Is my below computation correct (assuming flat volatlity Black Scholes model, flat interest rate curve):

$\mathbb{E}(\frac {S_{T_2}} {S_{T_1}}| \mathcal{F}_{T_0})$

$ = \mathbb{E}{\frac{S_{T_0}e^{(r-\frac{\sigma^2}{2})T_2+\sigma W_{T_2}}}{S_{T_0}e^{(r-\frac{\sigma^2}{2})T_1+\sigma W_{T_1}}}}$



$ = e^{r(T_2-T_1)}$

EDIT: Can anyone please re-confirm one of the steps above? $\mathbb{E}(e^{r(T_2-T_1)-\frac{1}{2}\sigma^2(T_2-T_1)+\sigma(W_{T_2}-W_{T_1})})$ $=e^{Mean(.) + \frac{1}{2}Variance(.)}$ $Mean(.) = r(T_2-T_1)-\frac{1}{2}\sigma^2(T_2-T_1)$ $Variance(.) = \mathbb{E}[\{\sigma(W_{T_2}-W_{T_1})\}^2]=\mathbb{E}[\sigma^2\{(W_{T_2})^2 +(W_{T_1})^2 -2W_{T_1}W_{T_2}\}]=\sigma^2(T_2+T_1-2T_1) = \sigma^2(T_2-T_1)$

I think I got it all correct, now! :-)

Related Question - Do we have an analytical formula (under standard Black Scholes) for -

$\mathbb{E}((\frac {S_{T_2}} {S_{T_1}}-K)^+| \mathcal{F}_{T_0})$ paid at $T_2$

My attempt .. basically using the Black Scholes pricing formula for call option -

$\mathbb{E}((\frac {S_{T_2}} {S_{T_1}}-K)^+| \mathcal{F}_{T_0}) = e^{r(T_2-T_1)}N(d_1)-KN(d2)$

where $d_1= \frac{\ln(\frac{e^{r(T_2-T_1})}{K})+\frac {\sigma^2(T_2-T_1)}{2})}{\sigma \sqrt(T_2-T_1)}$

$d_2= \frac{\ln(\frac{e^{r(T_2-T_1})}{K})-\frac {\sigma^2(T_2-T_1)}{2})}{\sigma \sqrt(T_2-T_1)}$

I would multiple with the discounting factor $e^{-r (T_2-T_0)}$ to the above formula to get the price at $T_0$.

  • 1
    $\begingroup$ The first computation is correct. For the second, look up forward start options and using the stock price as numeraire. $\endgroup$
    – Slade
    Commented May 27, 2019 at 6:44
  • 1
    $\begingroup$ Ok, thanks. I found this link for Forward Start Options - quant.stackexchange.com/questions/28133/… . So, looks like my attempt to the "Related Question" is also correct. Am I right to contend that $\frac {S_{T_2}} {S_{T_1}} $ is a random variable, which follows GBM with the forward $=e^{r(T_2-T_1)}$ and variance $=\sigma^2(T_2-T_1)$. $\endgroup$
    – bhutes
    Commented May 27, 2019 at 7:02

1 Answer 1


An expectation cannot exist for equity securities. The distribution of $$\frac{S_{T_2}}{S_{T_1}}$$ is the Cauchy distribution for equity securities, assuming no liquidity costs as with Black-Scholes, no mergers and no possibility of bankruptcy.

From auction theory, we know that there is no winner's curse in a double auction. Rational behavior for actors is to bid their expectations. If, as is assumed in models such as the CAPM or Black-Scholes, that there are many buyers and sellers, then it follows that the distribution of prices $S_{T_t}$ is the normal distribution by extending the central limit theorem.

If we then also assume equilibrium prices exist and that the security is in equilibrium except for random shocks, then we could treat the prices as normally distributed around the equilibrium price $S_{T_t}^*.$

Because the distribution is the distribution of the ratio of two random prices, the solution of which is well known in the statistical literature. It is complicated, slightly, by the fact that the integration should be around $(S_{T_1}^*,S_{T_2}^*)$ and that prices are truncated at -100%. In fact, the ratio of any two elliptical distributions will produce the same result.

When you factor in truncation, the distribution is $$\left[\frac{\pi}{2}+\tan^{-1}\left(\frac{\mu}{\gamma}\right)\right]^{-1}\frac{\gamma}{\gamma^2+(r-\mu)^2},\gamma=\frac{\sigma_2}{\sigma_1},r=\frac{S_2}{S_1}.$$ The expectation of that ratio does not exist as the integral diverges.

See http://mathworld.wolfram.com/NormalRatioDistribution.html

The reason that the Black-Scholes equations work is that the parameters are assumed to be known with perfect certainty and the distributions are assumed into existence.

If you have to do any form of parameter estimation, then nothing in Black-Scholes can hold as true.

Black-Scholes comes apart for several reasons in this case. This case is well known in the statistical literature. There are a wide range of issues. The distribution lacks a sufficient statistic for the parameter estimates, leaving you, in most cases to Bayesian statistics unless you are will to take the information loss. The estimate of the mean and therefore the variance is of zero power. A sample size of one million has the estimation power as a sample size of one.

  • $\begingroup$ Thanks for the above, but this is beyond me. In the limiting case, when $r=0, \sigma=0$, the ratio $\frac {S_{T_2}} {S_{T_1}} = 1$, as $S_T$ is a constant. I expect the general solution to collapse to the particular case I have cited. $\endgroup$
    – bhutes
    Commented May 29, 2019 at 4:04
  • $\begingroup$ If $r=0$ and $\sigma=0$ then $S_{T_t}$ is not a random variable. I found you a set of videos, the author makes a few mistakes but catches them. It is the full derivation. The first video is youtube.com/watch?v=9j398FbirQY It is followed on by youtube.com/watch?v=QLILi7oOTek youtube.com/watch?v=joF0RFx4_TE and youtube.com/watch?v=eqjg-FPUv3o It takes about an hour to watch. $\endgroup$ Commented May 29, 2019 at 14:29
  • $\begingroup$ Thanks, much appreciated. $\endgroup$
    – bhutes
    Commented May 29, 2019 at 15:44

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