2Making a Market

Answer to Question 2.1.

Quote 10 bid, 11 asked, and expect to earn 0.5. A quote of 11 bid, 12 asked also earns 0.5 on average.

Each die averages 3.5, so the sum is worth 10.5. If the customer sells to you at 10, you gain 0.5 on average, and if they buy at 11, you gain 0.5. At 11 and 12, buying at 11 loses 0.5 and selling at 12 gains 1.5, and the average of the two is again 0.5. A customer whose side is a coin toss pays you half your width wherever the quote sits.

Answer to Question 2.2.

Quote 6 bid, 8 asked, and lose $30/13 \approx 2.31$ on average. The narrowest whole-number width that loses less than 1 is 6, at 4 bid, 10 asked, which loses $12/13 \approx 0.92$.

The ranks are equally likely and centred on 7, so the symmetric quote is best. At 6 and 8, the ranks 1 to 5 sell to you and lose you 5, 4, 3, 2 and 1, and the ranks 9 to 13 buy from you and lose you 1 to 5. That totals 30 over 13 ranks. The best quote at each width loses:

Answer to Question 2.3.

Quote 14 bid, 16 asked. The contract is worth $91/6 \approx 15.17$, not $3.5^2 = 12.25$.

The expected value is $(1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6$. Of the half-steps, a mid of 15 is the nearest to 15.17 that fits a width of 2, so the quote is 14 to 16. The customer’s side is a coin toss, so the edge is 1 on average.

Answer to Question 2.4.

Quote the next card 5.75 bid, 6.75 asked, and the sum of the rest 263.5 bid, 264.5 asked. The first carries real risk, and the second carries none.

Each rank from 1 to 13 appears four times, so the whole deck totals $4 \times 91 = 364$. With 100 showing, the 42 unseen cards total exactly 264. Each is equally likely to be next, so the next card is worth $264/42 = 44/7 \approx 6.29$, and the nearest quarter gives a mid of 6.25. Your counterparty’s value lies inside both quotes, so they pass.

Answer to Question 2.5.

They sell you both $A$ at 0.55 and $B$ at 0.50, collecting 1.05 and owing exactly 1, for a sure profit of 0.05. To rule this out, the two bids must add up to at most 1, and the two asks must add up to at least 1.

Exactly one of $A$ and $B$ pays, because the two events cover every result of the die with no overlap, so a pair of one lot each is worth exactly 1 whatever happens. A pair sold to you for more than 1 is free money for the other side. A pair bought from you for less than 1 is free money too. Your asks here add up to 1.12, so that side is safe.

Answer to Question 2.6.

Quote 4 bid, 5 asked before the rolls, and 3.5 bid, 4.5 asked after the 3. After a 6, the contract is worth exactly 6, so you would show 5.5 bid, 6.5 asked.

The higher of two dice equals $k$ in $2k - 1$ of the 36 outcomes, so the value is $\sum k(2k - 1)/36 = 161/36 \approx 4.47$. Given a first die of 3, the second die gives a maximum of 3 in three cases, and otherwise 4, 5 or 6, so the value is $(3 + 3 + 3 + 4 + 5 + 6)/6 = 4$.

After the first die the values are:

Answer to Question 2.7.

Quote about 159 bid, 179 asked. There are 168 primes below 1,000, so that quote passes.

The prime number theorem says the count of primes up to $n$ is about $n/\ln n$, and a sharper form is about $n/(\ln n - 1)$. With $\ln 1000 = 3 \times 2.303 \approx 6.91$, the sharper form gives $1000/5.91 \approx 169$. The centre of the quote is 169, and 20 wide makes it 159 to 179.

If you do not recall the theorem, sieve. About $1000 \times \tfrac12 \times \tfrac23 \times \tfrac45 \times \tfrac67 \approx 229$ numbers have no factor of 2, 3, 5 or 7, and the exact count is 228. Those still include composites such as $11 \times 11$, so the answer is well below 229.

Answer to Question 2.8.

Quote 5 bid, 7 asked. Both sides earn 1 on average, and they differ completely in shape. If you are long at 5, you lose money with probability $1 - (5/6)^4 \approx 52\%$, and at most 4. If you are short at 7, you lose with probability $(5/6)^7 \approx 28\%$, and the loss has no limit.

