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.

  • $\begingroup$ 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? $\endgroup$ – James Feb 18 at 15:18
  • $\begingroup$ Yes it is. Interest rate has no effect on it. $\endgroup$ – MainCom Feb 18 at 16:19
  • $\begingroup$ sorry i don't understand. Shouldn't the risk-neutral prob be (e^rt - d)/(u - d)? $\endgroup$ – James Feb 18 at 16:28
  • $\begingroup$ yes it is, you can see easily it is the same. $\endgroup$ – MainCom Feb 18 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 $1-p$).

  • $\begingroup$ Shouldn't $U,D$ change accordingly when $r$ is changed, i.e. $e^{\Delta r}U$ and $e^{\Delta r}D$? $\endgroup$ – MainCom Feb 23 at 1:32
  • $\begingroup$ 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? $\endgroup$ – Kermittfrog Feb 23 at 5:19
  • $\begingroup$ In your notation, think about the case $\rho \to \infty$. $\endgroup$ – MainCom Feb 23 at 5:46
  • $\begingroup$ 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? $\endgroup$ – Kermittfrog Feb 23 at 7:03
  • $\begingroup$ 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). $\endgroup$ – MainCom Feb 23 at 7:20

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