Vanilla options as controls for estimating the conditional expectation of a European derivative payoff

Suppose Assumption 1 and Assumption 2 are satisfied. The conditional expectation

for all 𝑥 in 𝑆 uniformly as M → ∞ .

Proof

By Assumption 1 and Assumption 2, the conditional expectations of indicators

are continuous for all 𝑥 in 𝑆 which is compact. The conditional expectations of indicators are monotonic in 𝑀 and converge to 0 as M → ∞ for each 𝑥 in 𝑆. Hence, by Dini’s theorem, E ⁡ [ 1 { S T > M } | X = x ] → 0 for all 𝑥 in 𝑆 uniformly as M → ∞ . ∎

Suppose Assumption 1 and Assumption 2 are satisfied. For any non-negative real number 𝑎, we have

for all 𝑥 in 𝑆 uniformly as ϵ ↘ 0 .

Proof

The proof is similar to the proof of Lemma A.1. We consider a decreasing sequence of positive real numbers ϵ n such that ϵ n → 0 as n → ∞ and the sequence of conditional expectations of indicators

By Assumption 1 and Assumption 2, the conditional expectations of indicators are continuous for all 𝑥 in 𝑆 which is compact. The conditional expectations of indicators are monotonic in 𝑛 and converge to E ⁡ [ 1 { S T > a } | X = x ] as n → ∞ for each 𝑥 in 𝑆. Again, by Assumption 2, E ⁡ [ 1 { S T > a } | X = x ] is continuous for all 𝑥 in 𝑆. Hence, by Dini’s theorem, E ⁡ [ 1 { S T ≥ a + ϵ n } | X = x ] → E ⁡ [ 1 { S T > a } | X = x ] for all 𝑥 in 𝑆 uniformly as n → ∞ . ∎

Suppose Assumption 1 and Assumption 2 are satisfied. The conditional expectation

for all 𝑥 in 𝑆 uniformly as M → ∞ .

Proof

By Assumption 1 and Assumption 2, E ⁡ [ V T 4 | X = x ] exists and is continuous for all 𝑥 in 𝑆 which is compact. This implies that the image of E ⁡ [ V T 4 | X = x ] over 𝑆 is a compact set, which is bounded. Hence there exists a K > 0 such that E ⁡ [ V T 4 | X = x ] ≤ K for all 𝑥 in 𝑆.

By Cauchy–Schwarz inequality,

for all 𝑥 in 𝑆 uniformly as M → ∞ due to Lemma A.1. ∎

Suppose a real-valued function f ⁢ ( x ) is continuous on an interval [ a , b ] . There exists a sequence of functions ∑ i = 1 n p i ⁢ 1 { x ≥ q i } such that

uniformly on [ a , b ] as n → ∞ .

Proof

Note that [ a , b ] is compact since it is closed and bounded in ℝ. The function 𝑓 is continuous on this compact set [ a , b ] , so it is uniformly continuous on [ a , b ] . Therefore, for any ϵ > 0 , we may choose a δ > 0 such that | f ⁢ ( y ) − f ⁢ ( x ) | < ϵ for any x , y ∈ [ a , b ] satisfying | y − x | < δ .

Now consider a partition { q i } i = 1 , 2 , … , N ⊂ [ a , b ] such that

Define f N ⁢ ( x ) = f ⁢ ( b ) ⁢ 1 { x ≥ q N } + ∑ i = 1 N − 1 f ⁢ ( q i ) ⁢ 1 { q i ≤ x < q i + 1 } . For each 𝑥 in [ a , b ] , we have

Hence f N → f uniformly on [ a , b ] . Since 1 { q i ≤ x < q i + 1 } = 1 { x ≥ q i } − 1 { x ≥ q i + 1 } , the function f N may be written as a sum of step functions ∑ i = 1 N p i ⁢ 1 { x ≥ q i } . Finally, by the definition of f N , we have ∑ i = 1 N p i = f N ⁢ ( b ) = f ⁢ ( b ) . ∎

Suppose Assumptions 1–3 are satisfied. In addition, suppose the function 𝑔 stated in Assumption 3 is right-continuous everywhere. There exists a sequence of functions ∑ i = 1 n p i ⁢ 1 { S T ≥ q i } such that

for all 𝑥 in 𝑆 uniformly as n → ∞ .