The number of rolls is geometric with mean 6 and variance 30, so the standard deviation is about 5.5. Its median is 4, since $1 - (5/6)^3$ is below one half and $1 - (5/6)^4$ is above. Most results are below the mean, and a few are far above it.

So a long position usually loses a little and sometimes wins a lot, and a short position usually wins a little and sometimes loses a lot. It takes 20 rolls or more about 3% of the time, and then a short loses at least 13.

Answer to Question 2.9.

Quote 4 bid, 6 asked. The result lands outside the quote with probability about 29%. After a first red card, the value is $92/17 \approx 5.41$, and you quote 4.5 bid, 6.5 asked.

By symmetry each card is red with probability one half, so the expected number of reds is 5 whatever the dependence between cards. The count is hypergeometric, with variance

so the standard deviation is 1.43, below the 1.58 of ten independent coin tosses. Drawing without replacement pulls the count towards the middle. Summing the exact probabilities of 4, 5 and 6 gives 0.71.

After a red, the other nine cards come from 25 red in 51, so the value is $1 + 9 \cdot 25/51$. The first card was worth 0.5 and is now worth 1, a gain of 0.5. The other nine fall from 4.5 to 4.41, because a red has left the deck. Together the value rises by 0.41.

Answer to Question 2.10.

With $k = h/\sigma$, the expected loss is $2\sigma\left[\varphi(k) - k\,\Phi(-k)\right]$, where $\varphi$ and $\Phi$ are the standard normal density and distribution function. A 90% interval, $k = 1.645$, loses $0.042\,\sigma$. To lose less than $0.05\,\sigma$, you need $k \ge 1.57$, a width of about $3.1\,\sigma$.

The counterparty buys when $X > \mu + h$, costing you $X - \mu - h$, and sells in the mirror case. So the loss is

This is the normal loss function. At $k = 0$ it is $0.80\,\sigma$, and at $k = 1$ it is $0.167\,\sigma$.

Answer to Question 2.11.

Centre on about 254. The contract pays the number, so the fair value is the mean, and for a lognormal the mean is the median times $e^{\sigma^2/2}$. Here $\sigma = \ln 2$, so the factor is $e^{0.24} \approx 1.27$.

So $200 \times 1.27 \approx 254$. The true number exceeds 254 with probability $\Phi(-\sigma/2) = \Phi(-0.35) \approx 36\%$. The 90% range is $200 \cdot 2^{\pm 1.645}$, about 64 to 625.

Answer to Question 2.12.

Quote 94 bid, 106 asked. The break-even ask is the expected value given that someone buys, and the break-even bid is the expected value given that someone sells.

Use Bayes. If the value is 120, a buy comes with probability $0.30 + 0.70 \cdot \tfrac12 = 0.65$. If it is 80, a buy comes only from the coin toss, $0.70 \cdot \tfrac12 = 0.35$. So given a buy, the chance of 120 is $0.65/(0.65 + 0.35) = 0.65$, and the ask is $80 + 40 \times 0.65 = 106$. The bid is the mirror image, 94.

So the width is the share of knowing traders times the range, $0.30 \times 40 = 12$. This is the Glosten-Milgrom quote. Uninformed traders pay exactly what the informed ones take from you.

Answer to Question 2.13.

The quotes before the five traders are below, and the order does not matter, because each purchase multiplies the odds of 100 by 1.5, and each sale divides them by 1.5.

If the value is 100, a buy has probability $0.20 + 0.40 = 0.60$, and if it is 0, it has probability $0.40$. The likelihood ratio is 1.5, and a sale gives its reciprocal. So with $n$ net purchases the odds of 100 are $1.5^n$. The ask is the value given one more purchase, at odds $1.5^{n+1}$, and the bid is the value given one more sale, at odds $1.5^{n-1}$. With $n = 1$ the ask is $2.25/3.25$ of 100, which is 69.23. The final odds are $1.5^2$, so the chance of 100 is $9/13$.

Answer to Question 2.14.

The contract is worth 2.7 after the sale, and you quote 2.2 bid, 3.2 asked. After a pass, it is worth 3.5.

