Mutual Fund Mastery puzzles, solved step by step

Puzzles: 100
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037One investor earns 15% a year for 15 years and then 9% a year for 15 years; another earns 9% first and 15% later. With a lump sum they end with identical money. With a Rs 10,000 monthly SIP they do not. Who ends richer, and why does the order matter only when money arrives over time?Compounding and time valueHardIndian AMCsDistribution and sales
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With the monthly SIP, who ends the 30 years richer?
Show the worked solution
The investor who gets 9% first and 15% later ends richer, about Rs 3.62 crore against Rs 2.61 crore. A lump sum sees every year's return, and multiplication does not care about order: 1.15 to the 15 times 1.09 to the 15 is 29.64 either way. An SIP's money arrives over time, so the pot is small early and large late. Whichever return lands on the large late pot decides the result.
Why does order not matter for a lump sum?
A shop that marks a price up 15% and then 9% ends at the same tag as one that marks up 9% and then 15%; both multiply the price by 1.15 and by 1.09. A lump sum is one block of money exposed to every year's return, and because returns multiply, their order cannot change the product. Rs 10 lakh grows by 29.64 times in either order, to Rs 2.96 crore.
A Rs 10 lakh lump sum ends at Rs 2.96 crore in either order, but a Rs 10,000 monthly SIP does not: the 15%-first corpus leads at year 15 with Rs 61.6 lakh against Rs 36.9 lakh, is overtaken in year 24, and ends at Rs 2.61 crore against Rs 3.62 crore.
What breaks the symmetry in an SIP?
Think of a savings jar you add Rs 100 to every week. A bonus that doubles the jar is worth far more in December than in January, because in December the jar holds a year of savings. In an SIP, each instalment only experiences the returns after it is paid, so a return applied when the pot is large moves more rupees than the same return applied when the pot is small. After 15 years of Rs 10,000 a month, Rs 18 lakh has gone in and the pot holds Rs 61.6 lakh or Rs 36.9 lakh. The next 15 years apply to that pot plus Rs 18 lakh more, and it is the 15% that lands on it in one order and the 9% in the other.
The relationship
$\text{corpus} = \sum_{t=1}^{360} m \prod_{s=t}^{360} (1 + i_s)$
m the monthly instalment, Rs 10,000
i_s the return in month s, 15% or 9% a year as a monthly rate
t the month the instalment is paid
What it says in wordsEach instalment grows only by the returns from its own month onward, so late returns touch every instalment and early returns touch only the first few.
What does this mean for a real SIP investor?
For anyone still contributing, the returns in the final years carry the most weight, because that is when the most money is exposed. A poor decade at the start of an SIP mostly costs a little growth on a small pot; the same decade at the end hits the whole accumulated corpus. The investor who saw 15% first was ahead by Rs 24.7 lakh at year 15 and still finished behind. The limits: real returns do not arrive in two neat blocks, and nobody can choose the order. The point is what the order does, which is why a plan near its goal often moves money towards steadier assets.
Where candidates lose it
The trap is answering that order does not matter at all, because candidates remember that multiplication commutes and apply it to the SIP. That holds only when the same money sees every return. Say who is exposed to which year before you answer.
The second trap is picking the 15%-first investor because early growth compounds for longer. That is true per rupee, but there are very few rupees early on. The weight of the pot, not the length of compounding, decides it.
What the interviewer asks next
Which order wins for a retiree who is withdrawing rather than contributing?
If the SIP stopped at year 15 and the money stayed invested, would order matter for the rest?
How would a step-up SIP change the size of the gap between the two orders?
039Give an equation for the surface area of an n by n by n Rubik's cube, counted in small unit squares. Then give an equation for how many of the small cubes show at least one face.Logic and numeracy brainteasersHardT. Rowe PriceBaltimore · 2020
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For a standard 3 by 3 by 3 cube, how many small cubes show at least one face?
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The surface is 6n² unit squares, and n³ minus (n minus 2)³ cubes show at least one face. Six faces each carry n by n squares. For the cubes, count what is hidden: peeling one layer off every side leaves an (n minus 2) cube inside, so the visible ones are n³ minus (n minus 2)³, which expands to 6n² minus 12n plus 8. For a 3 by 3 by 3 cube that is 54 squares and 26 cubes.
Why are there two different counts here?
A house with a corner room has windows on two walls of that room, but it is still one room. Counting windows and counting rooms give different numbers. The surface counts squares, and a corner cube carries three squares and an edge cube two, so the square count is always larger than the number of cubes that show. Keeping the two counts apart is half the problem; candidates who blur them give 54 for both.
The surface is the easy part. Each of six faces is an n by n grid, so the area is 6n² unit squares: 54 for a standard cube. A quick check is to rebuild it from the cube types: 8 corners showing 3 squares, 12(n minus 2) edge cubes showing 2, and 6(n minus 2)² centre cubes showing 1. For n of 3 that is 24 plus 24 plus 6, which is 54 again.
On a 4 by 4 by 4 cube the skin carries 96 squares, but only 56 cubes show, because each corner cube shows three squares and each edge cube two; the dashed 2 by 2 by 2 core of 8 cubes is hidden, and 64 minus 8 is 56.
Why count the hidden cubes instead of the visible ones?
The hidden cubes form one clean block, (n minus 2) on every side, so counting them and subtracting from n³ avoids every double count. Counting the visible cubes directly means adding corners, edges and faces while remembering that each edge already lost its corners. Both routes give the same answer, and the second makes a good check.
The relationship
$V(n) = n^3 - (n-2)^3 = 6n^2 - 12n + 8 \qquad V(3) = 27 - 1 = 26$
n small cubes along one edge
(n-2)^3 the hidden core after peeling one layer from every side
V(n) cubes showing at least one face
What it says in wordsVisible cubes are all the cubes minus the core you cannot see; expanded, it is the surface area less a correction for corners and edges.
The expanded form is worth a sentence because it links the two answers. 6n² is the square count; the minus 12n plus 8 removes the extra squares that edge and corner cubes contribute. Say the edge case too: the formula needs n of at least 2. For n of 1 it gives 2, but a single cube is one cube.
Where candidates lose it
The trap is giving 6n² as the answer to both questions. The interviewer is checking whether you notice that one cube can show up to three squares. Name the two counts before you write anything.
The second loss is trying to count visible cubes from the surface and getting tangled in double counts at the edges. Go to the hidden core first, then offer the corner, edge and face breakdown as the check.
What the interviewer asks next
How many small cubes show exactly two faces, as a formula in n?
For which n is the hidden core larger than the visible shell?
How would the counts change for a 4 by 5 by 6 box?
Asked at T. Rowe Price, Equities, Baltimore, 2020 (Wall Street Oasis): Give an equation that yields the surface area of an n by n by n Rubic's cube based on number of blocks per side

