Maven Securities Interview Questions 2026

Practice Bank

Maven Securities Quant Interview Practice

↗

For a community event, an organiser buys five cases of bottled drinks. Each case contains eight drinks and costs $29.60, broken down on the invoice as $20 for the drinks plus eight refundable $1.20 bottle deposits. All 40 drinks are served.

Volunteers recover 31 empty bottles before the return deadline; the shop refunds $1.20 for each of those and nothing for the other nine. Ignore all other event costs. What is the final net cost per drink served? Reconcile both the cash paid and the refundable-deposit portion rather than treating every deposit as either a permanent expense or an automatic refund.

Get hint

Hint 1

Count the deposits refunded, not just the deposits paid. A bottle can be emptied and its drink served even if the empty is never returned.

Hint 2

Compute $5\times\$29.60-31\times\$1.20$, then divide by 40. Check by adding drink costs and forfeited deposits separately.

Show solution

The organiser pays $5(29.60)=\$148$ for 40 drinks and 40 refundable bottle deposits. The shop returns $31(1.20)=\$37.20$ because only 31 empties arrive by the deadline. The remaining nine deposits are forfeited even though their drinks were served. Thus the net cash cost is

$\$148-\$37.20=\boxed{\$110.80}$,

and the cost per served drink is $110.80/40=\boxed{\$2.77}$. As an independent reconciliation, the consumable drinks cost $5(20)=\$100$, while the nine lost deposits cost $9(1.20)=\$10.80$. Those sum to the same $110.80$. Treating the whole $48$ deposit payment as a loss overstates the cost, and treating it all as refundable understates the cost; consumption and bottle returns are distinct counts.

Assumptions

All five cases are purchased and all 40 drinks are consumed; the only refund is for each of the 31 returned bottles.
Deposits are exactly $1.20 per bottle, and the base drink price totals $20 per case.

Final Answer

The final net cost is $2.77 per served drink ($110.80 for all 40).

Numerical reasoning and estimation

↗

A visitor arrives at a reception desk at a time chosen uniformly from 9:00:00 am to 9:04:00 am. The exact arrival time is continuous, so every equally long interval within those four minutes has equal probability. The reception screen shows only the arrival time rounded to the nearest whole minute: 9:00, 9:01, 9:02, 9:03 or 9:04. If an arrival is exactly halfway between two minutes, the screen rounds upward.

Find the probability of each displayed time. Is a displayed arrival of 9:00 as likely as one of 9:02?

Get hint

Hint 1

Each displayed minute corresponds to a continuous interval of actual times that would round to that label.

Hint 2

Intersect the usual half-minute-to-half-minute rounding bins with the fixed 9:00–9:04 observation window; the first and last bins are shorter.

Show solution

Measure exact arrival time $T$ in minutes after 9:00, so $T\sim\operatorname{Uniform}(0,4)$, with density $f_T(t)=1/4$ on that interval. Ignoring zero-probability tie endpoints, the displayed minute offset $D$ has these preimages:
$\begin{array}{c|ccccc} D&0&1&2&3&4\\\hline \text{True-time interval}&[0,0.5)&[0.5,1.5)&[1.5,2.5)&[2.5,3.5)&[3.5,4]\\ \text{Length in minutes}&0.5&1&1&1&0.5 \end{array}$
The probability of a displayed label is its preimage length divided by the total four-minute window:
$P(D=0)=P(D=4)=\frac{0.5}{4}=\frac18, \quad P(D=1)=P(D=2)=P(D=3)=\frac14.$
The masses sum to $2(1/8)+3(1/4)=1$. Hence 9:00 is only half as likely as the interior label 9:02. The endpoints are not special because their exact time points are likely; each is special because its rounding interval is truncated by the observation window. The halfway rounding convention changes only measure-zero points under a continuous uniform law.

Assumptions

The true arrival time is uniform continuously over a four-minute interval, including endpoints of probability zero.
The display uses nearest-minute rounding with upward tie-breaking.

Final Answer

$P(9{:}00)=P(9{:}04)=1/8$; $P(9{:}01)=P(9{:}02)=P(9{:}03)=1/4$. The first displayed minute is half as likely as 9:02.

Probability foundations

↗

At a repair desk, an electronic device is either sound or faulty. Before testing, two out of every three devices are expected to be sound. The desk first runs a quick check, which shows green with probability $3/4$ for a sound device and $1/4$ for a faulty one.