Weigh each value by the probability of a sale. For values 1 and 2 the looker sells, so that is $\tfrac12 + \tfrac12 \cdot \tfrac12 = \tfrac34$. For values 3 to 6 only the coin toss can sell, so $\tfrac14$. The weights sum to $\tfrac52$, so the posterior is $0.3, 0.3, 0.1, 0.1, 0.1, 0.1$, and the mean is 2.7.

A pass can come only from a looker, and a looker passes only on 3 or 4. So the value is one of those two, with mean 3.5. Note that this leaves the mid where it was, while the range has shrunk to a single unit.

As a check, use the law of total expectation. A sale and a purchase each have probability $\tfrac{5}{12}$ and a pass $\tfrac16$, and the purchase value is 4.3 by symmetry. Then $\tfrac{5}{12} \cdot 2.7 + \tfrac{5}{12} \cdot 4.3 + \tfrac16 \cdot 3.5 = 3.5$, the original value.

Answer to Question 2.15.

Ask about 100.23 for one lot and 106.40 for 100 lots. A single ask for both sizes is about 106.0.

The value is 110 if the buyer knows, and it averages 100 if the buyer does not. So the ask is $100 + 10 \times$ the probability that the buyer knows, given the size. Size is the signal, and Bayes gives

A single ask must break even across lots traded, not across traders. Knowing traders send $0.1(0.2 + 80) = 8.02$ lots per arrival and the others send $0.9(0.95 + 5) = 5.355$, so knowing traders send 60% of the lots, and the ask is about 106.0.

Answer to Question 2.16.

The lowest ask is 99.21 and the highest bid is 99.19, so the centre is 99.20, and you show 98.20 bid, 100.20 asked. Short 40, you show 99.80 bid, 101.80 asked. The centre moves by 0.02 against each trade.

Selling one lot at $a$ from long 40 changes your score by $a - 100 + 0.01(40^2 - 39^2) = a - 100 + 0.79$, so you accept any $a \ge 99.21$. Buying one lot at $b$ changes your score by $100 - b - 0.01(41^2 - 40^2) = 100 - b - 0.81$, so you accept any $b \le 99.19$.

With position $q$ the centre is $100 - 2 \times 0.01 \times q$. The prices at which you are indifferent are that centre plus or minus 0.01. A purchase takes $q$ up by 1, and the centre down by 0.02. This shifting of both sides together is an inventory skew.

Answer to Question 2.17.

You expect to earn 9.125 over 10 rounds, against 10 with no limit. In the long run you earn $6/7 \approx 0.86$ per round. With a limit of $\pm L$, it is $2L/(2L + 1)$ per round.

Each trade earns 1, so the question is how many customers you turn away. A position at a limit turns away half of them. Starting from flat, you can first turn someone away in round 4, and stepping the position’s probabilities forward exactly gives $73/8$ trades in 10 rounds.

For the long run, the position is a random walk that stays put when it would cross a limit. Every row and every column of its transition matrix sums to 1, so its stationary distribution is uniform on the $2L + 1$ positions. So you are at a limit with probability $2/(2L + 1)$, and you turn away half the customers then, a share of $1/(2L + 1)$. For $L = 3$, that is one customer in seven.

Answer to Question 2.18.

They can lock in a profit if your bid is above 10.8 or your ask is below 10.2. If you hedge the sale at 10.7, you lock in a loss of 0.1. If you do not hedge, you earn 0.2 on average, with a standard deviation of about 2.96.

Three single-die contracts together pay exactly the same as $S$. So they cost 10.8 to buy and fetch 10.2 to sell. If you bid above 10.8, for example 10.9, the counterparty buys the three pieces at 10.8 and sells to you at 10.9, making 0.1 for sure. If you ask below 10.2, they do the reverse. Anywhere from 10.2 to 10.8 is safe from them.

A customer lifts your 10.7, and fair value is 10.5, so you have 0.2 of edge. Buying the pieces at 10.8 gives that up and costs 0.1 more. So a quote inside the pieces’ own bid and ask can make money only by carrying the risk for a while: you hold until a customer on the other side arrives, or you decide the risk is worth 0.2.

Answer to Question 2.19.

$A$ is now worth $28/3 \approx 9.33$ and $B$ is worth $49/6 \approx 8.17$. The best hedge is only $4/23 \approx 0.17$ of a lot of $B$, and it removes about 8.7% of the variance.

