Rho and Interest-Rate Sensitivity

What Rho Measures

Rho measures how an option price changes when the risk-free interest rate changes, with the other model inputs held fixed. In the Black–Scholes model, the rate affects both the risk-neutral growth rate of the stock and the present value of the strike. Rho is the net effect of those changes, not simply the derivative of the discount factor alone.

Use a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ and an equivalent risk-neutral measure $\mathbb{Q}$. Under $\mathbb{Q}$, let $B_t$ be Brownian motion and suppose the non-dividend-paying stock satisfies $dS_t=rS_t\,dt+\sigma S_t\,dB_t$, with constant $r$, constant $\sigma>0$, and $S_t>0$. The Ito integral here has progressively measurable integrand $\sigma S_t$, which satisfies $\mathbb{E}^{\mathbb{Q}}\int_0^T(\sigma S_t)^2\,dt<\infty$ for finite $T$: the geometric Brownian motion has finite second moments on every finite time interval. The measure $\mathbb{P}$ is the physical probability measure; pricing below uses $\mathbb{Q}$, not $\mathbb{P}$.

At time $t$, write $\tau=T-t$ for the time remaining, $S$ for the current stock price, $K>0$ for the strike, and $\sigma>0$ for volatility. For $\tau>0$, define

d_1=\frac{\ln(S/K)+(r+\tfrac12\sigma^2)\tau}{\sigma\sqrt{\tau}}, \qquad d_2=d_1-\sigma\sqrt{\tau}.

Let $\Phi$ and $\phi$ denote the standard normal cumulative distribution function and density. The Black–Scholes prices for European call and put options, with no dividends, are

C=S\Phi(d_1)-Ke^{-r\tau}\Phi(d_2), \qquad P=Ke^{-r\tau}\Phi(-d_2)-S\Phi(-d_1).

Definition: Call rho and put rho are the partial derivatives $\rho_C=\dfrac{\partial C}{\partial r}$ and $\rho_P=\dfrac{\partial P}{\partial r}$, evaluated with $S,K,\tau,\sigma$ fixed. Rho is the price change per unit change in the continuously compounded rate; a one-percentage-point rate change is $\Delta r=0.01$, not $1$.

Differentiating the Pricing Formulas

The key identity behind the calculation is $S\phi(d_1)=Ke^{-r\tau}\phi(d_2)$. To verify it, use $d_1=d_2+\sigma\sqrt{\tau}$ and the normal density formula:

\frac{\phi(d_1)}{\phi(d_2)} =\exp\!\left(-\frac{d_1^2-d_2^2}{2}\right) =\exp\!\left(-\sigma\sqrt{\tau}\,d_2-\frac{\sigma^2\tau}{2}\right) =\exp\!\left(-\ln(S/K)-r\tau\right) =\frac{Ke^{-r\tau}}{S}.

In the third equality, $\sigma\sqrt{\tau}\,d_2=\ln(S/K)+(r-\tfrac12\sigma^2)\tau$, which follows directly from the definition of $d_2$. Multiplying by $S\phi(d_2)$ gives the identity.

Theorem: For $S,K,\sigma,\tau>0$ and finite $r$, the Black–Scholes call and put rhos are $$ \rho_C=K\tau e^{-r\tau}\Phi(d_2), \qquad \rho_P=-K\tau e^{-r\tau}\Phi(-d_2). $$

Proof. Since $\partial d_1/\partial r=\partial d_2/\partial r=\sqrt{\tau}/\sigma$, differentiate the call formula term by term, including the derivative of the discount factor:

\begin{aligned} \frac{\partial C}{\partial r} &=S\phi(d_1)\frac{\sqrt{\tau}}{\sigma} +K\tau e^{-r\tau}\Phi(d_2) -Ke^{-r\tau}\phi(d_2)\frac{\sqrt{\tau}}{\sigma}\\ &=K\tau e^{-r\tau}\Phi(d_2). \end{aligned}

The first and third terms cancel by the density identity. For the put, differentiate both the discounted strike term and the term involving $\Phi(-d_2)$:

\begin{aligned} \frac{\partial P}{\partial r} &=-K\tau e^{-r\tau}\Phi(-d_2) -Ke^{-r\tau}\phi(d_2)\frac{\sqrt{\tau}}{\sigma} +S\phi(d_1)\frac{\sqrt{\tau}}{\sigma}\\ &=-K\tau e^{-r\tau}\Phi(-d_2). \end{aligned}

Here $\phi(-d_2)=\phi(d_2)$, and the last two terms cancel by the same identity. This proves both formulas. $\square$

The normal CDF satisfies $0<\Phi(x)<1$ for every finite real $x$. Thus, for positive strike and positive time to maturity, the call rho is strictly positive and the put rho is strictly negative. The sign is not a statement about every possible derivative of a financial claim; it applies to these standard European options under the stated no-dividend Black–Scholes assumptions.

Proposition: For $\tau>0$, $0<\rho_C<K\tau e^{-r\tau}$ and $-K\tau e^{-r\tau}<\rho_P<0$. At expiry, the option payoff does not depend on $r$, so its rate sensitivity is $0$.

Proof. For positive finite $S,K,\sigma,\tau$, the value $d_2$ is finite, and both $\Phi(d_2)$ and $\Phi(-d_2)$ lie strictly between zero and one. Multiplying these inequalities by the positive quantity $K\tau e^{-r\tau}$ gives the stated bounds and signs using the theorem. At expiry, $\tau=0$, the call and put values are respectively $\max(S-K,0)$ and $\max(K-S,0)$; neither payoff contains $r$, so the derivative with respect to $r$ is zero. $\square$

Worked Rho Calculations

Worked Example: Equal Call and Put Rho Magnitudes

Take $K=100$, $\tau=1$, $r=0$, $\sigma=0.2$, and $S=100e^{0.02}$. Calculate both rhos.

