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Delta of an option is defined as ratio of change in price of call option to change in price of underlying securities. If, $c_t$ is call option price at time $t$ and $S_t$ is the price of underlying securities, then the delta of call option is:

$$\Delta(t)=\frac{\partial c_t}{\partial S_t}$$

If change in $dS_t$ ie $(S_{t+dt} -S_t)$ is random, which by definition is true, then delta ($\Delta (t)$, computed at time t) must be random variable (instead of known constant) as it involve $dS_t$ in denominator. But under the Black-Scholes model, the delta of European of Call option (which is written as $N(d_1)$ or $\Phi (d_2)$) is deterministic variable (ie known with certainty) at time $t$.

I want to know why delta of a call option is deterministic quantity, why not it is random variable? If possible, please provide both logical reasoning and formal derivation.

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    $\begingroup$ My 2cents: It is defined as a partial derivative. It cannot be anything but deterministic as such. Remember that the call price knowing the filtration $\mathcal {F}_t $ is a deterministic function of $S_t $. $\endgroup$
    – Quantuple
    Mar 17, 2016 at 18:31
  • $\begingroup$ In other words: $\Delta(t) $ is random conditionally on $\mathcal {F}_0$, true... but not conditionally on $\mathcal {F}_t$ (since it is $\mathcal {F}_t$-measurable, because $S_t $ is adapted to $\mathcal {F}_t$). $\endgroup$
    – Quantuple
    Mar 17, 2016 at 20:41
  • $\begingroup$ @Quantuple: If at time $t$, $\Delta(t)$ is known and, of course, not random. However, we look the process $\{\Delta(t) \mid t \ge 0\}$ today, then it is a stochastic process. For future time $t$, we say that if we are there, but we are actually not, and we can have many paths $\omega$ to get there. This is how stochastic process coming into place. $\endgroup$
    – Gordon
    Mar 17, 2016 at 21:05
  • $\begingroup$ @Gordon: Pardon me, but I don't understand how this differs from/is in conflict with what I said. $\endgroup$
    – Quantuple
    Mar 17, 2016 at 21:06
  • $\begingroup$ @Quantuple: I was not saying that your statement is wrong. I meant to emphasis that $\Delta(t)$ is stochastic. $\endgroup$
    – Gordon
    Mar 17, 2016 at 21:11

2 Answers 2

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Since we are dealing with the change in the value of the call option in relation to the change in the value of the stock price, we are looking at the simple slope of the function. We do not need to know how much the stock price will change in order to estimate the slope.

In the absence of gamma, the stock price could change 2% or 1% and this wouldn't change the delta. Let's say we have a simple equation, y = mx + b. m is the impact that a change in x has on the change in y, which is the same idea behind delta. If we know how y responds to changes in x, we can deterministically define m and it is not a random variable.

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  • $\begingroup$ My initial question is why delta is deterministic quantity at time $t$, when $dS_t$ is stochastic? In your equation $y = mx + b$, if change in $x$ is random, then how $m$ can be known with certainty? $\endgroup$
    – Neeraj
    Mar 17, 2016 at 18:42
  • $\begingroup$ My short answer is because a derivative doesn't depend on how much the independent variable changes. I assume based on your question that you are not opposed to the Black Scholes equation. The derivative of the Black Scholes equation with respect to the stock price is delta. And, this is denoted by the partial derivative. A derivative does not rely on the amount of the change of the denominator because it is assumed to approach zero, and the derivative is simply the instantaneous slope immediately around the given point, or call value. $\endgroup$
    – RandyF
    Mar 17, 2016 at 19:17
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    $\begingroup$ @Neeraj: your question is equivalent to questioning Ito's lemma: $dC_t = \partial_t C dt + \partial_S C dS_t + 1/2 \partial_{SS} C d\langle S, S \rangle_t$ $\endgroup$
    – Quantuple
    Mar 17, 2016 at 21:59
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Note that, at time $t$, \begin{align*} d_1(t) = \frac{\ln \frac{S_t}{K} + (r+\frac{\sigma^2}{2}) (T-t)}{\sigma \sqrt{T-t}}, \end{align*} which is a function of $S_t$, and then, it is a random quantity. Consequently, the delta $N(d_1(t))$ is also a random quantity.

Note that, at time $t$, $N(d_1(t))$ is known. However, $\{N(d_1(t)) \mid t \ge 0\}$ is a stochastic process adapted to the filtration of the equity process $\{S_t \mid t \ge 0\}$, that is, $\mathcal{F}_t$.

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  • $\begingroup$ $S_t$ is known at time $t$. $\endgroup$
    – Neeraj
    Mar 17, 2016 at 17:59
  • $\begingroup$ Do you want to say that Delta also follow a distribution? If yes, can you please specify? $\endgroup$
    – Neeraj
    Mar 17, 2016 at 18:00
  • $\begingroup$ at time $t$, $S_t$ is known. However, we are looking at it from today. For a Brownian motion $W_t$, it is also known at time $t$, but we do not say it is deterministic, as we consider it from today. $\endgroup$
    – Gordon
    Mar 17, 2016 at 18:03
  • $\begingroup$ As the delta is random, then it has a distribution, which can be derived from the distribution of $S_t$.. $\endgroup$
    – Gordon
    Mar 17, 2016 at 18:04
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    $\begingroup$ For any process $\{X_t \mid t \geq 0\}$, $X_t$ is known at time $t$. $\endgroup$
    – Gordon
    Mar 17, 2016 at 18:16

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