They buy when $X + Y$ is at least 8, which has probability $15/36$. Then $\mathbb{E}[X + Y] = 140/15 = 9.33$. By symmetry $X$ takes half of the excess over 7, so $B$ rises by half as much as $A$: it goes from 7 to 8.17. The cross-update is the slope of $B$ on $A$, which is $\tfrac12$.

For the hedge, compute given what you now know. You are short $A$, and given $X + Y \ge 8$, the variance of $A$ is $14/9 \approx 1.56$, the variance of $B$ is $161/36 \approx 4.47$, and their covariance is $7/9$. The variance-minimising amount of $B$ is $(7/9)/(161/36) = 4/23$. It removes $(7/9)^2/\left(\tfrac{14}{9} \cdot \tfrac{161}{36}\right) \approx 8.7\%$.

Answer to Question 2.20.

$h = 1.6/10.8 \approx 0.148$. With a 50% chance of being hit, $h \approx 0.095$. With 5 customers, $h \approx 0.276$.

Over one cycle you earn $10h$ from customers. If the fast trader gets you, you sell at $f + h$ something now worth $f + 2$, or buy in the mirror image, and either way you lose $2 - h$. So break-even requires

with $n = 10$ customers, $p = 0.8$ and a move of $M = 2$. Each customer ends up paying for your share of the news: about $pM/n$ when $n$ is large.

Answer to Question 2.21.

Joining earns 0.40 per trader. Posting inside at half-width $h$ earns $h - 0.2$, so $h = 0.5$ earns 0.30, and $h = 0.9$ earns 0.70. If the rival stays put, 0.9 is best. If the rival also moves, quotes fall to a half-width of 0.3 to 0.4, just above the break-even half-width of 0.2.

Per trade at half-width $h$, an uninformed trader pays you $h$, and a knowing trader costs you $2 - h$. Together, $0.9h - 0.1(2 - h) = h - 0.2$. At 99 and 101 that is 0.8 per trade, and sharing gives 0.40. Posting inside beats joining when $h - 0.2 > 0.4$, which is $h > 0.6$, so small improvements win.

When both move, one step takes the whole flow rather than half. At a shared half-width $h$, you earn $(h - 0.2)/2$, and undercutting by 0.1 earns $h - 0.3$. Undercutting pays when $h > 0.4$. It ties at 0.4, and at 0.3 it loses. So the race stops there, one or two steps above break-even, because of the step size. At break-even, width is $2 \times 0.2 = 0.4$, which is the knowing share times the range, $0.1 \times 4$.

Answer to Question 2.22.

$h = 0.8$, a width of 1.6. With one quarter, $h = 0.375$. With three quarters, $h = 4/3 \approx 1.33$.

Someone who has seen a die $d$ values the sum at $d + 7$, so 8 to 13, or $\pm 0.5, \pm 1.5, \pm 2.5$ around 10.5. They trade only when that is outside your quote, and on average they then take the amount by which it exceeds $h$. For $h$ between 0.5 and 1.5, those amounts are 0, $1.5 - h$ and $2.5 - h$, each twice out of six, so the average loss to them is $(4 - 2h)/3$.

The coin toss pays you $h$ on average, so breaking even means

With $\alpha = \tfrac12$, this gives $h = 0.8$, and that lies in the range assumed. The other two cases solve the same way, each with its own range: $\alpha = \tfrac14$ gives $h$ below 0.5, where the loss is $(4.5 - 3h)/3$, and $\alpha = \tfrac34$ again uses the range from 0.5 to 1.5.

Answer to Question 2.23.

The quotes are below. The width first falls below 5 after four sales. It narrows as you learn, although you might expect it to grow as you take more losses. A run of four sales has probability $1/256$, about 0.4%, if the value is really 100.

A sale has probability $0.75$ if the value is 0 and $0.25$ if it is 100, so each sale divides the odds of 100 by 3. After $n$ sales the odds are $3^{-n}$. The ask is the value given one more purchase, at odds $3^{1-n}$, and the bid is the value given one more sale, at odds $3^{-n-1}$. So the ask after $n$ sales equals the bid after $n - 2$.

Width reflects how much a single trade can still tell you. At the start, the value is entirely unknown. After four sales, you are nearly sure, so a trade has little left to reveal.

