Analyse
47 articlesSine is differentiable everywhere and never leaves the band from minus one to one on the real axis, which makes it the counterexample everyone reaches for, and the modulus of sine at 10i is already 11013.23. The Cauchy estimate caps every Taylor coefficient by M over r to the n on a circle of radius r, so growing r kills every coefficient above the constant one and nothing but a constant survives. The same estimate with a polynomial growth bound gives more: an entire function bounded by C times one plus the modulus of z, all to the k, is a polynomial of degree at most k, and one corollary of that is the fundamental theorem of algebra.
A plain t sitting next to the t squared in a Gaussian exponent looks like a new function and is only a shift. Completing the square turns the integral of e to the minus a t squared over two plus b t, from x to infinity, into e to the b squared over 2a times the root of 2 pi over a times the standard normal at a rescaled and shifted argument, never at x itself. The worked case comes out as exactly half a bell, e root pi over two or 2.40901455, but only because its lower limit happens to land on the centre b over a.
The ratio and root tests both return 1 on the sum of e to the minus root n, which settles nothing, and the usual write-up of the problem then quotes 4 over e as the answer. That number is the floor rather than the cap: the sum is 1.6704068, which is 13.52 percent above it, and the usable bound comes from integrating from 0 instead of from 1, giving exactly 2. Where each bar of width one sits relative to its index is the single step that decides which way the inequality points.
An estimated line of expected return against market sensitivity that sits entirely above the theoretical one is not a market on sale, because pricing errors scatter above and below instead of lifting everything by the same amount. In a two-factor world where every asset carries the same 0.75 exposure to the second risk, the fitted single-factor line comes out exactly parallel to the theoretical one and exactly 3 percentage points above it, with residuals of zero, while a mispricing world engineered to have the same average lift leaves errors of both signs as large as 6.5 points. Let the second exposure grow with sensitivity and the slope moves too, at which point the two lines can cross inside an ordinary sample.
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.
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.
A futures settles up every day, so its fair price is a plain expectation, while a forward settles once, so its fair price is a discounted expectation renormalised. The difference between the two is exactly the covariance of the discount factor with the contract price divided by the expected discount factor, and for a deposit contract quoted as 100 minus the rate both fall together, so the fair forward price is 95.019999 against the futures' 95.000000. Long the forward and short the futures is worth 0.0195 points at inception, the gap reaches 39.48 basis points at ten years, and on an asset whose price rises with rates the whole answer reverses.
Give a pulled-back log price the same 20 percent instantaneous swing as a free-wandering one and its horizon variance stops being sigma squared times T: at one reversion time only 0.432332 of it survives, the volatility that prices a one-year call is 13.1504 percent, and the call falls from 7.9656 to 5.2425. The same pull makes consecutive returns fight each other, with a first-order autocorrelation of exactly minus half of one minus phi, and that is the independence the pricing model rests on. The formula still returns the right European price and has lost the hedging argument that justified it.
A six-month at-the-money call on a 50 dollar share sheds a cent and a half a night to time decay, and its expected price tomorrow is higher anyway. The deterministic total differential gives minus 0.66 cents and predicts the opposite of the truth, while Ito's third term, half the gamma times the squared move, adds plus 1.13 and runs on variance rather than direction. Substituting the pricing equation for theta cancels that term exactly and leaves an expected return of the riskless rate plus elasticity times the premium, which is 35.54 percent a year here and turns negative below a real drift of 4.18.
Daily, weekly and monthly returns give per-day variance estimates of 1.0000, 1.3225 and 1.4000, and the reflex is to average them into 1.2408, a figure no horizon produced. The variance ratio is a weighted sum of autocorrelations, so a forty percent overshoot at twenty periods measures dependence rather than noise, and the coefficient that reproduces it is 0.17554. With twenty years of daily data that ratio sits 4.6 standard errors above one and with five years only 2.3, which is why the number means nothing without the sample size attached.
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.
Let a fair coin decide at the start of the year whether the share runs at 15 or 35 percent, price calls in that world, then read the volatilities back out with the constant-volatility formula: 28.43, 25.91, 24.97, 25.68 and 27.16 percent across five strikes. The floor sits at the money and below the 25 percent average of the two regimes, which one second derivative settles without any numerics. The usual explanation for the wings is refuted here, because the coin-flip world is less likely to clear 130 than a flat 25 percent and its option is still worth 27 percent more.
A ticket paying a hundred dollars if a share finishes above its strike is squeezed between two ordinary call spreads at every width, so its value is pinned by prices already quoted with no distribution assumed anywhere. The limit is minus the derivative of the call price in the strike, which equals e to the minus rT times N(d2) because two density terms cancel exactly at every strike. Here that is 53.2325, against the 62.35 a real-world drift would give.
