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11 artículosA 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.
An American call that only wakes up at 80 and dies for good at 125 has no closed form, and it cannot be simulated either, because a path runs forward while the exercise decision looks back. Every path that avoids the ceiling either visited the floor or never did, so the contract is one knock-out minus another and both come off a standard tree. The identity is exact to machine precision for European exercise at all seven grids tested, and for American exercise only in the continuous limit: the finite-tree residual falls from 0.532 percent at 45 steps to 0.043 percent at 3,394.
A share at 100 that jumps to either 80 or 130 gives a call an exact price of 12, from two equations in two unknowns and no probability at all. Let the jump size be random, so 110 is also reachable, and that same hedge pays 18 where the option pays 10 while no other portfolio does better. The arbitrage-free prices then fill the whole interval from 20/3 to 12, and the obstruction turns out to be the kink in the payoff rather than the number of states.
An American put struck at 100 on a stock at 100, with no expiry date at all, is worth 23.21 when the rate is 5% and the volatility 30%. Removing the clock removes the time derivative from the pricing equation, which turns it into an ordinary differential equation solved by powers, and the exercise boundary collapses from a curve into the single level 1000/19 = 52.63. Its European twin, which cannot be exercised early, is worth exactly nothing, so every cent of the value is the right to stop.
Holding the share above the strike and nothing below it reproduces a short call's obligation on every single path, and it is still not a hedge: the residual has a standard deviation of 9.07 dollars against a premium of 11.9235, and monitoring four and sixteen times as often leaves it at 9.00 and 9.02. A real delta hedge on the same paths goes 1.24, 0.63, 0.31, halving each time the interval is quartered. Tanaka's formula says why the refinement cannot help, because the residual is exactly the premium minus half the share's local time at the strike, a random quantity that never mentions the monitoring interval and is bounded above by the premium with no floor below.
An option settling on the mean of a share's closes is strictly cheaper than one settling on the closing price, and the reason is convex order rather than any pricing model: for a martingale share every intermediate price is a forecast of the last one, so the average is dominated at every strike, for calls and for puts. Quantitatively the time average of a Brownian path carries variance T/3 against T, a swing ratio of 1/sqrt(3) = 0.57735, which turns 11.9235 into 6.9013 at a 30 percent volatility. A finite grid of 252 fixings sits at 0.33532 rather than 1/3, which accounts for most of the gap to the 6.918 measured by simulation on the true arithmetic average.
Three calls struck at 100: one plain, one that dies at 90, one that dies at 120. The knock-outs cost less, and their slopes can be ranked from the two ends of the picture instead of by differentiating a barrier formula. That argument only bounds an average slope, so the article also carries the exact pointwise gap, the strike times a normal tail at the reflected share price divided by the barrier, which comes to 0.138146 and turns 0.539828 into 0.677974.
The sharp statement is stronger than the usual one: on every path, missing the ceiling plus missing the floor counts the paths that miss both exactly twice, so the pair is twice the double plus the value of the one-sided survivors. That is a polynomial identity in indicators, so it holds under every pricing measure with no volatility anywhere, and one half is the tight bound. In a worked instance the pair is 10.317 against a double of 1.494, and the fastest refutation of the trap is that the pair exceeds the plain call with no barriers at all.
The integral of x dx is x squared over two, so the integral of W dW ought to be W(T) squared over two, and the only false step in that chain is the conclusion. A dissected square turns the Riemann sum into an identity exact at every partition, and the term that refuses to vanish is the total of the squared steps, which equals T rather than zero. The reflex answer is the exact value of the midpoint sum over the same partition, which is why it feels so solid.
A share sits at 75, the rate is zero, and a perpetual claim pays one dollar the first time the price ever touches 100. It is worth exactly 75 cents, and no volatility number is needed to say so. The reflex answer of a dollar assumes the barrier is always reached, which a price with a floor at zero never promises: a quarter of the paths fade away without paying anything.
Una recta con coeficientes racionales aplica ℚ sobre ℚ. Nada curvo lo logra jamás. Tres filtros — interpolación, forma, denominadores — dejan la clasificación completa.