Answer to Question 2.24.

On a scale of 0 to 1, the ask is $a = \dfrac{(1 + \alpha) - \sqrt{1 - \alpha^2}}{2\alpha}$, and the bid is $1 - a$. At $\alpha = 0.5$, the quote is 36.6 bid, 63.4 asked. As $\alpha$ approaches 1, the ask rises to 100 and the bid falls to 0, so no trade happens: the market breaks down.

The ask equals the expected value given a purchase. A purchase comes from a knowing trader when $V > a$, and from the coin toss at any $V$. So

Clearing the fractions gives $\alpha a^2 - (1 + \alpha)a + \frac{1 + \alpha}{2} = 0$, and the smaller root is the one in $[0, 1]$.

Small $\alpha$ gives $a \approx \frac12 + \frac{\alpha}{4}$, so the ask is about $50 + 25\alpha$. For example, $\alpha = 0.1$ gives 52.5, and $\alpha = 0.99$ gives 93.4.

Answer to Question 2.25.

The best half-width is $h = 250/21 \approx 11.9$, and it earns about 0.95 per trader. You only break even at $h = 50/7 \approx 7.14$. Above a knowing share of 25%, no quote is profitable.

A knowing trader takes $(V - 50 - h)^+$ on the upside, and the same on the downside, so with $V$ uniform that averages $(50 - h)^2/100$. An uninformed trader trades with probability $1 - h/20$ and pays $h$ on average. With share $\alpha$,

Setting the derivative to zero gives $h = 1\big/\left(\frac{1 - \alpha}{10} + \frac{\alpha}{50}\right)$. At $\alpha = 0.2$, this is $1/0.084 \approx 11.9$. Setting $\pi(h) = 0$ gives $h = 50/7$ for the smaller root.

So the two quotes tell different stories. A market maker with no competition takes 11.9. Competition pushes the half-width down to 7.14, where profit is zero. At $\alpha = \tfrac14$, the best half-width is 12.5, and it earns exactly zero, so at any larger share you should not quote at all.

Answer to Question 2.26.

Flat, you show 99.49 bid, 100.51 asked. Long 40, you show 98.69 bid, 100.00 asked. The ask reaches 100 from a position of 26.

The customer’s side is a coin toss, so choose each price separately. If a trade changes your penalty by $c$, you maximise $(\delta - c)\,e^{-2\delta}$, whose derivative vanishes at $\delta = \tfrac12 + c$.

Flat, either trade costs 0.01 in penalty, so $\delta = 0.51$. Long 40, a purchase costs $0.01(41^2 - 40^2) = 0.81$, so the bid depth is $1.31$. A sale saves $0.79$, so $c = -0.79$, and the formula says $\delta = -0.29$. But an ask through 100 gets no more fills than an ask at 100, and only lowers the price. So you stop at $\delta = 0$, and earn 0.79. This is the corner of the problem.

Your ask reaches 100 when $0.01(2q - 1) \ge \tfrac12$, so at $q = 26$. Your expected score from the trade is about 0.41, against about 0.18 for the flat quote at the same position.

Answer to Question 2.27.

With no signal, quote 97 bid, 103 asked. With the signal saying 110, quote about 103.66 bid, 107.63 asked, a width of 3.97 against 6. A signal correct only 60% of the time leaves about 98.94 to 104.72, a width of 5.78.

A public signal changes your prior, and the same Bayes step then gives the quote. Start with the signal, so the chance of 110 is 0.8, and treat a buy as a further observation with likelihoods $0.65$ and $0.35$. Then

which gives an ask of $90 + 20 \times 0.881 = 107.63$. A sale reverses the likelihoods, and that gives 103.66.

Note that the quote is not centred on the new fair value, which is $0.8 \times 110 + 0.2 \times 90 = 106$. The mid is at 105.64, because a trade can move you further towards 90 than towards 110.

Answer to Question 2.28.

At $m = 0.5$ you lose about 0.236 per deal. Break-even is $m = 35/36 \approx 0.97$. With three dealers, it is $35/24 \approx 1.46$.

