Using a call-spread to hedge a digital option

Question

I have a digital option that pays out \$1M at time $T$ if the price of the underlying stock is higher than \$1300 (with current price ~\$1000) and, obviously, zero otherwise. I am in the Black-Scholes setting and there are no dividends up to date $T$.

I have used the following to calculate the price of the digital option at $t=0$:

$C(0)=e^{-rT}N(d_2)$

where

$d_2 = \frac{\ln\big(\frac{S(0)}{Ke^{-rT}}\big) - \frac{\sigma^2}{2}}{\sigma\sqrt{T}}$

I am not given any of the parameters (apart from $K$ and $T$) and have been asked to make an educated guess as to what the call price should be. I have checked this using an online calculator and the price I get (~\$0.18) seems to be correct given the parameters I have fed into this model.

I have a couple of questions on this as it is not explained anywhere in my lecture notes and I cannot seem to find what I need online.

Q1: Why is the price of the call option not dependent on the pay-out? If this option was paying \$1 or \$0 then a price of \$0.18 seems reasonable, but not for \$1M. I would place that bet every week as it is cheaper and surely better odds than the lottery. I'm obviously missing some understanding here.

Q2: How would the trader who sold this digital option hedge against the potential losses? I have read online that this is possible using a call-spread but I'm not sure I understand how this would work. I understand that you need to go short on the call with the higher strike price and long on the other but I don't see how to apply this to this situation?

Answer 1

There are a few extra things to consider here where you'll get a different answer if you ask a quant or a trader.

If we have a european digital that pays \$1 if the underlying is above 120 ($S_0 = 100$) at expiry, then yes i can hedge it with a call spread. This can be approximated with a call spread (with a notional of $\frac{1}{\mathrm{d}K}$, this was missed out in previous comments/answers) as mentioned above where in the limit of $\mathrm{d}K \rightarrow 0$ the payoff approaches that of a digital. This is all wel land good from a theoretical standpoint, but the reality is you can't trade options at every strike, and if you reduce $\mathrm{d}K$ too much the notional becomes significantly large that there simply won't be enough liquidity to enable your hedge.

There is also the question of how you centre the strikes. If you do $K_\pm = K\pm\frac{1}{2}\mathrm{d}K$ then the value of the hedge and the digi is the same, but if the underlying expires in a small region $K<S_E<K+\epsilon$ where $\epsilon < \frac{\mathrm{d}K}{2}$ then you have to payout more than your hedge nets you. If on the otherhand you set the strikes as $K_{-} = K-\frac{1}{2}\mathrm{d}K$ and $K$, then regardless of where the underlying expires you make money. This is called an overhedge, it will cost more than the hedge centered on the barrier strike, since its payoff is at least higher everywhere.

The cost of a product not what it is worth, it is what someone is willing to sell it to you for. These are not the same thing. And if it costs me X to hedge something, i'm not going to sell it to you for less than X. Likewise, if i'm buying the digi i'll put the strikes on the other side. If the underlying is really illiquid, i won't be able to buy the strikes i need for the hedge, so it will cost more.

It's all well and good understanding the theoretical side, but the application is important too.

Answer 2

Isn't the 0.18 a percentage price? i.e. percentage of notional? if your notional is usd1 then you pay usd0.18. If the notional is usd100 you pay usd18?