The map x to 4x(1-x) contains no randomness and is still useless for prediction, because substituting x = sin squared of pi t turns it into angle doubling: one binary digit of your measurement is spent per step, so fifty steps eat fifteen decimal digits. The resulting series has autocorrelation exactly zero at every lag, proved by orthogonality of distinct cosine frequencies rather than measured. What that does not establish is anything about real return series, and the article says so.
Shade the region between a diffusing particle and the time axis over one second. The box is one wide and about one tall, so the eye guesses a variance of one, and the answer is one third because each increment counts only for the time remaining after it. No stochastic integration is needed to define the object, only continuity of the path, and the constant is pinned twice over: once by the weight (T minus t) and once by integrating the covariance min(s,t) across the square.
A flat belief about a coin's bias is an input to the calculation, not a conclusion of it, and a single head does not leave it standing. The density tilts to 2p, the cumulative law becomes p squared, the average bias moves to 2/3, and the old answer of one half is demoted to the lower quartile. The general update is the Beta conjugate family, which sends 750 heads in 1000 to Beta(751, 251) with mean 0.749501.
Traders treat opposite signs for theta and gamma as a law of the desk, but it is the pricing equation rearranged, and the equation names its own exceptions. At a zero rate the identity is exact and unbreakable; with a positive rate the interest on the bond leg buys the exception, and a deep in-the-money put has theta +7.0053 and gamma +0.0040317 together. The change of variables to the heat equation shows where the interest was hiding.
Both seats in the marble game average a dollar a play, and that arithmetic stays true to the last line. Seat A carries variance 3/2 against seat B's 1, and seat A's law turns out to be seat B's law with one prize smeared outward, so every concave utility prefers B without variance ever being mentioned. Once both players stop flipping coins, seat B is ahead on the average too, at 1 against 3/4.
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 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.
A holding pays $200 if a team wins four games first, you must take a symmetric position on every game, and committing the whole hundred to game one produces the right payoffs a week too early. Backward induction on the lattice fixes the amount at half the gap between the two successor values, $31.25. The same number is 5/16 of the holding, which is the chance the other six games split three each, and no win probability appears anywhere in the derivation.
Two stocks with equal expected returns, variances 0.10 and 0.40, and correlation 0.5: the reflex differentiates the portfolio variance and reports an interior weight. The minimum sits at 100% in the calmer stock, and the usual explanation for that, which blames the no-shorting rule, is wrong. The vertex of the variance parabola lands exactly on w = 1, so the constraint does no work at all and the answer survives dropping it.
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.
Turning all fifty-two cards lands on exactly zero, which makes zero the floor rather than the value. Backward induction over the grid of remaining cards gives the exact rational 41984711742427/15997372030584, and a two-line argument shows the optimal policy can never finish below zero in any deal. The article carries the small-deck ladder, the stopping boundary the table actually produces, and two plausible rules that lose money against it.
Twenty-four possible answers against 3^3 = 27 outcome sequences leaves just enough room, and four against four splits those answers into exactly 8, 8 and 8. Six against six always tips, so it wastes the level outcome and leaves twelve answers for nine remaining sequences, which makes halving impossible rather than merely slow. The counting argument bounds outcome sequences rather than strategies, so it rules out every adaptive continuation at once, and the explicit schedule closes the positive half.
There is no elementary antiderivative to evaluate, and the checkable slice of that is one line: if p is a polynomial then p' - 2xp has degree deg p + 1, which can never equal the degree of 1. Squaring the integral turns it into a rotationally symmetric integral over the plane, where the polar area element supplies the factor r that makes the radial integral elementary, so I squared equals 2 pi times one half. The same idea survives without polar coordinates via the substitution y = xt, and it fails for e to the minus x to the fourth because x^4 + y^4 is not a function of the radius.
The fraction that maximises long-run growth is exactly the edge, 2p-1, which is 0.2 on this coin, and one derivative gets you there. Double it and the growth rate is -0.0024469 a flip, negative on a game that leans your way three hundred times in a row, and the crossing happens at 0.3894 rather than at 0.4. The article carries the exact median over 300 flips, 25 dollars to 10504.19 at the optimum and to 12.00 at double, the reason about 48 percent of overbettors still finish ahead anyway, and the place where the textbook approximation mean minus half the variance returns the opposite sign.
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.
The standard deviation of 1, 2, 3, 4, 5 is either 1.4142 or 1.5811, and offering one of them without asking which question you are answering is the only wrong move. The sum of squared deviations is 10 either way, so everything turns on whether you divide it by 5 or by 4. Bessel's correction makes the variance unbiased and leaves the standard deviation biased low by about six percent at this sample size, and a third divisor beats both of them if you optimise for mean squared error instead.
