0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
0
$\begingroup$

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 $1-p$).

$\endgroup$
7
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.