Take a draw U, take its opposite 1 − U, and average the two results. If the quantity is monotone in U the two move in opposite directions, the covariance is negative, and the variance of the pair mean falls — to exactly zero in the best case. If the quantity is symmetric about 1/2 the two are identical, the covariance is maximally positive, and the pair costs two draws to produce an observation no better than one.
Core ideas
Three ideas. The first is the mechanism, the second is the condition, and the third is the accounting error that lets this technique take credit it has not earned.
The estimator is the average of the pair means, and each pair mean is one observation of it. That is the whole of the accounting, and it is the only way to compare fairly: a pair consumes two draws, so it must be judged against an independent pair, which also consumes two.
If g is non-decreasing in U then g(U) and g(1 − U) move in opposite directions and their covariance is negative — that is the whole argument. Inverse-transform sampling is monotone in U by construction, which is why the technique is available at all on this path. Where monotonicity fails the covariance can be positive, and then the pairing costs rather than pays.
Var((A + B)/2) = (Var A + Var B + 2 Cov(A, B))/4. With A and B identically distributed this is (Var A + Cov)/2, against Var A/2 for an independent pair. The entire effect is the covariance term, and it can take the variance anywhere from zero to twice the independent value.
The material
How pairing induces a covariance, what monotonicity has to do with its sign, and what the pair has to be compared against.
Given one uniform draw U, the antithetic pair is U and 1 − U. If U is uniform on [0, 1) then so is 1 − U, so each half of the pair is a legitimate draw and the estimator built on the pair mean is unbiased for the same quantity as the ordinary one.
For A = g(U) and B = g(1 − U) with the same distribution, Var((A + B)/2) = (Var A + Cov(A, B))/2. An independent pair has Cov = 0 and therefore variance Var A/2, so the antithetic pair is better exactly when Cov(A, B) < 0 and worse exactly when it is positive.
Var((A + B)/2) = (Var A + Var B + 2 Cov(A, B))/4, which is the expansion of the variance of a sum divided by four.
A and B have the same distribution, because U and 1 − U do, so Var B = Var A and the expression is (2 Var A + 2 Cov)/4 = (Var A + Cov)/2. Setting Cov = 0 gives the independent-pair value Var A/2, and the comparison is the sign of Cov.
□
So the technique has no magic in it and no tuning: it is a choice about which second draw to take, and the choice is good or bad according to the sign of one number. The panel prints that number.
Sixty-four pairs from the standard stream. Whenever U ≥ 1/2 the partner 1 − U is below 1/2, and conversely, so one of the two is always 1 and the other always 0: every pair mean is exactly 1/2. The sample variance of the pair means is exactly 0, the correlation within a pair is exactly −1, and an independent pair would have had variance 247/2016.
Look at what that does to the estimate. The sixty-four single draws came out 1 thirty-eight times, so an ordinary estimator on the same draws would report 19/32 = 0.59375 against a true value of 1/2. The antithetic estimator reports 1/2, exactly, because every pair mean is 1/2, exactly. The variance is not small here; it is zero, and the arithmetic says so rather than the picture.
|(1 − U) − 1/2| = |1/2 − U|, so g(1 − U) is g(U) identically: the two halves of the pair are the same number. The correlation is exactly +1, the pair mean has exactly the variance of a single draw — both 0.022439, exact fractions shown to six places — and an independent pair would have had 0.011219. The ratio is exactly 2.
Two draws were consumed to produce one observation that is no better than one draw. That is not a mild disappointment; it is exactly double the work for exactly the same width, and it is what happens when the condition fails. A reader who has only ever seen the technique work cannot tell the condition from the conclusion, which is why both cases are on the page rather than one of them being mentioned in a footnote.
The squared correlation is 1 in both columns, and that is the sharpest way to see why ρ² is not the whole story here. ρ² = 1 says the two halves of the pair determine each other; the sign of the covariance says whether that is worth nothing or worth everything. Later on this course ρ² is exactly the right summary, because there the sign has been optimised away; here it is not.
Inverse-transform sampling returns the first value whose cumulative probability reaches U, so a larger U never gives a smaller value: the sampler is monotone in U by construction. Any quantity that is a non-decreasing function of the sampled values is therefore monotone in U, and the pairing is guaranteed to have a negative covariance. That is a property of the method built in the opening lesson of this course, not a lucky feature of the examples.
It stops being guaranteed the moment a run uses more than one draw in a way that is not monotone in each of them, which is most interesting simulations. The queue runs earlier on this course are not monotone in the service draws in any useful sense, which is why the pairing is demonstrated here on functions of a single draw, where the claim can be exact, rather than asserted on a system where it would have to be hoped for.
Interactive
Pair U with 1 − U: a negative covariance halves the width, and a positive one wastes the draw
Method
Four steps, and the last is the one that decides whether the technique is being measured or flattered.
Not “roughly increasing” — monotone. If a larger U can give a smaller output anywhere, the covariance is not guaranteed negative, and the technique becomes something to measure rather than something to rely on.
Both halves are uniform, so both are legitimate draws, and each half on its own is an unbiased sample. Nothing about the model changes; only the second draw's provenance does.
The sequence of pair means is what the estimator is built from. Its length is half the number of draws, and that is the correct denominator for every variance computed afterwards.
A pair costs two draws either way. Comparing an antithetic pair mean against one draw makes the technique look like a factor-of-two improvement even when the covariance is zero, which is how it is routinely credited with a reduction it did not make.
The two quantities are functions of a single uniform draw, which is what makes both claims exact rather than measured. g(U) = 1 when U ≥ 1/2 is monotone and has mean 1/2; g(U) = |U − 1/2| is symmetric about 1/2 and has mean 1/4.
The 19/32 is the figure to keep. Thirty-eight of the sixty-four single draws landed above 1/2, which is a perfectly ordinary sixty-four-draw sample and would have produced an estimate of 0.59375 with a standard error of about 0.061. The same draws, paired with their opposites, produce 1/2 with no error at all — not a small error, none.
That happens because this g takes two values and the pairing balances them perfectly. On a monotone g with more values the covariance is still negative and the reduction is still real, but it is partial, and the number to report is the ratio of the two variances rather than the word “antithetic”.
For a faded rehearsal, halve the number of pairs from 64 to 32 in each case. The supplied first move is that in the monotone case nothing whatever happens to the variance of the pair mean: it is zero at every count, because every individual pair mean is 1/2. Say what happens to the symmetric case's ratio, and why that one is also insensitive to the count.
You should be able to derive Var((A + B)/2) = (Var A + Cov)/2 for identically distributed halves, say why monotonicity in U makes the covariance negative, state the correct comparison and why it is an independent pair, produce the exact zero in the best case and the exact factor of two in the worst, and explain why ρ² is not sufficient to tell those two cases apart.
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