040Two retirees each start with Rs 60 lakh and withdraw Rs 6 lakh at the start of every year. Both see the same ten yearly returns, averaging 6.8%: one meets minus 20% and minus 10% in the first two years, the other in the last two. After ten years one has about Rs 2 lakh left and the other about Rs 39 lakh. Why?Withdrawals and after-tax arithmeticHardWealth and advisoryDistribution and sales

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What explains the gap of about Rs 37 lakh?

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Because withdrawals turn the order of returns into a permanent difference. Without withdrawals both would end at the same Rs 108.1 lakh, since returns multiply in any order. But the first retiree takes Rs 6 lakh out of a pot that has just fallen, to Rs 33.5 lakh after two years, so later good years work on a small base and she ends with about Rs 2.0 lakh. The second takes withdrawals from a pot that has grown, and ends with about Rs 39.0 lakh.

Why would order not matter without withdrawals?

Returns multiply, and multiplication does not care about order. Rs 60 lakh left untouched through these ten years ends at Rs 108.1 lakh whichever year comes first. The withdrawals are what break the symmetry: a fixed rupee amount taken out of the pot is a bigger share of a small pot than of a large one.

What exactly happens in the first two years?

Think of a shopkeeper who must pay rent of Rs 6,000 on the first of every month by selling stock. If prices have just crashed, he sells many more items to raise the same Rs 6,000, and those items are gone when prices recover. A retiree withdrawing a fixed sum after a fall sells more units at low prices, and those units never take part in the recovery. The first retiree's Rs 54 lakh, after the first withdrawal, meets minus 20% and becomes Rs 43.2 lakh; Rs 37.2 lakh then meets minus 10% and becomes Rs 33.5 lakh. The second retiree's Rs 54 lakh meets plus 20% first and is Rs 64.8 lakh after year one.

Both retirees withdraw Rs 6 lakh at the start of each year and see the same returns averaging 6.8%. The one who meets minus 20% and minus 10% first is down to Rs 33.5 lakh after two years and ends with Rs 2.0 lakh, while the one who meets them last ends with Rs 39.0 lakh.