Proof

For any ϵ > 0 , by Lemma A.1 and Lemma A.3, there exists an M 1 > 0 and M 2 > 0 such that

for all 𝑥 in 𝑆. Taking M = max ⁡ ( M 1 , M 2 ) , we have

for all 𝑥 in 𝑆.

Next, by Assumption 3, we may write V T = g ⁢ ( S T ) and there exists h i ∈ [ 0 , ∞ ) , i = 1 , 2 , … , N J , such that

• for each h i , both lim x → h i − g ⁢ ( x ) and lim x → h i + g ⁢ ( x ) exist and lim x → h i − g ⁢ ( x ) ≠ lim x → h i + g ⁢ ( x ) ,

• 𝑔 is continuous on [ 0 , ∞ ) \ { h i } and is left-continuous or right-continuous at each h i , i = 1 , 2 , … , N J .

Denote

and define g ̃ as

which is clearly continuous for any 𝑥 not equal to any h i . For each h i , we have

Hence lim x → h i g ̃ ⁢ ( x ) = g ̃ ⁢ ( h i ) and g ̃ is continuous on [ 0 , ∞ ) . Applying Lemma A.4 to g ̃ on [ 0 , M ] , we have that there exists a function ∑ i = 1 N ′ p i ′ ⁢ 1 { x ≥ q i ′ } such that

for all 𝑥 in [ 0 , M ] . The last inequality implies

for all 𝑥 in [ 0 , M ] .

Now consider

The first term on the right-hand side of (A.1) is small since

for all 𝑥 in 𝑆.

Next, we break down the second term on the right-hand side of (A.1) and show that each part is small. Considering the inequality ( a + b + c ) 2 ≤ 3 ⁢ ( a 2 + b 2 + c 2 ) for any real numbers 𝑎, 𝑏 and 𝑐, we have

Then, by Cauchy–Schwarz inequality,

and so

for all 𝑥 in 𝑆.

Then note that

for all 𝑥 in 𝑆, where the second equality is due to the fact that q i ≤ M , i = 1 , 2 , … , N ′ .

Putting everything together, from (A.1),

for all 𝑥 in 𝑆. This proves the uniform convergence.

Finally, we may rewrite ∑ i = 1 N J J i ⁢ 1 { S T ≥ h i } + ∑ i = 1 N ′ p i ′ ⁢ 1 { S T ≥ q i ′ } as a function in the form of ∑ i = 1 N p i ⁢ 1 { S T ≥ q i } by suitably defining 𝑁, p i and q i , i = 1 , 2 , … , N . ∎

Suppose Assumptions 1–3 are satisfied. There exists a sequence of functions ∑ i = 1 n p i ⁢ 1 { S T ≥ q i } such that

for all 𝑥 in 𝑆 uniformly as n → ∞ .

Proof

By Assumption 3, we may write V T = g ⁢ ( S T ) and there exists h i ∈ [ 0 , ∞ ) , i = 1 , 2 , … , N J , such that

• for each h i , both lim x → h i − g ⁢ ( x ) and lim x → h i + g ⁢ ( x ) exist and lim x → h i − g ⁢ ( x ) ≠ lim x → h i + g ⁢ ( x ) ,

• 𝑔 is continuous on [ 0 , ∞ ) \ { h i } and is left-continuous or right-continuous at each h i , i = 1 , 2 , … , N J .

If 𝑙 is an empty set, 𝑔 is right-continuous at any h i , i = 1 , 2 , … , N J , and we conclude the proof by invoking Lemma A.5.

