# Which is riskier: a call option or the underlying?

From Joshi's Quant Interview Questions and Answers:

What is riskier: a call option or the underlying? (Consider a one day time horizon and compute which has bigger Delta as a fraction of value).

I find the answer he has given very confusing and not coherent. He is claiming the delta is as a fraction of the value, but this makes no sense to me. For me the delta is a component of the call option value so must be smaller.

• What was his answer? What confuses you particularly about it? – Bob Jansen Jul 2 '20 at 14:57
• ill update in a sec – Trajan Jul 2 '20 at 14:59
• Irrelevant on short terms, but for long-lasting contracts, you might consider that with the call option, there are 2 companies who could default / go bankrupt: The underlying value and the emmiter. – Zsolt Szilagyi Jul 3 '20 at 15:12

As @ir7 did, I only briefly want to add to @noob2's spot-on answer. He's of course right and $$\Lambda=\Delta\frac{S}{V}$$ decides how risky the option is compared to the stock.

Firstly, note that $$\Lambda=\frac{\frac{\partial V}{V}}{\frac{\partial S}{S}}=\frac{\partial V}{\partial S}\frac{S}{V}$$. An economist would call $$\Lambda$$ an elasticity. It tells you by how much percent $$V$$ changes if $$S$$ changes by one percent.

It is the key to determine the risk of an option. Cox and Rubinstein (1985) prove that the (CAPM) market beta (measures systematic risk) and the volatility (measures total risk) are linear in the elasticity, i.e. \begin{align*} \beta_V &= \Lambda \cdot\beta_S, \\ \sigma_V &= |\Lambda| \cdot\sigma_S. \end{align*} This way, you can compute the risk of your option (derivative) if you know the risk of your underlying.

In the simple Black-Scholes model, $$V=S_0\Phi(d_1)-Ke^{-rT}\Phi(d_2)\leq S_0\Phi(d_1)=S_0\Delta$$. Dividing by $$V$$ on both sides yields $$1\leq\frac{S_0\Delta}{V}=\Lambda$$. Thus, the elasticity of a call option is always greater or equal than one. Hence, a call option is always at least as risky as its underlying. (This observation holds, by the way, in a much more general setting than Black and Scholes)

For a put option, all we can say is that $$\Lambda=\frac{\Delta S_0}{V}<0$$ because $$\Delta<0$$. However $$P=Ke^{-rT}\Phi(-d_2)-S_0\Phi(-d_1)\geq -S_0\Phi(-d_1)=S_0\Delta$$ results in $$1\geq \frac{S_0\Delta}{P}=\Lambda$$. This merely tells us that a put's elasticity is at most 1. But we already know that the elasticity is below zero because $$\Delta<0$$. So, we can't make a similar argument as for call options. In general, it's possible to have a put option with $$\Lambda\in(-1,0)$$. As a conclusion, a put option has a negative elasticity and thus a negative market beta and a negative expected return (because it acts as an insurance from economically bad states).

But what else can we say about the magnitude of the elasticity of $$\Lambda$$ for a call option? Well, look at this plot.

The elasticity is particularly high (the option is particularly risky) if

• the stock price is low (option is OTM)
• volatility is low
• time to maturity is short

Thus, if the option has a low probability to be exercised and hence likely pays nothing (defaults''), the option is very risky. This makes intuitive sense. If you buy a deep OTM option expiring in one week's time, the option will be very cheap. If (due to some miracle) the option expires in the money, you make a large profit and your gamble paid off. But that's of course very unlikely. This is reflected in a higher riskiness. An ITM option on the other hand, is likely to be exercise and has a much lower risk (leverage).

• For a put option, all we can say is that $\Lambda<0$. Why? – Trajan Jul 12 '20 at 15:35
• @Permian Because $\Delta<0$ for a put, we have $\Lambda=\frac{\Delta S_0}{V}<0$. But try $P=Ke^{-rT}\Phi(-d_2)-S_0\Phi(-d_1)\geq -S_0\Phi(-d_1)=S_0\Delta$ and thus $1\geq \frac{S_0\Delta}{P}=\Lambda$. That tells us that a put's elasticity is at most 1. But we already know that the elasticity is below zero because $\Delta<0$. So, we can't make a similar argument as for call options. In general, it's possible to have a put option with $\Lambda\in(-1,0)$. – Kevin Jul 12 '20 at 16:08
• put it in the answer for others – Trajan Jul 12 '20 at 16:10

A better, clearer, answer is to compute Lambda (leverage) of the option (link) and see if it is bigger or smaller than 1. Lambda is $$\Delta \frac{S}{V}$$ so we test

$$\Delta \frac{S}{V} \lessgtr 1$$

which is what Joshi is saying in words: compare $$\frac{1}{S}$$ (delta to price for the stock, the delta of the stock is 1) to $$\frac{\Delta}{V}$$ (delta to price for the option) and see which is bigger. If $$\frac{\Delta}{V}$$ is bigger the option has leverage greater than the stock, so is risker in the sense of percent profit over the next day.

• Can you say anything in general about when this will or won't be the case? Depends on moneyness mostly I imagine? – Oscar Jul 2 '20 at 17:38
• I don't know... it is an interesting question. – noob2 Jul 2 '20 at 17:41
• @noob2 i agree with @'oscar, is there nothing more to say? – Trajan Jul 3 '20 at 17:03

Just a small addendum to @noob2's answer. The discrete shape of $$\lambda$$ is:

$$\lambda \approx \frac{V_1 - V_0}{S_1 - S_0} \times \frac{S_0}{V_0}$$

which can be rewritten as

$$\lambda \approx \frac{\frac{V_1 - V_0}{V_0}}{\frac{S_1 - S_0}{S_0}}$$

which is as @noob2 said just the ratio of the relative returns for option and stock.