Year	Return, bad first	Corpus, Rs lakh	Return, bad last	Corpus, Rs lakh
1	-20%	43.2	+20%	64.8
2	-10%	33.5	+13%	66.4
3	+18%	32.4	+5%	63.5
4	+15%	30.4	+7%	61.5
5	+11%	27.1	+9%	60.5
6	+9%	23.0	+11%	60.5
7	+7%	18.2	+15%	62.6
8	+5%	12.8	+18%	66.8
9	+13%	7.6	-10%	54.8
10	+20%	2.0	-20%	39.0

Year by year: Rs 6 lakh is withdrawn at the start of each year, then that year's return applies to what is left. The returns are the same set in reverse order.

The relationship

V_{t} = (V_{t-1} - W)(1 + r_t), \qquad V_0 = 60,\; W = 6

V_t	the corpus at the end of year t, Rs lakh
W	the withdrawal at the start of each year, Rs 6 lakh
r_t	the return in year t

What it says in wordsEach year the withdrawal comes out first and the return applies only to what is left, so a fall before a withdrawal makes the withdrawal bite harder.

What does an adviser do with this?

The years just before and just after withdrawals begin carry the most weight; the risk is called sequence risk, and it is managed, not predicted. Common responses are holding two or three years of withdrawals in a steadier, short-term bucket so that a bad year does not force selling equity at a low, or trimming withdrawals after a fall. Note the mirror image of an SIP: for a saver adding money, bad years early are the kinder order; for a retiree taking money out, bad years early are the cruel one. The returns here are illustrative, and a 10% withdrawal rate is high by most planning conventions, which is what makes the gap so stark.

Where candidates lose it

The trap is answering that the average return decides the outcome, or insisting the order cannot matter because returns multiply. Both are true only for money that sits untouched. The question tells you withdrawals happen; build from there.

The second loss is explaining it as bad luck in general. The interviewer wants the mechanism in one sentence: fixed withdrawals after a fall sell more units at low prices, and those units miss the recovery.

What the interviewer asks next

How large a cash bucket would have let the first retiree avoid selling equity in years one and two?
If withdrawals were 4% of the current corpus instead of a fixed Rs 6 lakh, would the order still matter?
How would you explain sequence risk to a client in two sentences?

