Why European Call Values Rise with Rates in a Binomial Model
Summary
The document sketches why a European call's value increases when the risk-free rate rises in a binomial setting. In the one-step illustration, increasing the rate raises the risk-neutral expected stock value while discounting the payoff. Rewriting the expression shows this is equivalent to lowering the strike in the payoff calculation, which cannot reduce a call's value.
A second derivation writes the risk-neutral up and down probabilities explicitly as functions of the rate. As the rate rises, the up-state probability increases and the down-state probability decreases; comparing the formulas gives the direction of each change. These arguments offer intuition for the positive rate sensitivity also known in continuous-time pricing. The document's derivations are sketches and do not fully develop the requested two-step model with distinct probabilities at each step, nor enumerate all assumptions needed for a valid tree.
Key ideas
- In a binomial model, a higher rate changes both discounting and the risk-neutral probabilities.
- The one-step expression can be rewritten as a call payoff with a lower effective strike.
- The risk-neutral probability assigned to the up state rises with the interest rate in the shown setup.
- The derivations provide intuition but do not fully prove the requested two-step case.
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# Answer by MainCom (score 1)
# Prove the Euro call option value has positive relationship with the risk-free rate under discrete time model (Binomial tree model)
Could anyone show me how to prove that the European call option value has a positive relationship with the risk-free rate in a two-step binomial model with strike price K and different risk neutral probability q between each step? I know that in continuous model, the rho(call) is positive which shows the positive relationship but I am only familiar with the equations and don't really know how to prove that. Thank you.
## Answer by MainCom (score 1)
https://quant.stackexchange.com/a/61240
Assume that the price of the next step are $u,d$, with probability $p,1-p$. The the discounted payoff is $e^{-r}(p(u - K)^+ + (1-p)(d-K)^+)$. Now suppose the interest rate $r$ is increased by $\Delta r$. Then the new discounted payoff would be $e^{-r - \Delta r}(p(e^{\Delta r}u - K)^+ + (1-p)(e^{\Delta r}d-K)^+) = e^{-r}(p(u - e^{-\Delta r}K)^+ + (1-p)(d-e^{-\Delta r}K)^+)$.
Since $e^{-\Delta r}K < K$, the new discounted payoff would be greater than the original one.
## Answer by Kermittfrog (score 0)
https://quant.stackexchange.com/a/61307
I was not able to reproduce @MainCom's example on a spreadsheet, hence I gave it a shot myself:
$$ \begin{align} C(r)&:=e^{-r\Delta t}\left(\frac{e^{r\Delta t}-D}{U-D}\left(U-K\right)^++\frac{U-e^{r\Delta t}}{U-D}\left(D-K\right)^+\right)\\ \Rightarrow C(r+\rho)&=e^{-r\Delta t-\rho\Delta t}\left(\frac{e^{r\Delta t + \rho\Delta t}-D}{U-D}\left(U-K\right)^++\frac{U-e^{r\Delta t + \rho\Delta t}}{U-D}\left(D-K\right)^+\right)\\ &=e^{-r\Delta t}\left(\frac{e^{r\Delta t}-De^{-\rho\Delta t}}{U-D}\left(U-K\right)^++\frac{Ue^{-\rho\Delta t}-e^{r\Delta t }}{U-D}\left(D-K\right)^+\right)\\ \end{align} $$ Let us now compare the first probability term $p(r):=\frac{e^{r\Delta t}-D}{U-D}$. As the denominator is independent of $r$, we simply compare
$$ \left(e^{r\Delta t}-De^{-\rho \Delta t}\right) \quad -\quad \left( e^{r\Delta t}-D\right) $$ Clearly,this is $D(1-e^{-\rho\Delta t})>0$, i.e. this term is increasing (and decreasing for $1-p$).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.