Valuing a Zero-Strike Perpetual Call Like Its Underlying Stock

# Equivalent call option to the underlying stock


# Equivalent call option to the underlying stock












Would there be any benefit to an investor to purchase an American Call option on a single share with no expiration date and a strike price of 0 as opposed to purchasing the underlying zero dividend stock instead, assuming both are the same price? I see a potential benefit to owning the stock itself as having a (negligible) influence in the direction of the company as it may entitle you to vote in board elections and at the annual meetings, but is there any benefit to taking a step away from actual ownership while taking on an identical risk from a pure monetary standpoint?

Assuming this is in the US, are there any tax advantages to holding an option as opposed to holding the underlying stock?

## Answer by Neeraj (score 1)

https://quant.stackexchange.com/a/27961

In your question, you reduced your derivative to just a stock. No doubt, it is unlikely to have a derivative with zero strike price and never ending expiry. But let's assume it exists.

Keeping aside ownership advantage of owning of a stock [as mentioned by OP], from here, it is appear to be a dilemma between buying a stock or call. Theoretically, derivative prices are driven by underlying value, not the vice-versa. Therefore, price of derivative would only be influenced by volume in underlying, not from volume in derivative. It means that you can always buy or sell unlimited quantity of option at existing price [assuming no changes in underlying value] without having any influence on stock price (and, also at option price) but such thing is unlikely to occur for stock.

In such a scenario, call option would assure more price certainty than the underlying stock.

## Answer by Quantuple (score 0)

https://quant.stackexchange.com/a/27963

As of time $t$, the price of an American call option struck at $K$ and expiring at $T $ is: $$ V_t = \text{sup}_{\tau} E^\mathbb{Q} \left [ e^{-r(\tau-t)} (S_\tau - K)^+ \vert \mathcal {F}_t \right] $$ where $\tau$ figures a family of stopping times with values in $[t,T]$.

Now setting $K=0$ and letting $T \rightarrow \infty$ we have: $$ V_t = \text{sup}_{\tau} E^\mathbb{Q} \left [ e^{-r(\tau-t)} S_\tau \vert \mathcal {F}_t \right] $$

Assuming no dividends: $e^{-rt} S_t = \frac {S_t}{B_t}$ is a $\mathbb{Q}$ martingale in the absence of arbitrage opportunities. In that case, the optimal stopping theorem states that: \begin{align} E^\mathbb{Q} \left [ e^{-r(\tau-t)} S_\tau \vert \mathcal {F}_t \right] &= e^{rt} E^\mathbb{Q} \left [ e^{-r\tau} S_\tau \vert \mathcal {F}_t \right] \\ &= e^{rt} e^{-rt} S_t \\ &= S_t \end{align} Such that: $$ V_t = S_t $$ in other words, holding the option is equivalent to holding the stock.

Now the real question is: what happens when you consider dividends? This is the game changer IMHO (holding the stock allows you to cash in the dividends, which you cannot do when holding the option). Remember that in that case $e^{(q-r)t} S_t$ is the $\mathbb {Q} $-martingale assuming a continuous div yield. If you consider discrete divs you should additionally be careful with your modelling assumptions (so that $(S_t)_{t \geq 0}$ remains positive, as it should)