Suppose 𝑙 is non-empty. We consider an auxiliary function g r defined by removing all h i where 𝑔 is left-continuous. Denote

Define g r as

Then g r is clearly continuous for any 𝑥 not equal to any h i . For each i ∈ l , we have

and, by the definition of 𝑙,

Hence lim x → h i g r ⁢ ( x ) = g r ⁢ ( h i ) . That is, g r is continuous at h i for each i ∈ l . This implies that g r is right-continuous at h i for each i = { 1 , 2 , … , N J } \ l and continuous elsewhere.

Consider any ϵ > 0 . By Lemma A.5, there exists a function ∑ i = 1 N p i ⁢ 1 { S T ≥ q i } such that, for all 𝑥 in 𝑆,

implying

From Lemma A.2, for each i ∈ l , there exists a δ i > 0 such that

for all 𝑥 in 𝑆.

Considering the Cauchy–Schwarz inequality, we have

for all 𝑥 in 𝑆, where N L is the number of elements in the set 𝑙. This concludes the proof of the uniform convergence.

Finally, we may rewrite ∑ i ∈ l J i ⁢ 1 { S T ≥ h i + δ i } + ∑ i = 1 N p i ⁢ 1 { S T ≥ q i } as a function in the form of ∑ i = 1 N ′ p i ′ ⁢ 1 { S T ≥ q i ′ } by suitably defining N ′ , p i ′ and q i ′ , i = 1 , 2 , … , N ′ . ∎

Suppose Assumption 1 and Assumption 2 are satisfied. Then, for any K > 0 ,

uniformly as ϵ → 0 .

Proof

Consider a decreasing sequence of positive real numbers ϵ n such that ϵ n → 0 . Denote

First, note that

tends to 0 as n → ∞ and Δ n ⁢ ( x ) is bounded by 1 for all 𝑛 and all 𝑥. Hence, by (conditional) bounded convergence theorem, for each 𝑥 in 𝑆, we have f n ⁢ ( x ) → 0 . Then note that Δ n ⁢ ( x ) ≥ Δ n + 1 ⁢ ( x ) for all 𝑛 and all 𝑥, so { Δ n } is monotonic. Applying the conditional expectation operator E [ ⋅ | X = x ] to Δ n ⁢ ( S T ) , we have that { f n } is monotonic. Finally, by Assumption 1, 𝑆 is compact, and by Assumption 2, { f n } are continuous. Hence, by Dini’s theorem, the convergence of f n to 0 is uniform on 𝑆. ∎

Suppose Assumptions 1–3 are satisfied. There exists a sequence of functions ∑ i = 1 n b i ⁢ ( S T − K i ) + such that

for all 𝑥 in 𝑆 uniformly as n → ∞ .

Proof

For any ϵ > 0 , by Theorem A.6, there exists a function ∑ i = 1 N p i ⁢ 1 { S T ≥ q i } such that

for all 𝑥 in 𝑆.

By Lemma A.7, for each I { S T ≥ q i } , i = 1 , … , N , we may choose a (small) constant r i > 0 such that

for all 𝑥 in 𝑆.

Putting everything together, we have

for all 𝑥 in 𝑆, where the first inequality is due to the inequality ( a + b ) 2 ≤ 2 ⁢ ( a 2 + b 2 ) for any real numbers 𝑎 and 𝑏 and the second inequality is due to Cauchy–Schwarz inequality. This proves the uniform convergence.

Finally, we may write

in the form of

by suitably defining N ′ and the constants b i , K i , i = 1 , 2 , … , N ′ . ∎

Proof of Theorem 4.1

For any ϵ > 0 , by Theorem A.8, there exists a function ∑ i = 1 N p i ⁢ 1 { S T ≥ q i } such that

for all 𝑥 in 𝑆.

By Cauchy–Schwarz inequality, we have

for all 𝑥 in 𝑆.

Therefore, the conditional variance

for all 𝑥 in 𝑆. This proves the uniform convergence. ∎

Proof of Theorem 4.2

Denote r ⁢ ( x ) = ρ ⁢ ( V T , ∑ i = 1 n b i ⁢ ( S T − K i ) + | X = x ) . For all 𝑥 in 𝑆, we have