049A Rs 10,000 monthly SIP runs for 20 years at 12% a year, taken as 1% a month. What share of the final corpus appears only in the last five years? Most people guess about a quarter.Compounding and time valueHardIndian AMCsDistribution and sales
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What share of the 20-year corpus is added in years 16 to 20?
Show the worked solution
About half: 49.5%. At 1% a month the corpus is about Rs 50.5 lakh after 15 years and Rs 99.9 lakh after 20, so the final five years add Rs 49.5 lakh. Only Rs 6 lakh of that is new instalments; about Rs 41.2 lakh is growth on the pot already built. An SIP's corpus is back-loaded, which is why stopping early costs far more than the missed instalments.
Why is the guess of a quarter so far off?
A quarter assumes the corpus grows by the same amount each year, like a piggy bank filled with the same coins every month. Compounding is not a piggy bank. Each year's growth is a percentage of a pot that is already larger than the year before, so the rupee gains get bigger every year and the corpus is heavily back-loaded. After 15 years of Rs 10,000 a month the pot holds about Rs 50.5 lakh, and five more years at 1% a month multiply that by 1.01 to the 60th, about 1.82, before a single new instalment is counted.
The SIP corpus reaches Rs 23.2 lakh at year 10 and Rs 50.5 lakh at year 15, then adds Rs 49.5 lakh in the last five years to end at Rs 99.9 lakh; of that addition, Rs 41.2 lakh is growth on the year-15 pot and only Rs 6 lakh is new instalments.
Where exactly does the last five years' money come from?
Split the Rs 49.5 lakh three ways. The 60 new instalments put in Rs 6 lakh. Those instalments earn about Rs 2.2 lakh of growth by year 20. The year-15 pot of Rs 50.5 lakh earns about Rs 41.2 lakh. The engine of the final years is the money already invested, not the money still to come. For contrast, the first ten years build only about Rs 23.2 lakh, 23% of the final corpus, though half the instalments are paid in them.
The relationship
$\text{share}_{16\text{-}20} = 1 - \frac{FV_{180}}{FV_{240}}, \qquad FV_n = m\,\frac{(1.01)^n - 1}{0.01}\,(1.01)$
m the monthly instalment, Rs 10,000
FV_n the corpus after n monthly instalments paid at the start of each month
1.01 one plus the 1% monthly return
What it says in wordsThe last five years' share is one minus the year-15 corpus divided by the year-20 corpus.
What does this mean for an investor thinking of stopping at year 15?
It depends what stopping means. Redeeming at year 15 gives up about Rs 49.5 lakh, half the eventual corpus, to save Rs 6 lakh of instalments. Pausing the instalments but staying invested is far cheaper: the year-15 pot alone grows to about Rs 91.7 lakh, so the cost is about Rs 8.2 lakh. The 12% is an assumption for the arithmetic; real returns vary year to year, and a bad last five years would cut the back-loaded gain sharply, which is the sequence risk of an SIP.
Where candidates lose it
The trap is answering a quarter because five years is a quarter of twenty. It treats the corpus as if it grew in a straight line. Name the compounding before you name a number.
The second loss is crediting the last five years' gain to the last five years' instalments. Only Rs 6 lakh of the roughly Rs 49 lakh is new money; most of it is growth on the pot built in the first fifteen years.
What the interviewer asks next
What share of the corpus appears in the last five years of a 30-year SIP at the same return?
At 8% a year instead of 12%, is the corpus more or less back-loaded?
Why does starting an SIP five years earlier matter more than adding five years at the end?
051A Rs 800 crore debt fund writes a Rs 40 crore bond down to zero and segregates it into a separate portfolio. An investor holds 10,000 units bought at a NAV of Rs 25. What does she hold after segregation, at what NAVs, and what does she receive if 60% of the bond is later recovered?NAV, units and fund mechanicsHardFund operationsFixed income desks
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Right after segregation, what does her holding look like?
Show the worked solution
She holds 10,000 main units at Rs 23.75, worth Rs 2,37,500, plus 10,000 segregated units valued at zero. The fund has 32 crore units; Rs 760 crore of healthy bonds over those units gives Rs 23.75. If 60% of the bond comes back, Rs 24 crore is spread over 32 crore segregated units, Rs 0.75 each, so she receives Rs 7,500. Her total is Rs 2,45,000 against Rs 2,50,000 before.
Why split the fund instead of simply marking the bond to zero?
Picture a housing society that lends money to a builder who then stops paying. If the society writes the loan off and lets members leave with their share of what is left, the members who leave first give up their slice of any money later recovered, and the ones who stay collect it all. A plain write-down hands the value of a future recovery to whoever is still in the fund when the cash arrives, not to whoever owned the fund when the loss happened. A segregated portfolioA separate pool, created on a credit event, that holds only the troubled bond. Everyone who held the fund that day gets matching units in it. fixes that by giving every holder on the day a separate, frozen claim on the bad bond.
Here the fund has Rs 800 crore at a NAV of Rs 25, so it has 32 crore units. The bond of Rs 40 crore moves out; Rs 760 crore of healthy bonds stays behind. Divide by the same 32 crore units and the main NAV is Rs 23.75. Her 10,000 units there are worth Rs 2,37,500, and she also receives 10,000 segregated units, valued at nothing today.
On the credit event the fund splits: the main portfolio keeps Rs 760 crore at a NAV of Rs 23.75, the defaulted bond moves into a segregated portfolio at zero with matching units, and a 60% recovery later pays Rs 0.75 a unit, taking her total to Rs 2,45,000 against Rs 2,50,000 before.
What does the recovery pay, and who would have got it without segregation?
A 60% recovery brings back Rs 24 crore. Spread across 32 crore segregated units it is Rs 0.75 a unit, so she receives Rs 7,500 whether or not she has since redeemed her main units. Without segregation, the same Rs 24 crore would land on whoever held the single fund on recovery day. Suppose half the units redeem at Rs 23.75 after the write-down: the recovery then lands on only 16 crore units, Rs 1.50 each. The holders who stayed collect Rs 15,000 per 10,000 units, double their fair share, and the holders who left get nothing.
