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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.

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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.

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  • $\begingroup$ How do you obtain such simple expressions... I did the calculation twice and each time it gave me a horrible mess. $\endgroup$ – Vim Mar 25 '17 at 13:10
  • $\begingroup$ See the reference I provided above. For a specific derivation, you may need some careful algebraic manipulations $\endgroup$ – Gordon Mar 25 '17 at 13:12
  • $\begingroup$ At time $t$, $S_t$ is a known quantity, we do not treat it as a function of $t$. At time $0$, $S_t$ depends on $r$, but is also a random number. $\endgroup$ – Gordon Mar 25 '17 at 13:15
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    $\begingroup$ I added some details for how to convert $\phi(d_2)$ to $\phi(d_1)$. $\endgroup$ – Gordon Mar 25 '17 at 13:48
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    $\begingroup$ @Gordon: I think the sign for the call is wrong..? $\endgroup$ – LocalVolatility Mar 25 '17 at 15:47

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