The quantile function is defined as $Q(u)= \inf \{x: F(x) \geq u\}$.

It is well known the distribution function characterizes the probability distribution in the following sense

Theorem Let $X_{1}$ and $X_{2}$ be two real valued random variables with distribution functions $F_{1}$ and $F_{2}$ respectively. If $F_{1}(x)=F_{2}(x)$, $\forall x\in \mathbb{R}$ then $X_{1}$ and $X_{2}$ have the same probability distribution.

I want to prove that the quantile function also characterizes the probability distribution, this fact is stated as a corallary of the above theorem in this book (see Corallary 1.2 on p19):

Corallary: Let $X_{1}$ and $X_{2}$ be two real valued random variables with quantile functions $Q_{1}$ and $Q_{2}$ respectively. If $Q_{1}(u)=Q_{2}(u)$, $\forall u\in\left(0,1\right)$ then $X_{1}$ and $X_{2}$ have the same probability distribution.

The proof in the book is based on the following facts

Fact i: $Q(F(x)) \leq x$.

Fact ii: $F( Q(u) ) \geq u$.

Fact iii: $Q(u) \leq x$ iff $u \leq F(x)$.

Fact iv: $Q(u)$ is nondecreasing.

But I think it is wrong. Assuming $F_{1}(x_0) < F_{2}(x_0)$ for some fixed $x_0$ the author sets out to prove that this leads to a contradiction. Using facts (i) and (iv) he shows $Q_{2}(F_1(x_0)) < Q_{2}(F_2(x_0)) \leq x_0$. Then he applies fact (iii) to obtain $F_{1}(x_0) \leq F_{2}(x_0)$. The author claims that this is a contradiction. But clearly its not and the argument proves nothing.

Am I missing something here? Does anyone know the correct proof to the corallary?


2 Answers 2


Changed since version discussed in first five comments:

The key line in the proof of Corollary 1.2 in Severini's Elements of Distribution Theory book is

Hence by part (iii) of Theorem 1.8, $F_2(x_0) \ge F_1(x_0)$ so that $F_1(x_0) \lt F_2(x_0)$ is impossible.

As you say, $F_1(x_0) \lt F_2(x_0)$ in fact implies $F_1(x_0) \le F_2(x_0)$, i.e. $F_2(x_0) \ge F_1(x_0)$, so this is not a contradiction.

  • $\begingroup$ I have not miscopied part iii of the theorem. You have got one of the inequalities reversed in the above. But indeed if this was true it would prove the theorem. $\endgroup$
    – AUK1939
    Commented Oct 10, 2012 at 22:07
  • $\begingroup$ @aukie: Try rereading theorem 1.8 from your Google link. Carefully. $\endgroup$
    – Henry
    Commented Oct 10, 2012 at 22:44
  • $\begingroup$ @@Henry: This is what the book says. "$Q(t) \leq x$ iff $F(x)\geq t$". This is what you have written "$Q(t) \leq x$ iff $t \geq F(x)$". You have misread the second inequality. $\endgroup$
    – AUK1939
    Commented Oct 10, 2012 at 23:02
  • $\begingroup$ @aukie - so I cannot read either; sorry. But the proof of Corallary 1.2 still says $F_{1}(x_0) \ge F_{2}(x_0)$ $\endgroup$
    – Henry
    Commented Oct 10, 2012 at 23:07
  • 1
    $\begingroup$ Answer changed radically. $\endgroup$
    – Henry
    Commented Oct 10, 2012 at 23:40

I believe this to be the correct proof. It is a direct proof as opposed to the incorrect proof by contradiction given in the book.

$Q_1(F_2(x_0))=Q_2(F_2(x_0)) \leq x_0$ (by part i)


$F_1(x_0) \geq F_2(x_0)$ (by part iii)


$Q_2(F_1(x_0))=Q_1(F_1(x_0)) \leq x_0$ by (by part i)


$F_2(x_0) \geq F_1(x_0)$ (by part iii)

and ofcourse the only way both these conclusions can be true is if $F_1(x_0)=F_2(x_0)$.


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