The relationship
$\text{NAV}_{\text{main}} = \frac{800 - 40}{32} = 23.75 \qquad \text{recovery per unit} = \frac{0.6 \times 40}{32} = 0.75$
800 fund assets before the default, Rs crore
40 the defaulted bond, Rs crore
32 units outstanding, crore, fixed on the day of segregation
0.6 the share of the bond later recovered
What it says in wordsBoth portfolios divide by the same unit count, because every holder on the day gets one segregated unit per main unit.
Say the limit too. Segregation does not reduce the loss; she is still Rs 5,000 down on Rs 2,50,000. It only makes sure the loss and any recovery fall on the same people. Indian rules allow it only after a defined credit event such as a rating downgrade, with conditions set in SEBI circulars that should be checked before quoting them.
Where candidates lose it
Most candidates stop at the main NAV of Rs 23.75 and forget the second set of units. The interviewer is testing whether you know the investor keeps a claim on the bad bond, which is the whole reason segregation exists.
The second loss is dividing the recovery by the wrong number. The segregated units match the units outstanding on the day of the event, 32 crore, not the units left in the main portfolio after later redemptions or new purchases.
What the interviewer asks next
A new investor buys the main portfolio the day after segregation. Does she get any segregated units?
Why might a fund manager prefer to hold a defaulted bond at a small positive value instead of zero?
The recovery arrives in three instalments over two years. How is it paid out?
052A company trades at 10 times EBITDA of Rs 200 crore. It has net debt of Rs 500 crore, depreciation of Rs 40 crore, interest of Rs 50 crore and a 25% tax rate. What P/E is the market paying?Valuation riddlesHardHoulihan LokeyChicago · 2026
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Before you work it: which is closest to the P/E?
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About 18.2x. Enterprise value is 10 x 200, Rs 2,000 crore. Take off Rs 500 crore of net debt and shareholders own Rs 1,500 crore. Earnings for shareholders are EBITDA of 200 less depreciation of 40 and interest of 50, which is 110 before tax and 82.5 after 25% tax. Rs 1,500 crore over Rs 82.5 crore is 18.2x.
Why can you not read the P/E straight off the EV multiple?
Think of a house worth Rs 1 crore with a Rs 40 lakh home loan. The owner's stake is Rs 60 lakh, and the rent left for the owner is the rent after the loan interest is paid. An EV multiple compares the whole business with profit before lenders and the tax office are paid; a P/E compares the shareholders' slice with the profit left for them. So you must take both the value and the earnings down to the shareholder line before you divide. Converting one side and not the other is the usual slip.
Enterprise value of Rs 2,000 crore less Rs 500 crore of net debt leaves Rs 1,500 crore of equity, and EBITDA of Rs 200 crore less depreciation, interest and tax leaves Rs 82.5 crore of net profit, so the P/E is 18.2x.
Why does the P/E come out higher than the EV multiple here?
Both sides shrink on the way down, but not by the same share. Value falls from 2,000 to 1,500, a quarter lost to lenders. Earnings fall from 200 to 82.5, well over half lost to depreciation, interest and tax, so the denominator shrinks faster and the multiple rises. Heavy depreciation, heavy interest or a high tax rate all push the P/E above the EV/EBITDA; a debt-free, low capex business has the two much closer together.
The relationship
$\text{P/E} = \frac{10 \times 200 - 500}{(200 - 40 - 50)(1 - 0.25)} = \frac{1{,}500}{82.5} \approx 18.2$
10 x 200 enterprise value, EV/EBITDA times EBITDA
500 net debt, the lenders' claim
200 - 40 - 50 profit before tax after depreciation and interest
1 - 0.25 the share of pre-tax profit kept after 25% tax
What it says in wordsTake the lenders out of the value and the lenders, the asset wear and the tax out of the earnings, then divide.
A quick check catches the common half-conversion. If you remember the tax but forget the interest, earnings are 160 x 0.75, which is 120; divide the full EV of 2,000 by that and you get 16.7x, a number that mixes the whole business with a shareholders' profit line. State the assumption that net debt is the only claim between EV and equity: minorities or preference capital would sit there too.
Where candidates lose it
The fast wrong answer is 10x, treating the two multiples as the same thing with a different name. The second wrong answer converts only one side: equity value over EBITDA, or EV over net profit, each of which pairs a value with an earnings line that belongs to someone else.
Say the matching rule before you calculate: enterprise value goes with profit before interest, equity value goes with profit after interest and tax. Then the arithmetic takes thirty seconds.
What the interviewer asks next
The company repays Rs 200 crore of debt from cash. What happens to the P/E if the EV multiple stays at 10x?
Which is the better multiple for comparing two companies with very different debt levels, and why?
Interest falls to zero and the tax rate rises to 30%. What is the P/E now?
Asked at Houlihan Lokey, Private Funds Advisory, Chicago, 2026 (Wall Street Oasis): Valuation ratio questions like if EV/EBITDA is 10x, what is
053You backtest 20 fund-selection rules, none of which has any real skill. Each rule's measured alpha is pure noise with a standard deviation of 2% a year. What alpha does the best rule show in the backtest, and what should you expect from it out of sample?Statistics, correlation and diversificationHardFund research and ratingsGlobal asset managers
Try it first
Roughly what alpha does the best of the 20 rules show in the backtest?
Show the worked solution
The best rule shows about 3.7% of alpha in the backtest, and you should expect about zero from it afterwards. The largest of 20 normal draws sits about 1.87 standard deviations above the mean, and 2% noise times 1.87 is 3.7%. Picking the winner selects the luckiest draw, and luck does not repeat, so its expected alpha out of sample is the true alpha of every rule: zero.
Why does the best rule look good when none of them is?
