left(|X_n - X| > \frac \varepsilon 2 \right) < \eta/2 + \eta/2 = \eta

which proves that $\{X_n\}_{n \ge 1}$ is cauchy in probability as desired.

The other direction is much harder. I’d produce the proof as it is done by our Professor with some commentary as the proof is quite involved. Assume that $\{X_n\}_{n \ge 1}$ is cauchy in probability.

Step 1 – Construct a rapidly Cauchy subsequence.

Since

\{X_n\}

is Cauchy in probability, for each

j\in\mathbb{N}

we can choose an index

n_j

(with

n_1 < n_2 < \dots

) such that

\forall\, r,s \ge n_j, \qquad \mathbb{P}\bigl(|X_r - X_s| > 2^{-j}\bigr) < 2^{-j}.

In particular, for every $j$ we have

\mathbb{P}\bigl(|X_{n_{j+1}} - X_{n_j}| > 2^{-j}\bigr) < 2^{-j}.

Step 2 – Apply Borel–Cantelli.

Because

\sum_{j=1}^{\infty} 2^{-j} < \infty

, the first Borel–Cantelli lemma gives

\mathbb{P}\bigl( |X_{n_{j+1}} - X_{n_j}| > 2^{-j} \text{i.o. in $j$} \bigr) = 0.

Define the event

A = \bigl\{ \omega : |X_{n_{j+1}}(\omega) - X_{n_j}(\omega)| > 2^{-j} \text{ for only finitely many } j \bigr\}.

Then $\mathbb{P}(A) = 1$ .

Step 3 – Almost sure convergence of the subsequence.

For any

\omega \in A

, there exists

J(\omega)

such that for all

j \ge J(\omega)

|X_{n_{j+1}}(\omega) - X_{n_j}(\omega)| \le 2^{-j}.

For $k > \ell \ge J(\omega)$ we have the telescoping estimate

|X_{n_k}(\omega) - X_{n_\ell}(\omega)| \le \sum_{i=\ell}^{k-1} |X_{n_{i+1}}(\omega) - X_{n_i}(\omega)| \le \sum_{i=\ell}^{\infty} 2^{-i} = 2^{-\ell+1}.

Hence $\{X_{n_j}(\omega)\}$ is a Cauchy sequence in $\mathbb{R}$ and therefore converges. Define

X(\omega) = \begin{cases} \displaystyle \lim_{j\to\infty} X_{n_j}(\omega), & \omega \in A,\\[4pt] 0, & \omega \notin A. \end{cases}

Then $X$ is a random variable (the limit of measurable functions on $A$ , and constant on $A^c$ ). Moreover, $X_{n_j} \xrightarrow{\text{a.s.}} X$ , which implies $X_{n_j} \xrightarrow{p} X$ .

Step 4 – From the subsequence to the whole sequence.

Fix

\varepsilon > 0

and

\eta > 0

. Because

X_{n_j} \xrightarrow{p} X

, there exists

J

such that for all

j \ge J

\mathbb{P}\bigl(|X_{n_j} - X| > \varepsilon/2\bigr) < \eta/2.

Since $\{X_n\}$ is Cauchy in probability, we can also choose $K$ such that for all $m,n \ge K$ ,

\mathbb{P}\bigl(|X_m - X_n| > \varepsilon/2\bigr) < \eta/2.

Now take $N = \max(n_J, K)$ . For any $n \ge N$ , pick a subsequence index $n_j$ with $n_j \ge N$ (possible because $n_j \to \infty$ ). Then

\begin{aligned} \mathbb{P}\bigl(|X_n - X| > \varepsilon\bigr) &\le \mathbb{P}\bigl(|X_n - X_{n_j}| > \varepsilon/2\bigr) + \mathbb{P}\bigl(|X_{n_j} - X| > \varepsilon/2\bigr) \\ &< \eta/2 + \eta/2 = \eta. \end{aligned}

Thus $X_n \xrightarrow{p} X$ , completing the proof. $\quad\square$

Remark: Note that here, we just defined a map $X$ from $\Omega\to \RR$ and we claimed that $X_n \overset{p}{\to} X$ , but for this we need to make sure that $X$ is not just some arbitrary map, but a random variable indeed. For the curious beginners, this is covered in the the new blog post on Random Variables.

Returning back to the problem, let $\{Y_n\}_{n \ge 1}$ be a sequence of random variables with $Y_n \in \mathcal{T}$ for all $n \in \mathbb{N}$ which $d-$ cauchy(that is cauchy with the metric $d$ ). We will prove that $\{Y_n\}_{n \ge 1}$ is _cauchy in probability. Define $a_{m,n} \coloneqq d(Y_m,Y_n)$ . Fix $\varepsilon,\eta > 0$ , then by definition there exists some $N_1, N_2 \in \mathbb{N}$ such that

\begin{align*} a_{m,n} &= d(Y_m,Y_n) < \varepsilon \quad \text{for all } m,n > N_1;\\ a_{m,n} &= d(Y_m,Y_n) < \frac{\eta}{2} \quad \text{for all } m,n > N_2. \end{align*}

For all $m,n > \max(N_1, N_2)$ , we must have

\mathbb{P}(|Y_m-Y_n| > \varepsilon) \le \mathbb{P}(|Y_m-Y_n| > a_{m,n}) \le a_{m,n} \le \frac{\eta}{2} < \eta

where the second inequality is by definition of $a_{m,n}$ . This shows that $\{Y_n\}_{n \ge 1}$ is cauchy in probability. Invoking the cited theorem, the proof to Part D is ‘complete’. $\blacksquare$

Footnotes

We have excluded the non-negativity condition which is given in a lot of books. Regardless, if you assume all these conditions, then the non-negativity of $d$ follows. You can see this stackexchange thread. In particular, see the reply to the question by Hagen von Eitzen. In our case, proving non-negative is easy, since $D_{X,Y}$ is bounded below by $0$ (!) ↩
Union bound simply means that for any two measurable events $A,B$ we have $\mathbb{P}(A \cup B) \le \mathbb{P}(A) + \mathbb{P}(B).$ ↩

The Ky-Fan Metric

Table of Contents

🔍 Problem Statement

💡Solution

Part A

Part B

Part C

Part D

Footnotes

Footnotes