This document discusses some problems in mathematical finance. It presents a proposition showing that if Expected Shortfall (ES) is used to monitor risk, the "old" Value at Risk (VaR) model test may not be sufficient to prove the soundness of the ES number. Specifically, it is possible to construct two profit and loss distributions that have very close VaR at all confidence levels but arbitrarily large differences in ES at a predefined confidence level. The document also mentions some open problems and includes references.

1. Some problems in math finance
Hong Xu2
2
EXGD Estimate INC
Richmond Hill, ON, Canada xxx-xxx
E-mail address: wuchang69@yahoo.com
Abstract
Discussion of some math finance problems.
1 Introduction
2 VaR and ES
The following proposition shows that if ES (short for Expected Shortfall) is used to monitor risk, then the
“old” VaR model test may be not sufficient to prove the soundness of ES number.
Proposition 2.1 Let f be a non-negative continuous function such that f(x) > 0 if x < 0;
∞
−∞
f(t)dt = 1.
For any α ∈ (0, 1), let q(α) be the quantile function of F(x) =
x
−∞
f(t)dt, namely F(q(α)) = α. Given
any α ∈ (0, 1), we have ∀ > 0 ( arbitrarily small), ∀M > 0 (M arbitrarily large), there exists an
integrable function g(x) ≥ 0, ∀x ∈ R, such that
∞
−∞
g(t)dt = 1; G(x) :=
x
−∞
g(t)dt = F(x), ∀x > q(α);
supx∈(−∞,q(α)) |G(x) − F(x)| < ; but 1
α
q(α)
−∞
t(g(t) − f(t))dt < −M.
Proof
We will construct a function p(x) such that p(x) = 0, ∀x ≥ q(α) and g = f +p is the function satisfying
all the above properties. Because f is continuous and f(x) > 0, ∀x < 0, there exists an interval [c, d] such
that d < min(0, 2q(α)), m0 := minx∈[c,d] f(x) > 0. Let h2 ∈ (0, m0) such that h2(d − c) < 2 . Let (h1, a, b)
be a triple such that a < b < c, C1 := (b − a)h1 = (d − c)h2. Note that we will still be able to impose
other conditions on (h1, a, b). Let us define p(x) := h1I[a,b](x) − h2I[c,d](x) and g(x) = f(x) + p(x). It is
obvious that
∞
−∞
g(x)dx =
∞
−∞
f(x)dx = 1; supx∈(−∞,q(α)] |G(x) − F(x)| ≤ C1 < 2 . Let us look at
1
α
q(α)
−∞
t(g(t) − f(t))dt
=
1
α
h1
b
a
tdt − h2
d
c
tdt
=
1
α
h1
2
(b2
− a2
) −
h2
2
(d2
− c2
)
=
C1
2α
(b + a − d − c)
(1)
So the last condition to impose on (h1, a, b) is that C1
2α (b + a − d − c) < −M, which is possible because
while maintaining b − a as constant, one can make b + a as negative as possible.
Remark 2.2 1. The construction of g essentially says that under some mild conditions, one may
find two PnL distributions that have very close VaR at all confidence levels, but at any predefined
confidence level its Expected Shortfall may have arbitrarily large difference.
2. It is technically possible to make p(x) ∈ C∞
(R).
3. Requiring that f is continuous and f(x) > 0 when x < 0 is to simplify the proof. The essential
requirement is that for any preset q(α), there exist an interval [c, d] such that d < q(α) and
m0 := minx∈[c,d] f(x) > 0.
1

2. mathFin01.tex 2
3 Open problems
References