An at-the-money call with a zero interest rate is worth one over the root of two pi, which is 0.39894, times the absolute swing of the terminal price, so a 10 dollar standard deviation prices it at 3.9894 and the closest of the offered 1, 5 and 10 is 5. The two-step estimate is exact under a symmetric terminal law, where the option finishes above the strike exactly half the time and the average gain when it does is 7.979. Calibrate a lognormal to the same 10 dollar swing and those two factors become 0.4801 and 8.285, whose product is still 3.98, which is why the qualifier about half cannot be cut.
A call whose payoff is the square of the stock minus 100 does not start paying at 100, it starts paying at 10, because the square clears the strike exactly when the stock clears its square root. Placing the kink at the written strike prices the option at essentially zero on a stock at 12, when its real value is 53.8168. The closed form is ordinary Black-Scholes on the transformed asset with a growth term of 4 percent and a strike leg that still discounts at the riskless rate, and the value curve sits above intrinsic everywhere while being shallower than it at the spot in question.
A share at 100, a one-year call struck at 100, a zero rate and a twenty percent swing return 0.539828 and 0.460172. The first is the number of shares in the replicating portfolio, and reading it as the chance of finishing in the money hands you the complement of the right answer, since at the money with a zero rate the two numbers sum to one. The article carries the density identity that makes the share count exact, the measure under which the first number is a probability after all, and the four cents the straight line misses over a two-dollar move.
At the money the shorter maturity always wins, and a single negative derivative settles it for every volatility and every maturity: curvature runs 0.06907 against 0.02814 for one month against six. Ten percent out of the money the order reverses, 0.01854 against 0.02352, and the two curves cross 8.845 percent above the strike. What forces a crossover to exist is a conservation law, since every option in the family carries exactly the same total curvature and can only choose how to spread it.
Moving the average inside the exponential returns 1, and 1 happens to be the exact median and the exact geometric mean of e^X, which is why the mistake survives every re-check of the arithmetic. Completing the square in the exponent gives the true value e^(sigma squared over two), or 1.6487 at unit spread, because multiplying a Gaussian density by e^x slides its centre and scales its mass. Convexity settles the direction before any integral is set up, and on a heavy-tailed variable the quantity stops being finite at all.
A share swinging twenty dollars a year gives an at-the-money call that looks like it should cost ten, half the swing collected half the time. It costs 7.98, because the upper half of a bell curve averages 0.798 of a standard deviation rather than a whole one. The article derives the general arithmetic-Brownian price, checks both limits, and quantifies the negative-price defect that got the model retired and then rehabilitated.
Acceptance restricts the value to below your bid, where a uniform variable averages half of it, and doubling half your bid returns exactly your bid. The expected profit is therefore identically zero at every bid up to 100 and 100 minus b above it, so there is no optimal bid to find. With a general multiplier the profit is b squared times k minus 2, over 200, making doubling the exact break-even multiple, and the article shows a value distribution starting at 50 where the same bidder profits.
The put reaches its strike more often, 0.3348 against 0.2821, and the call is still worth more, 4.2920 against 3.5891. The mechanism is not the unbounded-upside story, which would predict a gap at the money where put-call parity provably gives none; at a zero rate the 110 call equals 1.1 times a put struck at 90.909, and the put on offer is struck lower than that. The article also records two circulating claims that fail at these strikes, since the in-the-money chances at r = sigma^2/2 are 0.3168 and 0.2992 rather than equal, and the price ratio is 1.63 rather than 2.
At the money the log term in d1 vanishes and what remains is strictly positive for every non-negative rate and every volatility, so the delta always beats 0.5 and is 0.6554 at twenty percent. A square rather than a derivative gives the sharp floor: at six percent over a year the delta can never fall below 0.6355. The article also kills the sentence that sounds like a restatement of the answer, since the chance of finishing in the money falls to 0.4801 at forty percent volatility.
Six months of a sixty dollar year carries 60 over root two, which is 42.43 rather than 30, because variances add over disjoint intervals and standard deviations do not, so the digital is worth exactly $239,750. The figure of $250,000 in circulation comes from rounding the z score 0.7071 up to 0.75 and then reading the tail at 0.75 as 0.25, but Phi(0.75) = 0.773373, so even the rounded chain gives 0.2266. Rounding z upward has to make the tail smaller, and 0.25 is larger, which is the tell that a symbol changed meaning mid-calculation.
An at-the-money call has no ceiling on its payoff and an at-the-money put is capped at the strike, yet at a zero interest rate the two cost exactly the same. The reason is put-call parity and it uses no model at all: the difference of the two payoffs is a straight line, so pricing it needs only the risk-neutral mean. The equality was checked on five terminal distributions with mean at the strike, and on a sixth whose mean is 120, where the gap is exactly 20.
Two people arrive at random inside the same hour and each waits fifteen minutes, so the reflex answer is a quarter. Drawing both arrival times as one point in a 60 by 60 square turns the question into an area, and the two corner triangles it leaves out have legs of 45, giving 7/16 rather than 1/4. The general formula n(2T-n)/T squared shows why the first minutes of patience buy the most.
Draw X and Y uniformly from the unit interval and their product beats a half with probability (1 - ln 2)/2, about 15.3 percent. The reflex answer of a quarter counts a condition that is genuinely necessary and treats it as sufficient, which is why 0.8 times 0.6 sits inside the quarter square and still loses. The hyperbola y = 1/(2x) cuts the winners down to a sliver, and one integral measures it.