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How will the probability of an option ending up in the money change if the volatility of the underlying stock increases?

Intuitively, I think the answer to this is that if volatility goes up the chance of being in the money at expiry increases?

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    $\begingroup$ Hopefully someone else will provide the gory details but it's not a symmetric thing because the price can't go below zero. $\endgroup$ – mark leeds Nov 10 '20 at 15:29
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Call option:

$$\mathbb{P}\left(S_t\geq K\right)=\mathbb{P}\left(S_0e^{(rt-0.5\sigma^2t+\sigma W_t)}\geq K\right)=\\=\mathbb{P}\left(W_t\geq \frac{ln\left(\frac{K}{S_0}\right)-rt+0.5\sigma^2t}{\sigma}\right)=\\=\mathbb{P}\left(Z\geq \frac{ln\left(\frac{K}{S_0}\right)-rt+0.5\sigma^2t}{\sigma\sqrt{t}}\right)=\mathbb{P}(Z\leq d2)$$

So we have shown the well-known result that the (risk-neutral) probability of the call option ending up in the money is $N(d_2)$.

I might want to differentiate with respect to $\sigma$ to see where the derivative is positive and where it is negative:

$$\frac{\partial}{\partial \sigma}\mathbb{P}(Z\leq d2)=\frac{\partial}{\partial \sigma}\left(\int_{-\infty}^{d2} f_Z(h) dh \right)=\\=\frac{\partial}{\partial d2}\left(\int_{-\infty}^{d2} f_Z(h) dh \right)\frac{\partial d2}{\partial \sigma}=\\=f_Z(d2)\left(\frac{-ln\left(\frac{S_0}{K}\right)-rt}{\sigma^2\sqrt(t)}+\sqrt{t}\right)$$

In all honesty, from the above expression, it is not immediately obvious and it is far easier to plot $N(d_2)$ vs $\sigma$ for OTM, ATM and ITM call options (I set all options to 1 year expiry, rates are set to 0.01, strikes are 80, 100 & 120 respectively, spot is set to 100). Plotting, I get the below:

enter image description here

The graph above makes sense to me for OTM and ITM: OTM calls do like higher volatility as one would intuitively expect (up to about 0.6), whilst ITM calls dislike higher volatility (again, as one would expect).

I am a bit puzzled (intuitively) as to why ATM calls dislike increasing vol across the entire domain with regard to (risk-neutral) Probability of ending up in-the-money. With the downside limited at zero and unlimited upside, I would have intuitively thought that ATM Call options would like increasing $\sigma$ with regard to ending up in-the-money at expiry.

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    $\begingroup$ @Kermittfrog: you mean it's straightforward to see why ATM calls dislike higher vol? I agree it's easy to see it from the $N(d2)$ equation, but does it make sense intuitively? I can see intuitively why ITM calls dislike higher vol and OTM calls like higher vol. For ATM calls, if somebody woke me up at midnight and asked me: my first intuitive answer would be that ATM calls like higher vol with respect to probability of "being in the money"... (because of larger potential upside...). $\endgroup$ – Jan Stuller Nov 10 '20 at 16:33
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    $\begingroup$ @JanStuller I gave this some more thought and I agree with you: It feels counterintuitive that the probability of ending ITM decreases monotonically as soon as $\sigma$ is large enough. I could not think of an economic reason for this. Maybe it has something to do with the drift under $Q$ being $r-\frac{1}{2}\sigma^2$? $\endgroup$ – Kermittfrog Nov 10 '20 at 19:00
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    $\begingroup$ The probability of ending ITM under the risk-neutral measure should tend to 0. But the same probability under the stock measure tends to 1. $\endgroup$ – Daneel Olivaw Nov 11 '20 at 12:04
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    $\begingroup$ I'm glad you enjoy the books Jan! As for my comment, I don't really have a clear intuition. I too was surprised about $\Phi(d_2)$ tending to 0 for ITM options (I rationalized it with the absorbing barrier at 0 of a GBM), then I realized that $\Phi(d_1)$ actually tends to 1 which is the probability under the stock measure. I mean of course as discussed in an older question of yours the drift of $S$ under measure $\mathcal{Q}$ is $r$ whereas under $\mathcal{S}$ it is $r+\sigma^2$ but then I am not sure whether that tells us anything meaningful in a financial or economic sense. $\endgroup$ – Daneel Olivaw Nov 13 '20 at 16:21
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    $\begingroup$ My intuition tells me the stock price measure is the "odd" case: we observe the proba going to 1 merely because the vola is also in the drift. The term $d_\pm$ is actually equal to (log-moneyness + total drift $\pm$ total vol drag)/(total volatility). Hence generally we should have the proba tending to 0. It's just that under the stock measure that breaks down because the vola ends up being also in the drift, cancelling the vol drag. Ignoring the stock measure case then, this seems to be yet another case of things boiling down to volatility drag. $\endgroup$ – Daneel Olivaw Nov 13 '20 at 16:28

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