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077An equity index stands at 20,000 and its implied volatility is 18% a year. Where do you think it closes in four months? Give a central value and a 90% range you would be willing to make a market around.Market making, betting and sizingHardMorgan StanleyTokyo · 2025
Try it first
Roughly how wide is a 90% range for the index four months out?
Show the worked solution
Centre on today's level, about 20,000, with a 90% range of roughly 16,800 to 23,600. Four months is a third of a year, so one standard deviation is 18% x sqrt(1/3), about 10.4%. In log terms the 90% band is 1.645 of those either side, which gives 16,767 and 23,601. The median sits a little below 20,000 and the upside tail is longer than the downside.
Why is a single number the wrong answer?
Ask a cab driver how long the airport run takes and a good one says forty minutes, maybe an hour in traffic. The range is the useful part, because you plan your flight around it. A trading interviewer asking where an index closes wants a distribution, because a market maker quotes against the spread of outcomes, not against a guess. Your central value should not be a view on the economy either: with no edge, the best central estimate of a traded index is roughly its forward, which for four months is close to today's 20,000 once financing and dividends roughly offset.
The width comes from the implied volatility the market already quotes. Volatility grows with the square root of time, because independent daily moves add their variances, not their standard deviations. Four months is a third of a year, so one standard deviation is 18% x sqrt(1/3) = 10.4%, about 2,078 index points.
With 18% volatility over four months, the index's middle 90% runs from about 16,767 to 23,601, which is 3,233 points below today's level and 3,601 points above, because a lognormal distribution stretches further up than down.
The relationship
$S_T = S_0\, e^{-\sigma^2 T/2 + \sigma\sqrt{T} Z} \qquad \text{5th, 95th pct} = S_0\, e^{-\sigma^2T/2 \mp 1.645\,\sigma\sqrt{T}}$
S_0 today's level, 20,000
sigma implied volatility, 0.18 a year
T time in years, 1/3
Z a standard normal draw
What it says in wordsLog returns are normal with a standard deviation of sigma times root T, and a small drift correction keeps the mean at today's level.
Why is the range lopsided, and where does the median sit?
A fall of 10% and a rise of 10% are not mirror images in log space. Normal log returns make the upside tail longer: the 90% band stretches 3,601 points up but only 3,233 points down. The same convexity pushes the median below the mean: if the mean is 20,000, the median is 20,000 x exp(-sigma squared T / 2), about 19,892. That gap of about 108 points is small here, but it grows with volatility and time, and a candidate who names it shows they know the difference between the most central outcome and the average one.
What would you add before quoting a market on it?
Two honest caveats. Implied volatility is a price, not a forecast: it tends to sit above the volatility that is later realised, because option sellers charge for bearing crash risk, so the band built from it is usually a little wide. Against that, real index returns have fatter tails than the lognormal, so the 5% tails are more likely to hold a larger move than the curve suggests. Say both, then give your market: a tight two-way price around 20,000 if asked for the level, and the 90% band as the range you would sell outside of.
Where candidates lose it
The first loss is scaling volatility linearly with time: a third of 18% is 6%, which gives a band far too narrow. Volatility scales with the square root of time, so four months is about 10.4%, not 6%.
The second loss is answering with a macro story and a point forecast. The interviewer wants you to use the price the market already gives you, implied volatility, and to say that the honest answer is a distribution with a lopsided shape.
What the interviewer asks next
What 90% range would you give for one week out?
How would the range change if implied volatility jumped to 30%?
If you had to bet on the index finishing above 22,000, what fair probability would you quote?
Asked at Morgan Stanley, Sales and Trading, Tokyo, 2025 (Wall Street Oasis): What do you think this index will close at by the end of the year (4 months from now)
099A stock is worth either 100 or 110, with equal probability. 20% of the traders who arrive know the true value: they buy if it is 110 and sell if it is 100. The other 80% buy or sell at random, half and half. Where should a market maker set its ask so that it breaks even, on average, when someone buys from it?Market making, betting and sizingHardJane StreetNew York · 2025
Try it first
Where should the ask be?