First, $\ln(S/K)=\ln(e^{0.02})=0.02$. Thus

d_2=\frac{0.02+(0-\tfrac12(0.2)^2)(1)}{0.2} =\frac{0.02-0.02}{0.2}=0.

Also $e^{-r\tau}=e^0=1$, so substitution into the rho formulas gives

\rho_C=100(1)(1)\Phi(0)=50, \qquad \rho_P=-100(1)(1)\Phi(0)=-50,

because $\Phi(0)=\tfrac12$. Check: the call and put have opposite sensitivities, and their magnitudes agree here because $\Phi(d_2)=\Phi(-d_2)=\tfrac12$. The calculation is for a one-unit change in $r$; it does not mean a one-percentage-point change moves the call price by $50$.

Worked Example: Approximation for a One-Percentage-Point Rate Shock

Use the same inputs as in the previous example. Estimate the effect of increasing the rate from $0$ to $0.01$ on the call and put prices.

For a small rate change, the first-order approximation is $\Delta C\approx\rho_C\Delta r$, and likewise $\Delta P\approx\rho_P\Delta r$. With $\Delta r=0.01$, the estimates are

\Delta C\approx 50(0.01)=0.5, \qquad \Delta P\approx -50(0.01)=-0.5.

These are local approximations, not exact finite-change identities. To write the exact repricing, at the new rate $r=0.01$ the inputs give $d_1=(0.02+0.01+0.02)/0.2=0.25$ and $d_2=0.25-0.2=0.05$. The exact call change is therefore

100e^{0.02}\bigl[\Phi(0.25)-\Phi(0.2)\bigr] -100\bigl[e^{-0.01}\Phi(0.05)-\Phi(0)\bigr].

The exact put change is

100\bigl[e^{-0.01}\Phi(-0.05)-\Phi(0)\bigr] -100e^{0.02}\bigl[\Phi(-0.25)-\Phi(-0.2)\bigr].

Check: the approximate changes have opposite signs, as the rho formulas require. The exact expressions use the new $d_1,d_2$ and the new discount factor; replacing them by their values at the original rate would incorrectly treat a finite change as if the local derivative were constant.

Worked Example: Rho Away from the At-the-Money Case

Let $S=K=100$, $r=0$, $\sigma=0.2$, and $\tau=1$. Find call and put rho in terms of the standard normal CDF, and approximate their values.

Here $\ln(S/K)=0$, so

d_2=\frac{0+(0-\tfrac12(0.2)^2)(1)}{0.2} =\frac{-0.02}{0.2}=-0.1.

Consequently $\rho_C=100\Phi(-0.1)$ and $\rho_P=-100\Phi(0.1)$. Using $\Phi(-0.1)\approx0.4602$ and $\Phi(0.1)\approx0.5398$, the rhos are approximately $46.02$ and $-53.98$. Check: these values have the required signs. Their magnitudes differ because $d_2\ne0$; the identity $\Phi(-0.1)+\Phi(0.1)=1$ also gives $46.02+53.98=100$ before rounding.

Using Rho Carefully

For a rate shock $\Delta r$, the first-order Taylor approximation is

C(r+\Delta r)\approx C(r)+\rho_C(r)\Delta r, \qquad P(r+\Delta r)\approx P(r)+\rho_P(r)\Delta r.

The approximation is most reliable for small $|\Delta r|$; for a larger shift, recompute $d_1,d_2$ and the discounted strike at the new rate. Rho is a partial derivative: it holds $S,K,\tau,\sigma$ fixed. In particular, it does not account for a simultaneous change in volatility, stock price, or time to maturity.

Because the rate is continuously compounded, a one-percentage-point increase means adding $0.01$ to $r$. If rho is quoted in currency units per one-unit rate change, multiply it by $0.01$ to estimate the price change for that shock. A quoted rho may instead be scaled by convention, so check the units before applying it.

The opposite signs have a useful economic interpretation. A higher rate reduces the present value $Ke^{-r\tau}$ of the fixed strike. That supports the call and reduces the put value. In risk-neutral pricing the rate also changes the distribution of the future stock price, which is why the derivative calculation includes terms involving $d_1$ and $d_2$. Those terms cancel in the final formulas; omitting them during differentiation would give an incomplete argument.

Common Pitfall

A frequent error is to differentiate only the discounted strike in the call formula and conclude that call rho is $-K\tau e^{-r\tau}\Phi(d_2)$. This has the wrong sign and misses the rate dependence of both $d_1$ and $d_2$. For the call, the full derivative contains $S\phi(d_1)\sqrt{\tau}/\sigma$ and $-Ke^{-r\tau}\phi(d_2)\sqrt{\tau}/\sigma$; the density identity shows that these cancel, leaving $+K\tau e^{-r\tau}\Phi(d_2)$. The corresponding put derivative leaves $-K\tau e^{-r\tau}\Phi(-d_2)$. Differentiating the complete pricing formula, rather than only its explicit exponential factor, is essential.

Check Your Understanding

In each question involving a rate derivative, hold $S,K,\tau,\sigma$ fixed and assume $\tau>0$.

For a call with $K=80$, $r=0$, and $\tau=2$, what is the largest possible upper bound on its rho from the proposition?
Explain why the call rho is positive even though $d_2$ can be negative.
If a put has rho $-24$, estimate its price change for a rate decrease of $0.005$.
State which quantities must be recomputed to obtain an exact price after a finite rate change.
Why is the derivative of the expiry payoff with respect to $r$ zero?