You sell only when your own die is the lower one. Your ask is $d_1 + 3.5 + m$, and the contract pays $d_1 + d_2$, so your profit on a sale is $3.5 + m - d_2$, and winning selects a high $d_2$. If your die is lower you win everything, and if they tie you win half. So given $d_2$, your share is $(d_2 - \tfrac12)/6$, and

This vanishes at $m = 35/36$. Among the deals where your die is strictly lower, the rival’s die averages $14/3$, not 3.5. This is the winner’s curse: you win the trade when you have underestimated.

With three dealers, the same count over 216 outcomes gives $m = 35/24$. More rivals make it worse, because winning says more.

Answer to Question 2.29.

Quote 25 bid, 75 asked in round 1. After a purchase, quote 50 bid, 83.33 asked. After a purchase and then a sale, quote 50 bid, 50 asked. With a new person each round, the same three quotes are 25 to 75, 50 to 90, and 25 to 75.

There are four states: knows or does not, and 0 or 100, each of probability $\tfrac14$. A purchase has likelihood 1 from a knower at 100, 0 from a knower at 0, and $\tfrac12$ otherwise. So after one purchase the states have probabilities: knower at 100, $\tfrac12$; no knowledge at 100, $\tfrac14$; no knowledge at 0, $\tfrac14$. The value is 75.

The next ask is the value given a second purchase, which reweights those to $\tfrac12 : \tfrac18 : \tfrac18$, giving $100 \times \tfrac{5/8}{6/8} = 83.33$. The next bid is the value given a sale. A knower who has just bought cannot sell, so a sale proves they know nothing, and the value goes back to exactly 50. Since a person who reverses is uninformed and their trades now say nothing, you need no width at all.

Answer to Question 2.30.

The unconditional value is $33/4 = 8.25$. The best quote is 7.5 bid, 9.5 asked, which loses 0.05 on average. It is centred on 8.5, above the value. If they buy, the contract is worth 10, and if they pass, it is worth $145/18 \approx 8.06$. They never sell.

Their value depends only on the card they saw:

Low cards barely matter, because one of the other two is usually higher. The values are bunched at the bottom and spread at the top.

A quote centred on 8.25, at 7.25 to 9.25, has no seller either, but sells to card 10 at 9.25, losing $0.75/10 = 0.075$. Moving up to 7.5 to 9.5 cuts that to 0.05. At 7.75 to 9.75, cards 1 to 4 begin to sell to you, and the loss returns to about 0.056. With a skewed distribution of their value, the best quote is not centred on the mean. As a check, $0.9 \times 145/18 + 0.1 \times 10 = 8.25$.

Answer to Question 2.31.

If they bought $S$, then $S$ is worth 9, and $D$ stays at 7. If they buy both, both are worth 9, and they saw $P$. If they buy $S$ and sell $D$, then $S$ is worth 9 and $D$ is worth 5, and they saw $Q$. You lose 2 on average per deal.

Seeing $P = p$, they value both contracts at $p + 3.5$. Seeing $Q = q$, they value $S$ at $q + 3.5$ and $D$ at $10.5 - q$. Either way they buy $S$ when the seen die is 5 or 6. So a purchase of $S$ raises its value by $5.5 - 3.5 = 2$, whichever die it was. For $D$, it is up 2 or down 2 with equal chance, so unchanged on average.

Their trade on $D$ then tells you which die. Buying both means $P$ was the seen die, so $D$ rises to 9. Buying $S$ and selling $D$ means it was $Q$, so $D$ falls to 5.

If the seen die is 3 or 4, they pass on both, which happens with probability $\tfrac13$. Otherwise you lose 1.5 on each of two trades, and $\tfrac23 \times 3 = 2$.

Answer to Question 2.32.

Quote through the middle each time, which is a binary search: start at 5 bid, 6 asked, then split what remains. The expected total loss is $24/10 = 2.4$. With numbers 1 to 16, it is $4.75$. Walking up from 1 bid, 2 asked costs 7.

Start at 5 and 6. If they pass, the number is 5 or 6, and you never lose again, so you show 5 bid, 6 asked forever. If they sell, the number is from 1 to 4, and you have lost $5 - x$. Next quote 2 bid, 3 asked. If the number is 1, they sell again and you lose 1, and then you know it. If it is 4, they buy, and you lose 1. So the total losses for the numbers 1 to 10 are:

That sums to 24. The first quote costs 2.0 on average, and the second only 0.4 more. A dynamic program over all sets of remaining numbers confirms that no other quoting does better, and that more rounds cost nothing once you have the number.