Show the worked solution
Set the ask at 106, and by the same logic the bid at 104. If the stock is worth 110, a buy arrives with probability 0.2 + 0.8 x 0.5 = 0.6; if it is worth 100, with probability 0.4. Given a buy, Bayes puts the chance of 110 at 0.6, so the stock is worth 106 to the market maker selling it. The spread of 2 is the price of trading against informed flow.
Why can't the market maker just quote the expected value of 105?
A second-hand car dealer who pays the average price for every car will find that the owners of good cars go elsewhere and the owners of bad ones queue up. Who chooses to trade with you is information. A market maker does not care what the stock is worth on average; it cares what the stock is worth given that someone has just chosen to buy from it. At an ask of 105, noise buyers are harmless, a loss of 5 when the stock is worth 110 and a gain of 5 when it is worth 100. Informed buyers only appear in the 110 world, and they cost 0.5 per arriving trader on average, so 105 is a losing quote.
Tracing who sends a buy order in each world, buys come with probability 0.30 from the 110 world and 0.20 from the 100 world, so a buy lifts the chance of 110 from 0.5 to 0.6 and the break-even ask is 106.
The relationship
$P(110 \mid \text{buy}) = \frac{0.5 \times 0.6}{0.5 \times 0.6 + 0.5 \times 0.4} = 0.6, \qquad \text{ask} = \mathbb{E}[V \mid \text{buy}] = 100 + 0.6 \times 10 = 106$
0.6 the chance of a buy when the stock is worth 110: 0.2 informed plus 0.8 x 0.5 noise
0.4 the chance of a buy when the stock is worth 100: noise only
V the stock's true value
What it says in wordsSet the ask at the value of the stock conditional on being bought from, which Bayes' rule gives directly.
What sets the width of the spread?
The share of informed traders and the size of what they know. With a share alpha informed, a buy is alpha + (1 - alpha)/2 likely in the high world and (1 - alpha)/2 in the low world, and the ask works out to 105 + 5 alpha. The spread is a fee for adverse selection: it is zero when nobody is informed and widens to the full 100 to 110 range when everyone is. The table runs the formula for a few shares. Order processing and inventory costs add to this in real markets, but the information component is what makes spreads jump around earnings and news.
Informed share Ask Bid Spread
0% 105 105 0
10% 105.5 104.5 1
20% 106 104 2
50% 107.5 102.5 5
100% 110 100 10
The break-even spread equals the informed share times the 10-point value gap, so it is 2 at 20% informed and 5 at 50% informed.
What happens after the first trade?
The market maker updates. After one buy, the chance of 110 is 0.6, and if a second buy arrives the same Bayes step lifts it to 0.692, so the next ask is about 106.92. Each order moves the quotes towards the true value, which is how prices come to reflect what the informed traders know. The limitation is that the model has one share size, no inventory risk and no competition between market makers; real desks also skew quotes to manage position, which this puzzle leaves out.
Where candidates lose it
The fast wrong answer is 105, the unconditional expected value. It ignores that the act of buying is evidence: informed traders buy only when the stock is worth 110, so a market maker at 105 loses on every informed buyer and only breaks even on noise traders.
The second loss is overreacting and quoting 110 because some buyers are informed. Most buyers are noise traders, and a quote at 110 drives them away. Bayes gives the exact weight, 0.6 on the high value, and the ask of 106.
What the interviewer asks next
Where should the bid be, and why is the spread symmetric here?
After one buy at 106, where is the next ask?
How does the spread change if half of the traders are informed?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): It was a probability theory based quant trading style market making questions which were intense

S_0	today's level, 20,000
sigma	implied volatility, 0.18 a year
T	time in years, 1/3
Z	a standard normal draw

0.6	the chance of a buy when the stock is worth 110: 0.2 informed plus 0.8 x 0.5 noise
0.4	the chance of a buy when the stock is worth 100: noise only
V	the stock's true value

Informed share	Ask	Bid	Spread
0%	105	105	0
10%	105.5	104.5	1
20%	106	104	2
50%	107.5	102.5	5
100%	110	100	10

Try it first

Why is a single number the wrong answer?

Why is the range lopsided, and where does the median sit?

What would you add before quoting a market on it?

Where candidates lose it

What the interviewer asks next

Try it first

Why can't the market maker just quote the expected value of 105?

What sets the width of the spread?

What happens after the first trade?

Where candidates lose it

What the interviewer asks next