Skip to content
All library documents

Bounding a Convex Derivative’s Price from Its Endpoint Payoffs

Article Quant Q&A · Author: Paris K. Patsogiannis

Summary

The document derives an upper bound for the no-arbitrage price of a nonnegative convex payoff that depends on a single risky asset’s terminal value. The asset’s minimum and maximum possible values bracket its risk-neutral expected terminal value, which is fixed by the initial price and risk-free growth. Any intermediate terminal value can be expressed as a weighted average of those two endpoints; convexity then bounds the payoff by the corresponding weighted average of endpoint payoffs.

Taking the risk-neutral expectation turns that pointwise inequality into a price bound, with weights determined by the asset’s expected terminal value relative to the endpoints. The argument illustrates how convexity and no-arbitrage pricing work together without requiring the full distribution of terminal prices. It assumes finite lower and upper asset-value bounds and a risk-neutral measure satisfying the pricing relation. The displayed exercise formula appears to omit the payoff at the lower endpoint; the answer’s derivation includes both endpoint payoffs, so the formula as printed follows only if that lower-end payoff is zero.

Key ideas

  • Convexity bounds an intermediate payoff by a weighted average of its endpoint payoffs.
  • The asset’s risk-neutral expected terminal value determines the weights on the endpoints.
  • Discounting the expected payoff bound yields an upper bound on the derivative price.
  • The result uses bounded terminal asset values and the no-arbitrage pricing relation.

Tags

Full text
# Hedging Value-Financial Mathematics


# Hedging Value-Financial Mathematics












EXERCISE

> We consider a free from arbitrage financial market $(Ω,F,P,S_0,S_1)$ with $α<S_0^{1}\cdot(1+r)<β$,where $$0<α:=min_{ω \in Ω} S_1^{1}(ω), β:=max_{ω \in Ω}S_1^{1}, α<β$$ Let C be a financial derivative of the form $C:=h(S_1^{1})$ where $h\geq0$ is a convex function.Show that the hedging value $\overline π(C)$ of derivative $C$ is given by the formula $$\overline π(C)=\dfrac{h(β)}{1+r}\cdot \dfrac{(1+r)S_0^{1}-α}{β-α}+\dfrac{α}{1+r}\cdot \dfrac{β-(1+r)S_0^{1}}{β-α}$$

QUESTIONS

We have that : $$α<S_0^{1}\cdot(1+r)<β$$ and $$0<min_{ω \in Ω} S_1^{1}(ω)<S_0^{1}\cdot(1+r)<max_{ω \in Ω} S_1^{1}(ω)$$

We have also that $$α<β\Longrightarrow min_{ω \in Ω} S_1^{1}(ω)<max_{ω \in Ω} S_1^{1}(ω)$$

We have a financial market with no-arbitrage so we have the form: $$π(C)=E_Q\bigg[\dfrac{c}{1+r}\bigg]<\infty$$ for $Q\subset P$

So,I am new in financial Mathematics and i don't have the experience to understand how to proceed with this data!Can anyone help me with this?How can i use the fact that function $h \geq 0$ is a convex function.Did i miss any data from what the exercise gives me?How can i start so to estimate $\overline π(C)$

I would really appreciate any hints/thorough solution because I don't have any experience in this type of exercise.

Thanks, in advance!

## Answer by Freelunch (score 3, accepted)

https://quant.stackexchange.com/a/42649

Note that we can write $S_1(\omega)$ as a convex combination of $\alpha$ and $\beta$ with

\begin{equation} S_1(\omega) = \frac{\beta-S_1(\omega)}{\beta-\alpha} \alpha + \frac{S_1(\omega) - \alpha}{\beta-\alpha} \beta \end{equation}

Since $h$ was a convex function then by definition \begin{equation} h(S_1(\omega)) \leq \frac{\beta-S_1(\omega)}{\beta-\alpha} h(\alpha) + \frac{S_1(\omega) - \alpha}{\beta-\alpha} h(\beta) \end{equation}

Taking the $\mathbb{Q}$-expected value, while noting that $E^\mathbb{Q}[S_1] = (1+r)S_0$, \begin{equation} E^\mathbb{Q}[h(S_1)] \leq \frac{\beta-(1+r)S_0}{\beta-\alpha} h(\alpha) + \frac{(1+r)S_0 - \alpha}{\beta-\alpha} h(\beta) \end{equation} which gives you an upper bound on the price of the derivative.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.