Walking upwards loses far more. If the number is 10, you lose 8, then 6, 4 and 2, for a total of 20, and the average over all numbers is 7. Every quote is on the wrong side of the same number.

Answer to Question 2.33.

Take a side down when one more lot would add more penalty than it can repay. The largest position you will go to, by rounds left from 1 to 10, is 2, 3, 3, 4, 4, 5, 5, 5, 6, 6. The best expected score is 8.41, against 8.00 for never taking a side down, and 8.35 for the best fixed limit, which is $\pm 3$.

Work back from the end by backward induction. With one round left, a purchase at position $q$ gains $1 - 0.2(2q + 1)$, which is positive for $q = 1$ and exactly zero for $q = 2$. So you buy up to 2 and no further. With two rounds left, compute the value of each position with one round left, and compare again. A purchase now is worth more than it would be at the end, because one more trade may take you back down. So you buy up to 3.

Each step back adds time for flow to return you towards flat, so the allowed position grows with the rounds left.

Never taking a side down earns one per round, or 10, minus $0.2 \times \mathbb{E}[q^2] = 0.2 \times 10$, so 8.00. A fixed limit of $\pm 3$ gets most of the gain but is too tight early and too loose at the end.

Answer to Question 2.34.

$\lambda = \sigma_v/(2\sigma_u) = 0.01$ per share. The insider’s order is $x = (\sigma_u/\sigma_v)(V - 100) = 50(V - 100)$, and the expected profit is $\sigma_v\sigma_u/2 = 100$. After the price is set, half the variance of $V$ remains, so 2 out of 4. This is the one-period Kyle model.

The insider chooses $x$ to maximise $x(V - 100 - \lambda x)$, since the noise averages to zero. That gives $x = \beta(V - 100)$ with $\beta = 1/(2\lambda)$. You set the price to the expected value given the flow, so $\lambda$ is the regression slope of $V$ on $y$:

Substituting $\beta = 1/(2\lambda)$ gives $4\lambda^2\sigma_u^2 = \sigma_v^2$, which is the result.

The insider’s order has the same variance as the noise, because $\beta\sigma_v = \sigma_u$. It hides in the crowd exactly. So half of the variance of $y$ is signal, and the price moves half of the way to the truth. The insider’s expected profit is $\mathbb{E}\left[\beta (V-100)^2\right]/2 = 50 \times 4/2 = 100$, and the noise traders pay for it: each order $u$ moves the price against itself by $\lambda u$, which costs them $\lambda\sigma_u^2 = 100$ on average.

Answer to Question 2.35.

The bid is about 98.35, and the ask is about 100.05. The width is $\gamma\sigma^2 + \frac{2}{\gamma}\ln\left(1 + \frac{\gamma}{k}\right) \approx 0.40 + 1.29 = 1.69$, and the centre is $100 - \gamma\sigma^2 q = 99.2$, with $\gamma = 0.1$, $\sigma^2 = 4$, $k = 1.5$ and $q = 2$. As $\gamma$ goes to 0, each side goes to $1/k$ from 100, so about 99.33 to 100.67.

With no fill, exponential utility of a normal gives $U_0 = -\exp\left(-\gamma(x + 100q) + \tfrac12\gamma^2\sigma^2 q^2\right)$, where $x$ is your cash. If your bid at depth $\delta$ is taken, you pay $100 - \delta$ and hold $q + 1$, so

Here $d_b$ is the inventory cost of that purchase. Expected utility on that side is

Since $U_0$ is negative, you want the bracket small. Setting $u = \delta - d_b$, the first-order condition gives $e^{-\gamma u} = k/(k + \gamma)$, which is

For the ask, the same steps give $d_a = \tfrac12\gamma\sigma^2(1 - 2q)$. So $\delta_b = 1.0 + 0.645$ and $\delta_a = -0.6 + 0.645 = 0.045$. Holding 2 shares, you barely charge to sell, and you charge a lot to buy.

This is the one-period form of the Avellaneda-Stoikov quote. The shift depends on your position and the risk, and the width on the risk and the customers’ sensitivity to price.