Why Option Rho Changes with Moneyness

# How does Rho behaves with moneyness of option?


# How does Rho behaves with moneyness of option?












I was trying to find the relationship between nature of Rho and moneyness of the option.

After finding certain values I found that Rho Value keep increases as the option gets further in the money. But I can't think of a reason to justify this behaviour.

Can you please explain this behaviour of Rho?

Thanks in advance.

## Answer by Daneel Olivaw (score 4)

https://quant.stackexchange.com/a/39805

Note: I have amended my previous answer as it was misleading, as pointed out by @byouness.

In terms of sensitivity, there are two "channels":

- The discounting channel, which as shown below has a negative effect; and

- The asset's drift channel under the risk-neutral measure (which is equal to the rate), which has positive effect for call options only.

Indeed, because under Black-Scholes $dS_t = \color{blue}{rS_t}dt+\sigma S_tdW_t^Q$, the greater the current asset value, the greater we can expect it to be at the option's maturity $T$:

$$ E^Q_t(S_T)=S_te^{r(T-t)}$$

Therefore, for a European call option, sensitivity to rates under the asset's drift channel increases with moneyness because the absolute impact of a rate increase to the risk-neutral expected asset price is higher for higher prices $S_t$ hence we can expect the payoff $[S_T-K]^+$ to be larger. On the other hand, the inverse is true for a put by similar arguments.

To understand sensitivity through the discounting channel, it is useful to think in terms of absolute sensitivity $|\rho|$ and consider a zero-coupon bond with principal $N$ maturing in $T$ in a market with a constant interest-rate $r$. Letting $D(t,T)$ be the discount factor, the price of the bond at $t$ is:

$$ P(t,T) = Ne^{-r(T-t)} = ND(t,T)$$

The absolute rho of the bond is (note that rho is equal to Macaulay duration scaled by the bond's value):

$$ |\rho|=\left|\frac{\partial P(t,T)}{\partial \, r}\right|=(T-t)Ne^{-r(T-t)}=(T-t)ND(t,T)$$

Hence the greater the principal (i.e. the payoff), the greater the sensitivity to interest rates. This is because a change in the discounting curve will have a bigger absolute impact on larger payoffs than on smaller ones. With an option the mechanism is the same: the more the option is in-the-money, the greater the likelihood of a big payoff and therefore the greater the sensitivity to the interest rate.

Given $\rho$ is positive for calls, we conclude the asset's drift channel has a dominant impact(1). See below a plot for $\rho$ against moneyness under Black-Scholes model with $K=100$, $r = 4.1\%$, $\sigma=40\%$ and $T-t=0.25$, for equally-spaced values of $S_t$:

(1) [Edit 23/11/2018] Note this can be seen by looking at the expected discounted payoff at maturity if the option expires in the money: $$\eta P(t,T)E^T_t[S_t-K]=\eta\big(S_0-P(t,T)K\big)$$ where $\eta=1$ for calls and $-1$ for puts. The value of the discounted call payoff increases with rates, whereas the opposite is true for puts.

## Answer by Anton Tsches (score 3)

https://quant.stackexchange.com/a/39804

Let it be call. If strike K is fixed, then the moneyness means bigger S. The dependency of Rho from moneyness can be described by $\frac{dRho}{dS}= \frac{d}{dS} \left(\frac{dC}{dr}\right)$ which is the same as $ \frac{d}{dr} \left(\frac{dC}{dS}\right) = \frac{d}{dr}N(d_1)=N'(d_1)\frac{\sqrt T}{\sigma} > 0$

So the derivative is stricly positive which means that Rho increases when S increases. The same calculations can be made for put option where moneyness means smaller S.

## Answer by Rodolfo Oviedo (score 1)

https://quant.stackexchange.com/a/42745

If you knew that the option will expire out of the money (OTM), the value of the option would be 0, whichever the value of the local interest rate r. Therefore Rho would be 0.

If you knew that a Call with strike K will expire in the money (ITM), its value would equal that of a (long) forward with delivery price K. (The delivery price is that agreed at inception to be paid at expiration.) The payoff of a Call that will be exercised for sure is S(T)-K, which equals that of the aforementioned forward.

The value of a forward and a call, assuming the latter will expire ITM, is

```
f = S(t)*exp(-q(T-t)) - K*exp(-r(T-t))
```

where q is the yield of the underlying, for example, the dividend yield of a stock index, a foreign exchange rate, or a convenience yield of a commodity.

Differentiating with respect to r yields

```
Rho of a Forward = K * exp(-r(T-t)) * (T-t)
```

which is POSITIVE.

In our uncertain world, the Rho of a Call will be between 0 (the one of a call sure to expire OTM) and the Rho of a Forward (the one of a call sure to expire ITM).

Rho tends asymptotically to the upper limit when the call becomes deeper ITM. This is because a higher S(t) indicates that the scenarios of high S(T) at expiration become more likely, which turn the option more likely to expire ITM, and its rho closer to that of a forward.

Rho tends to 0, the lower limit, as the call becomes deeper OTM. This is because a lower S(t) indicates that scenarios of low S(T) at expiration become more likely, which in turn makes the Rho closer to that of a call sure to expire OTM.

The analysis for a Put is similar. A put sure to expire ITM is equivalent to a Short Forward, whose value is

```
-f = - S(t)*exp(-q(T-t)) + K*exp(-r(T-t))
```

Differentiating with respect to r,

```
Rho of a Short Forward = -K * exp(-r(T-t)) * (T-t)
```

which is NEGATIVE.

In our uncertain world the Rho of a Put before expiration is between the Rho of a Short Forward (the lower limit, a negative number) and 0 (the upper limit, the Rho of a put sure to expire OTM).

When S(t) tends to 0, the Rho of a Put tends to the Rho of a Short Forward. When S(t) tends to be large, the Rho of a Put tends to 0 asymptotically.

There is also another Rho, the derivative of the premium with respect to q. But that is another story because the Rho of the Forward depends on S(t).