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Says X follows a driftless geometric brownian motion(GBM) given a volatility ($\mu = 0$). It gives the expected value of its initial spot. (Source: https://en.wikipedia.org/wiki/Geometric_Brownian_motion)

$E(X) = X_0$

Since Black Scholes Pricing Model assumes spots following GBM,

$Binary\ Cash \ or \ Nothing \ Call = e^{rt}N(d_2)$

and

$Binary\ Cash \ or \ Nothing \ Put = e^{rt}N(-d_2)$

My question is by referring to Black Scholes formula, why would cash or nothing put is supposed to be more expensive than call, provided that both were on driftless GBM? Would Black Scholes assumed downside probability has higher than upside probability?

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This is due to the asymetry in the lognormal distribution.

You have a higher probability of being below the mean, but since values of $X$ are lower that compensates, in the expectation computation, for the lower probability of being above the mean with higher values of $X$.

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  • $\begingroup$ It is the expected answer, thanks. Meaning to say the term "driftless" is only referring to return of spots instead of probability up/down? $\endgroup$
    – Ang Yiwei
    Commented Dec 18, 2019 at 3:20
  • $\begingroup$ There is a quite subtle distinction here. In your GBM $dX_t = \sigma X_t dW_t$, you can have a symmetric (same probability of up and down) shock to the asset price through the Brownian motion; a symmetric instantaneous, arithmetic return. However, B. m. paths are (a. s.) continuous: as you shrink your time intervals, $dW$ will shrink too. It results from $s\rightarrow t, dX_{s} = \sigma X_{s}dW_s$ that if $dX_t >0$, i.e. your stock price grew (infinitesimally) the shock is applied to a slightly higher level, and conversely if the stock dropped. Is it clearer where the asymmetry comes from? $\endgroup$
    – siou0107
    Commented Dec 18, 2019 at 8:27

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