Ask twenty people to toss a coin ten times and one of them will probably get eight heads. Nobody concludes that person has a talent for heads. When you choose the best of many attempts, you are choosing the luckiest draw, and the more attempts you make, the luckier the winner looks. One rule on its own shows more than 2% alpha only about one time in six. Among 20 rules, the best one shows more than 3% about 75% of the time.
Twenty rules with no skill scatter around zero with a 2% spread, the best of them shows about 3.7% in the backtest, and the same rule's expected alpha in the years after selection is zero.
The relationship
$E\left[\max_{i \le 20} \alpha_i\right] \approx 1.87 \times 2\% \approx 3.7\% \qquad E[\alpha_{\text{next}}] = 0$
alpha_i the backtest alpha of rule i, pure noise
1.87 how many standard deviations the largest of 20 normal draws sits above the mean, on average
2% the spread of the noise in each rule's alpha
What it says in wordsThe winner's backtest alpha measures how many rules you tried, not how good the winner is.
What should you expect out of sample, and how would you check a real rule?
Out of sample the noise is drawn again, fresh, and the rule has no skill, so its expected alpha is zero. The gap between 3.7% and zero is the price of searching, sometimes called selection biasThe distortion that comes from reporting the best of many tries as if it were the only try. The winner looks better than its true quality.. The honest checks follow from that. Hold back data the rules never saw and test only the winner on it. Raise the bar with the number of rules tried: a one-rule test might accept 2 standard deviations, but after 20 tries the best result is expected to reach 1.87 on luck alone. Ask whether the rule has an economic reason to work.
This is also why fund ranking tables are a weak guide on their own. A category with 20 funds and no skill still produces a fund that beat its peers by about 3.7% a year over the backtest window, and the marketing for that fund writes itself. The limit of the arithmetic: real rules are correlated, which shrinks the effective number of tries and the size of the winner's luck.
Where candidates lose it
The trap is answering zero to the first half. Every rule averages zero, but the question asks about the best one, and the maximum of 20 draws is far from the average draw. Candidates who say zero have not noticed the selection step.
The mirror trap is answering 3.7% to the second half and believing the winner will keep it. The interviewer wants both halves: a large number in the backtest and zero afterwards, with the reason.
What the interviewer asks next
How would the best backtest alpha change if you tested 200 rules instead of 20?
Your rules are strongly correlated with each other. Does the winner look more or less lucky?
How many years of out-of-sample data would you need to tell a true 2% alpha from zero at this noise level?
054At an 8% yield, which has the highest duration: a 10-year zero-coupon bond, a 10-year 9% coupon bond, or a 15-year 12% coupon bond?Bond maths and durationHardFixed income desksIndian AMCs
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Gut call first: which bond carries the most interest rate risk?
Show the worked solution
The 10-year zero, with a duration of 10.0 years. A zero pays everything at maturity, so its duration equals its maturity. The 15-year 12% bond has a duration of about 8.6 years, because its large coupons bring value forward, and the 10-year 9% bond is about 7.1. The zero moves most when rates move, even though it is not the longest bond.
Why is maturity the wrong ruler?
Think of two loans to a friend. One friend repays everything in one go after ten years. The other repays a large slice every year for fifteen years. Most of the second loan is back in your hands long before the first friend pays a rupee. Duration measures when, on average, a bond's value comes back to you, weighted by the present value of each payment, and rate risk follows that average, not the final date. A zero returns all of its value at one date, so its duration is exactly its maturity.
At an 8% yield the 10-year zero balances at 10.0 years, the 10-year 9% bond at 7.1 years and the 15-year 12% bond at 8.6 years, because the large coupons of the longest bond pull its balance point inside that of the zero.
How do you rank them without a calculator?
Use two rules and one check. A zero's duration is its maturity. A coupon bond's duration is always below its maturity, and the higher the coupon, the further below. So the only real contest is between the zero at 10 and the 15-year bond, and a 12% coupon pulls hard: about 60% of that bond's present value arrives in the first ten years. The 10-year 9% bond cannot beat the 10-year zero, because it has the same final date and pays some value earlier.
The relationship
$D = \frac{\sum_t t \cdot \frac{CF_t}{(1+y)^t}}{\sum_t \frac{CF_t}{(1+y)^t}}$
t the year a cash flow arrives
CF_t the cash flow in year t, coupon plus face at maturity
y the yield, 8% here
What it says in wordsDuration is the average arrival time of a bond's cash flows, each weighted by what it is worth today.
Add the sizing, because desks price risk in rupees. Modified durationMacaulay duration divided by one plus the yield. It gives the approximate percentage price change for a one point move in yield. is Macaulay duration over 1.08, so the zero loses about 9.3% of its price for a one point rise in yield and the 15-year bond about 7.9%. The limit: this is a small-move estimate, and it ignores the curve changing shape. Even a 30-year 12% bond only reaches about 11.5 years, because a coupon bond's duration can never exceed that of a perpetuity, 13.5 years at 8%.
Where candidates lose it
The instinct is to pick the 15-year bond because it is the longest. That confuses the last payment date with the average one, and on a desk it means hedging the wrong book.
The quieter miss is treating coupon size as a detail. At 12% the coupons are large enough to pull the duration down to 8.6 years; at a 2% coupon the same 15-year bond would have a duration well above 10.
What the interviewer asks next
What coupon on the 15-year bond would make its duration equal to the zero's?
Why does a floating-rate bond have a duration close to its next reset date?
Yields rise from 8% to 10%. Which of the three loses the most in rupees per Rs 100 of face?

