Black–Scholes Rho Signs for European Calls and Puts
Summary
The document derives the sensitivity of European option prices to the risk-free interest rate in the Black–Scholes model. Differentiating the call price and simplifying with the relationship between the normal density values at d1 and d2 produces call rho as discounted strike, time to expiry, and the cumulative normal probability at d2. These terms are nonnegative under the standard model assumptions, so the call price increases with the rate.
For a European put, the corresponding rho is negative, making its price decrease as the rate rises. The derivation addresses the question of proving these directions mathematically, rather than relying on plotted prices. The result is specific to the standard Black–Scholes setup with other inputs held fixed; the note does not discuss dividends, alternative rate conventions, or more general pricing models.
Key ideas
- In the standard Black–Scholes model, a European call has nonnegative sensitivity to the risk-free rate.
- Call rho simplifies to a product of discounted strike, time to expiry, and a cumulative normal probability.
- A European put has nonpositive rho under the same model assumptions.
- The signs follow from the analytic derivatives rather than numerical plots.
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Full text
# From BS formula how to show Euro-call values go up and Euro-put values go down with interest rates?
# From BS formula how to show Euro-call values go up and Euro-put values go down with interest rates?
The BS formula gives, as quoted from Wikepdia:
> $${\displaystyle {\begin{aligned}C(S_{t},t)&=N(d_{1})S_{t}-N(d_{2})Ke^{-r(T-t)}\\d_{1}&={\frac {1}{\sigma {\sqrt {T-t}}}}\left[\ln \left({\frac {S_{t}}{K}}\right)+\left(r+{\frac {\sigma ^{2}}{2}}\right)(T-t)\right]\\d_{2}&=d_{1}-\sigma {\sqrt {T-t}}\\\end{aligned}}}$$
in which all notations are standard (I don't think any notational ambiguity exists; otherwise please refer to this article directly.)
I want to show that when everything else is fixed, $C$ will go up with $r$. Or alternatively, that $\partial C/\partial r\ge 0$. But it turns out the sign of the partial derivative is not that obvious.
To convince myself somehow, I drew a couple of $C$-$r$ plots on my computer using various sets of parameters and found out the curve always went upwards as expected. But I still want a rigorous proof in the mathematical sense. So am I missing something here? Is there anybody who can help? Thanks.
## Answer by Gordon (score 2, accepted)
https://quant.stackexchange.com/a/33231
For a call, the rho is given by \begin{align*} \frac{\partial C_t}{\partial r} &= \frac{\sqrt{T-t}}{\sigma}\left(S_t\phi(d_1) -Ke^{-r(T-t)}\phi(d_2)\right)+K\,(T-t)e^{-r(T-t)}N(d_2)\\ &=\frac{\sqrt{T-t}}{\sigma}\left(S_t\phi(d_1) -Ke^{-r(T-t)}\phi(d_1-\sigma\sqrt{T-t})\right)+K\,(T-t)e^{-r(T-t)}N(d_2)\\ &=\frac{\sqrt{T-t}}{\sigma}\left(S_t\phi(d_1) -Ke^{-r(T-t)}\phi(d_1)e^{d_1\sigma \sqrt{T-t}-\frac{\sigma^2}{2}(T-t)}\right)+K\,(T-t)e^{-r(T-t)}N(d_2)\\ &=K\,(T-t)e^{-r(T-t)}N(d_2). \end{align*} where \begin{align*} d_1 &= \frac{\ln \frac{S_t}{K}+\left(r+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}, \ d_2 = d_1-\sigma\sqrt{T-t}. \end{align*} Similarly, for a put, the rho is given by \begin{align*} \frac{\partial P_t}{\partial r} =- K\,(T-t)e^{-r(T-t)}N(-d_2). \end{align*} It is now obvious that the call is an increasing function of the interest rate, while the put is a decreasing function. See https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_model for a reference.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.