Interest Rate Sensitivity of Call and Put Options

# A few questions about signs of the Greek letters


# A few questions about signs of the Greek letters












Rho is the partial derivative of the value of call option, $C$, w.r.t the riskfree interest rate $r$: $$\rho \equiv \frac{\partial C}{\partial r}$$

In the standard B-S formula this term is positive, but what's the intuition? I understand that two forces are at hand: one is that as $r$ increases future exercise price $K$ values less, so $C$ becomes more valuable. But on the other hand, increased $r$ also diminishes present value for future payoffs from the option, so $C$ becomes less valuable.

Another question I'd like to know is how general could this result be for arbitrary distributions? Since B-S formula is derived under certain assumptions about distributions of the price of the underlying asset (such as geometric Brownian motion with constant drift and volatility, etc).

Edit: @Quant, I agree with you on BSM, for which the particular distribution of the underlying allows one to perfectly duplicate the distribution of the call by shorting the riskfree bond and longing the underlying appropriately. But for arbitrary distributions, this may not be possible, so $C$ need not increase as $r_f$ increases. Consider a two period example: $S_0=1$, $S_1=1, 2, 4$ each with some strict positive probabilities (say $1/3, 1/3, 1/3$), strike price $K=3$ and gross riskfree rate $r_f=2$. In this case, no combination of $S$ and the bond would perfectly duplicate the call, and any $C \in (0,1/6)$ would be permissible. Hence an increase of $r_f$ need not increase the value of $C$.

It seems that only in binomial tree model (BSM being BTM in the limit) can we pin down the value of $C$ by no-arbitrage criterion alone.

## Answer by Brian B (score 2, accepted)

https://quant.stackexchange.com/a/3221

The main reason that rho term is positive is that we are using arbitrage-free pricing theory. In particular, regardless of model, the value of a forward contract (for an asset paying no dividends) is

$$ F_T = S_0 e^{rT} $$

Therefore, in whatever option pricing model you choose, the center of its forward distribution for the asset price $S_T$ at time $T$ increases with increasing $r$.

The same increase does not of course apply to the strike $K$, so for a call this increase in distributional center results in higher option prices under all the common option pricing models, Black-Scholes included.

## Answer by Ram Ahluwalia (score 4)

https://quant.stackexchange.com/a/3199

Here's the relationship of rho on calls and puts.

When you buy call options instead of the the underlying, you are effectively buying an indirect leveraged position in the underlying. A simple way to see this is buy re-arranging the terms of the Put-Call parity equation solving for the call price. The value of the call is equal to a synthetic position consisting of: i) long the underlying, ii) short a zero-coupon bond that matures at T with strike price K, and iii) long a put.

When interest rates are higher, buying the call instead of financing a direct leveraged position in the underlying is more attractive. Also, the investor by using call options saves more money by not paying for the underlying until a later date.

However, for put options, the higher interest rates are are disadvantageous. In this case, investors lose more interest while waiting to sell the underlying when using puts. Said another way, the opportunity cost of waiting is higher when interest rates are higher.

Therefore the sign of rho is positive for calls, and negative for puts, although the effect is very weak especially compared to the impact of volatility on option prices.

These arguments hold true for arbitrary distributions (so long as interest rates are positive).