The two correlations you are handed do not pin the third one down, but they fence it into exactly [−1/50, 1], and that interval dips below zero. The fence falls out of a 3×3 determinant read as a quadratic in the unknown, and out of a picture: 0.7 is an angle of 45.573°, both stocks live on a cone of that half-angle around the index, and putting them on opposite sides opens 91.146° between them. Also here: why the real tipping point is ab ≥ 0 together with a² + b² ≥ 1 rather than "both above 0.707", why 0.9 and 0.5 force a positive answer while 0.9 and −0.9 allow −1, why standing on the floor costs a rank, and why three Bernoulli(0.5) indicators with the same two correlations are confined to [0.40, 1] instead.
Take the centre, then mirror every move through it, and you place the last coin. The proof has three requirements and only one of them needs that opening move, which is the step a one-line answer skips. Central symmetry alone is not the condition: an annulus is centrally symmetric and the first player loses on it.
Nine slots hold the digits 1 to 9 and four overlapping windows each add to 20. One subtraction pins the centre digit to 5 with no search at all, and the same structure classifies every arrangement: there are 96 of them, falling into 48 mirror pairs. Change the target and the centre is forced to an odd digit that sometimes admits nothing.
Raise your right hand at a mirror and the reflected hand stays on the same side of the room, which means the usual question has a false premise. A plane mirror is the matrix diag(1, 1, -1): it fixes both axes lying in the glass and reverses only the direction you look along. Its determinant is -1, so no rotation reproduces it, and the sideways flip everyone reports belongs to the half turn you perform in your head.
Every family averages exactly one girl and contains exactly one boy, so the ratio of expected counts is exactly one half and a large town splits evenly. The expected share inside a single family is not one half but ln 2, and it is still 0.5249 across ten families, with the excess falling off like one over four m.
A pebble climbing four boxes on coin flips needs 18/5 flips on average, and the two-line renewal argument that gives 4 is wrong. Its premise is true, since half of all games really do end on flip two, but the non-finishing half is two different states: tails-tails sends the pebble home while heads-heads leaves it on box 3, one flip from the exit and worth only 14/5.
You hit one time in ten, your two opponents three and six, and you shoot first. Firing into the air is worth 965/4736 = 20.376%, which beats removing the strongest player by 0.195 percentage points, because a landed hit drops you into the duel you must enter second at 7/37 rather than first at 10/37. The article carries all three option values, the fixed points they solve, and the single Nash equilibrium that turns the usual assumption into a conclusion.
Deux cassures au hasard, trois morceaux, et trois inégalités qui n'en font qu'une. Le quart tombe d'un carré sans la moindre intégrale, et le morceau le plus long, 11/18 en moyenne, explique pourquoi la réponse paraît trop basse sans l'être.
A driver who eats one apple per loaded mile delivers 833 of 3000 across a thousand miles, because the price of a mile is the ceiling of the stock over the truck's capacity: three apples, then two, then one. The continuous optimum is 2500/3, but rounding the switch point down to mile 333 leaves 2001 apples needing three passes and produces a fake 834. A lower bound on loaded traversals proves 833 optimal without a dynamic programme, and the free return legs are the clause that separates 833 from 533.
En devinant au hasard, les prisonniers valent 7,9 x 10^-31. En suivant le papier qu'ils viennent de trouver, ils valent 0,311828, et l'écart de trente ordres de grandeur tient dans une règle d'une phrase. La stratégie ne rend personne plus chanceux : elle ne fait que rendre les échecs simultanés, et c'est là toute la leçon.
On the unit interval the indicator of the rationals has upper sum 1 and lower sum 0 for every partition ever written, so the two never meet and Riemann returns nothing at all. Give the k-th rational an interval of width ε/2^k and the whole countable set sits inside a total length of ε, which puts its measure at 0 and its Lebesgue integral at 0. Lebesgue's criterion turns that into the general law — a bounded function on a compact interval is Riemann integrable exactly when its discontinuities have measure zero — which is why Thomae's function, discontinuous on the same dense set, is integrable and this one is not. The trade is not free: sin x over x on the half line has an improper Riemann value of π/2 and no Lebesgue integral at all.
A counterfeit coin that might be heavy or light, a balance that only reports which side falls, and a hundred dollars a weighing. Counting rules out four; only a construction gets you five.
Une droite à coefficients rationnels envoie ℚ sur ℚ. Aucune courbe n'y parvient jamais. Trois filtres — interpolation, forme, dénominateurs — laissent la classification complète.