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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.

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I'm not shure I follow your first question: As you say, C can go up or down following an increase in $r_f$, depending on the features of the option. But $\rho$ gives you the result by the definition of $\rho$, it will be the change in $C$ relative to the change in $r_f$. –  Owe Jessen Apr 6 '12 at 15:37
    
@Owe: I was not claiming that $C$ can go up or down (it can only go up in the BS model) following an increase in $r_f$. I meant when $r_f$ increases, there are two contradicting forces on the movement of $C$. But in the BS model the latter force always prevails. I want to know why, and I want to know whether it holds true for any distributions of prices of the underlying. –  Eric Apr 7 '12 at 3:08
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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.

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I think what you meant is that under risk neutral probability, $F_T=E(S_T)=S_0{e}^{rT}$ is increasing in $r$. There's no dispute about that. But notice the ${e}^{-rT}$ term in $C={e}^{-rT}E[max(S_T-K,0)]$. So even if you can conclude that $E[max(S_T-K,0)]$ is increasing in $r$, can you really conclude ${e}^{-rT}E[max(S_T-K,0)]$ is increasing in $r$ from that? –  Eric Apr 11 '12 at 9:02
    
Well consider just the zero-vol value of an in-the-money option. Here $C=e^{-rT}(S_0 e^{rT} - K)^+$ or $S_0 - e^{-rT}K$ and so the rho is simply $T e^{-rT} K$, very positive. Basically, the increase in forward value has overwhelmed the decrease due to discounting. Pathological setting of parameters may give negative rho in some cases, but this is the gist of why rho of calls is generally positive. –  Brian B Apr 11 '12 at 17:08
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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).

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By put-call parity $C=S-K/{e}^{rT}+P$. But the value of the call is always equal to that of the leverage on the RHS. Why is buying the call more attractive? And what do you mean by "not paying for the underlying until a later date"? –  Eric Apr 6 '12 at 12:01
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The value of the call is not equal to the leverage. The equation you show indicates 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. Suppose you purchase a call at the arbitrage-free price of C. An instant later the interest rate increases. Since you are short a zero-coupon bond that matures "at a later date" the value of your bond liability diminishes because bond prices fall when rates rise. Think of "later date" as settling the zero-coupon bond at maturity. –  Quant Guy Apr 6 '12 at 12:22
    
I understand your point. But as I emphasized in the question, this is not the whole story. As $r$ increases, present value of future payoff also gets discounted more. Why is the gain from the bond always outweighs the loss from discounted payoff? –  Eric Apr 6 '12 at 14:32
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The simplest way to see this is to note that in your equation where you have solved for C, the call value payoff is at expiry. More technically, if you look at the BSM model for pricing a call option you will note that coefficient on the interest rate term in d1 is positive. You could look at the derivation of BSM to scratch that itch a bit further, but one intuition is that expected spot prices are calculated by compounding the current spot price by the risk-free rate. So if interest rates are higher, futures prices are higher as well ceteris paribus. –  Quant Guy Apr 6 '12 at 15:52
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