Deltas on Barrier options vs Vanilla options

Question

In "Heard on the Street" it states that

$$\Delta_{\text{up and out call}} \leq \Delta_{\text{standard call}} \leq \Delta_{\text{down and out call}}$$

Is there an intuitive explanation for why this is true?

Answer 1

Standard call options are trivially more expensive than up/down and out call options.

However, for high strikes, down and out options will very likely never be knocked out, therefore their prices should be close to standard call options. For low strikes, down and out call options are almost worthless, therefore , the down and out call options curve price wrt strike should be very steep to catch up with the standard call price curve, hence a higher delta than standard's

For up and out, for high strikes, the up and out call option are worthless, and for low strikes, up and out call options are as valuable as standard call option as they will very likely never be exercised. Since the strikes are low, they are also worthless. The transition from high strikes to low strikes is relatively flat as we go from worthless to worthless, therefore, one would expect the delta to be smaller than standard's

Answer 2

for an intuitive answer,

if we start with a vanilla call as our base, then with an up & out call, we would like the underlying to go up in price yes. But as the price increases, we also increase the probability of kicking out and losing our payout - so we don't want it to go up too much. If the barrier is so far away that the probability of reaching it is vanishingly small, then we can basically think of it as not being there, and so the up & out converges to being the same as the vanilla.

For the down & out call, the same argument applies, but the other way around - when the underlying price decreases, not only does out option decrease in value, but also the chance of losing our option increases - i.e. we have an extra reason to want the underlying price not to decrease. So we have more delta for a down & out call.