# 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.

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.

• Hi! For the p and 1-p, are you referring to the risk-neutral probability? If so, did you consider the interest rate effect on it? Feb 18, 2021 at 15:18
• Yes it is. Interest rate has no effect on it. Feb 18, 2021 at 16:19
• sorry i don't understand. Shouldn't the risk-neutral prob be (e^rt - d)/(u - d)? Feb 18, 2021 at 16:28
• yes it is, you can see easily it is the same. Feb 18, 2021 at 16:33

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 $$1-p$$).

• Shouldn't $U,D$ change accordingly when $r$ is changed, i.e. $e^{\Delta r}U$ and $e^{\Delta r}D$? Feb 23, 2021 at 1:32
• no I don’t think so, due to the distributive property I would just apply the factor once and I chose to apply it at the probability. Does that make sense? Feb 23, 2021 at 5:19
• In your notation, think about the case $\rho \to \infty$. Feb 23, 2021 at 5:46
• My math checks out. The case $\rho \to \infty$ needs to be analysed quite carefully as with $e^{r\Delta t+ \rho \Delta t} > U$, we have arbitrage opportunities, no? Feb 23, 2021 at 7:03
• Exactly. That's why I assume $U,D$ change together with $r$ while spot price is unchanged. (or we can assume spot price change together with $r$ but $U,D$ unchanged). Feb 23, 2021 at 7:20