058A fund's yearly returns average 12% with 18% volatility, and are roughly normal and independent from year to year. What is the chance of losing money in any one year, and what is the chance that its average return over ten years is negative?Probability and expected valueHardFund research and ratingsIndian AMCs

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Which pair is closest?

Show the worked solution

About 25% in any one year, and about 2% over ten years. One year: zero sits 12/18 = 0.67 standard deviations below the mean, which leaves 25.2% of outcomes below it. The ten-year average keeps the 12% mean but its spread falls to 18 divided by the square root of 10, 5.7%, so zero is 2.11 standard deviations away and the chance is 1.8%.

How likely is a losing year?

Picture the daily commute. Any one day might be twenty minutes late because of rain or a breakdown; your average over a month is almost never more than a few minutes off. Single outcomes are noisy; averages of many independent outcomes are much less noisy, because the bad days and the good days partly cancel. For one year, the question is how far zero sits below a 12% mean when the spread is 18%. That distance is 12/18 = 0.67 standard deviations, and the normal table puts 25.2% of the curve below it. One year in four is a loss, which matches what investors in equity funds actually live through.

The one-year return curve, centred on 12% with an 18% spread, has 25% of its area below zero, while the ten-year average curve has the same centre but a 5.7% spread and only 1.8% of its area below zero.

Why does the ten-year chance collapse to about two percent?

Averaging ten independent years keeps the centre at 12% but divides the spread by the square root of ten. The standard errorThe spread of an average. For independent draws it equals the spread of one draw divided by the square root of the number of draws. of the ten-year average is 18 / 3.16 = 5.7%, so zero is now 2.11 standard deviations below the mean instead of 0.67. The tail beyond 2.1 standard deviations is 1.8%. Five years sits in between: a spread of 8.0% and a chance of about 7%.

The relationship

P(\bar r_{10} < 0) = \Phi\!\left(-\frac{12}{18/\sqrt{10}}\right) = \Phi(-2.11) \approx 1.8\%

r bar 10	the average yearly return over ten years
12	the mean yearly return, per cent
18 / sqrt(10)	the spread of the ten-year average
Phi	the share of a normal curve below a given number of standard deviations

What it says in wordsDivide the distance to zero by the spread of the average, not by the spread of one year, then read the tail.

Now the limits, because a sharp interviewer will push. A negative arithmetic average is not quite the same as losing money: compounding knocks roughly half the variance off growth, so the fund compounds nearer 10.4% than 12%, and the chance of ending ten years below the starting amount is a little higher than 1.8%. Real returns also have fatter tails than a normal curve and are not fully independent, since bad years cluster. The direction survives all of that: time narrows the spread of the average, not the risk of a bad single year.

Where candidates lose it

The first trap is saying the risk of loss is the same at every horizon, or that it falls in proportion to time. It falls with the square root of time, which is why ten years cuts 25% to about 2%, not to zero and not to 2.5%.

The second is overselling the answer. Say what the two percent assumes: normal returns, independent years and a fixed mean, and that compounding and fat tails push the true figure somewhat higher.

What the interviewer asks next

What volatility would make the one-year chance of loss exactly one in three?
Why is the chance of ending below your starting value higher than the chance of a negative arithmetic average?
Over how many years does the chance of a negative average fall below 1%?

060A fund has a tracking error budget of 4% a year and runs 20 active positions, each carrying the same amount of active risk, with the positions uncorrelated. How much active risk can each position carry?Risk, volatility and drawdownHardRisk and complianceIndian AMCs
Try it first
How much active risk can each position carry?
Show the worked solution
About 0.89% each. Uncorrelated risks add in squares. Twenty positions of active risk s give a total variance of 20 s squared, which must equal 4 squared, 16. So s squared is 0.8 and s is 0.89%. The straight split of 4% over 20, 0.20%, would use under a quarter of the budget, because it assumes every position fails at the same time.
Why can each position carry more than a twentieth of the budget?
Think of twenty friends each guessing the weight of a cake. Each guess is off by about 100 grams, but in random directions, so the errors partly cancel and the total of their errors is nowhere near 2 kilograms. Independent errors grow with the square root of how many there are, not in proportion, because some push up while others push down. Tracking errorThe standard deviation of the gap between a fund return and its benchmark return, usually quoted per year. works the same way. Twenty uncorrelated bets of equal size s give a total of s x sqrt(20), about 4.47 s.
Twenty uncorrelated positions of 0.89% active risk each have squares that add to 16, whose square root is the 4% budget, while the straight-line split of 0.20% each would use only about 22% of the risk allowed.
How do you solve it, and what happens if the bets are not independent?
Set the total equal to the budget and solve. 4 = s x sqrt(20), so s = 4 / 4.47 = 0.89%: each position may carry almost four and a half times the naive answer. The catch is the word uncorrelated. If every pair of bets has a correlation of 0.2, the variance picks up 20 x 19 cross terms, each worth 0.2 s squared. The total variance becomes 96 s squared and each position can carry only 0.41%. Correlated bets behave more like one large bet.
The relationship
$\sigma_{TE}^2 = N s^2 \;\Rightarrow\; s = \frac{\sigma_{TE}}{\sqrt{N}} = \frac{4\%}{\sqrt{20}} \approx 0.89\%$
sigma TE the tracking error budget, 4% a year
N the number of independent active positions, 20
s the active risk of each position
What it says in wordsWith independent bets the budget is shared out in variance, so each bet's risk is the budget divided by the square root of the count.
This is why risk teams care about the true number of independent bets more than the number of line items. Twenty stocks that are all quietly a bet on falling interest rates are close to one position, and they would breach the budget at a fraction of the size the arithmetic above allows. The limit: tracking error is a one-number summary of normal-times behaviour, and correlations between bets tend to rise in a sell-off, exactly when the budget matters.
Where candidates lose it
Almost everyone says 0.20%. It assumes risks add like rupees, which happens only when every position moves together. The interviewer wants to hear the word variance before any number.
The second trap is answering 0.89% and stopping. Say that it rests on zero correlation, and show how a modest correlation of 0.2 shrinks the allowance to 0.41%.
What the interviewer asks next
The fund adds 20 more uncorrelated positions. What can each one carry now?
One position is twice the size of the others in risk terms. How does that change the allowance for the rest?
Why do correlations between active bets tend to rise in a market sell-off?

n	small cubes along one edge
(n-2)^3	the hidden core after peeling one layer from every side
V(n)	cubes showing at least one face

800	fund assets before the default, Rs crore
40	the defaulted bond, Rs crore
32	units outstanding, crore, fixed on the day of segregation
0.6	the share of the bond later recovered

10 x 200	enterprise value, EV/EBITDA times EBITDA
500	net debt, the lenders' claim
200 - 40 - 50	profit before tax after depreciation and interest
1 - 0.25	the share of pre-tax profit kept after 25% tax

t	the year a cash flow arrives
CF_t	the cash flow in year t, coupon plus face at maturity
y	the yield, 8% here

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Why does order not matter for a lump sum?

What breaks the symmetry in an SIP?

What does this mean for a real SIP investor?

Where candidates lose it

What the interviewer asks next

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Why are there two different counts here?

Why count the hidden cubes instead of the visible ones?

Where candidates lose it

What the interviewer asks next

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Why would order not matter without withdrawals?

What exactly happens in the first two years?

What does an adviser do with this?

Where candidates lose it

What the interviewer asks next

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Why is the guess of a quarter so far off?

Where exactly does the last five years' money come from?

What does this mean for an investor thinking of stopping at year 15?

Where candidates lose it

What the interviewer asks next

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Why split the fund instead of simply marking the bond to zero?

What does the recovery pay, and who would have got it without segregation?

Where candidates lose it

What the interviewer asks next

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Why can you not read the P/E straight off the EV multiple?

Why does the P/E come out higher than the EV multiple here?

Where candidates lose it

What the interviewer asks next

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Why does the best rule look good when none of them is?

What should you expect out of sample, and how would you check a real rule?

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What the interviewer asks next

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Why is maturity the wrong ruler?

How do you rank them without a calculator?

Where candidates lose it

What the interviewer asks next

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How likely is a losing year?

Why does the ten-year chance collapse to about two percent?

Where candidates lose it

What the interviewer asks next

Try it first

Why can each position carry more than a twentieth of the budget?

How do you solve it, and what happens if the bets are not independent?

Where candidates lose it

What the interviewer asks next