A green result sends the device to scanner A; a red result sends it to scanner B. Scanner A reports clear with probability $4/5$ if the device is sound and $1/5$ if it is faulty. Scanner B reports clear with probability $1/2$ if sound and $1/10$ if faulty. Conditional on the device's actual condition and which scanner is used, the follow-up result is independent of the quick check.

A supervisor sees only the final label clear, not the earlier colour or which scanner was used. What is the probability that this device is sound? Explain why using the accuracy figures for scanner A alone is not sufficient.

Get hint

Hint 1

A final clear label can come from two distinct test routes. Account for both green→A and red→B.

Hint 2

Calculate $P(\text{clear}\mid\text{sound})$ and $P(\text{clear}\mid\text{faulty})$ by routing probabilities, then apply Bayes’ rule with prior 2/3.

Show solution

The final label is a mixture of two tests selected by the earlier check, so we must first compute how often the complete procedure reports clear in each true condition.

For a sound device, the check is green with probability $3/4$ and red with probability $1/4$. The selected scanner's clear probabilities are therefore weighted as
$P(C\mid S)=(3/4)(4/5)+(1/4)(1/2)=3/5+1/8=\frac{29}{40}.$
For a faulty device, the quick check is green with probability $1/4$ and red with probability $3/4$, so
$P(C\mid F)=(1/4)(1/5)+(3/4)(1/10)=1/20+3/40=\frac18.$
With $P(S)=2/3$ and $P(F)=1/3$, the probability of a final clear label is
$P(C)=(2/3)(29/40)+(1/3)(1/8)=\frac{29}{60}+\frac1{24}=\frac{21}{40}.$
Bayes' rule now gives
$P(S\mid C)=\frac{P(S)P(C\mid S)}{P(C)} =\frac{29/60}{21/40}=\boxed{\frac{58}{63}}.$
The two routes are mutually exclusive, and all four combinations of true condition and route are included. Scanner A's individual clear rate cannot replace $P(C\mid S)$ or $P(C\mid F)$, because devices sent to B also contribute to the merged final report.

Assumptions

Prior sound probability is 2/3, and the device’s true condition does not change during inspection.
The green/red check determines exactly one of the two scanners.
The scanner’s stated conditional clear rates apply on its assigned route, independent of the quick check given quality and route.
Only the merged clear label is observed by the supervisor.

Final Answer

$P(\text{sound}\mid\text{final clear})=\frac{58}{63}\approx92.06\%$. The full routing procedure gives $P(\text{clear}\mid\text{sound})=\frac{29}{40}$ and $P(\text{clear}\mid\text{faulty})=\frac18$.

Conditional probability

↗

Three customers arrive together at a small café. Their orders take different amounts of time to prepare: Alice's takes 3 minutes, Ben's 5 minutes and Cleo's 8 minutes. The café picks their service order uniformly from all six permutations and prepares one order at a time.

To apologise for waiting, the café pays each customer cash only for the minutes before their own service starts: Alice receives $2 per waiting minute, Ben $4, and Cleo $1. There is no compensation for the minutes during their own preparation, and nobody leaves the queue.

What is the expected total waiting compensation paid by the café?

Get hint

Hint 1

A customer waits for the preparation times of those ahead of them, not for their own order. Consider a pair consisting of an earlier customer and a later one.

Hint 2

In a uniformly random order, each pair of distinct customers has probability 1/2 of appearing in either order. Sum duration of the earlier person's service times the later person's dollar-per-minute rate.

Show solution

Denote customers by $i$, with preparation times $d_i$ and waiting compensation rates $r_i$. Total compensation in a particular order is

$W=\sum_{i\ne j} d_i r_j\,\mathbf1\{i\text{ is served before }j\}.$

Every ordered pair of distinct customers has probability 1/2 of appearing in that order under a uniform random permutation. Taking expectations and using linearity,

$E[W]=\frac12\sum_{i\ne j}d_i r_j =\frac12\left[\left(\sum_i d_i\right)\left(\sum_j r_j\right)-\sum_i d_i r_i\right].$

Here $d=(3,5,8)$ minutes and $r=(2,4,1)$ dollars per minute. Therefore

$E[W]=\frac12[(3+5+8)(2+4+1)-(3\cdot2+5\cdot4+8\cdot1)] =\frac{112-34}{2}=\boxed{\$39}.$

A full six-order check gives compensation totals of $20 (Alice–Ben–Cleo), $47 (Alice–Cleo–Ben), $18 (Ben–Alice–Cleo), $31 (Ben–Cleo–Alice), $60 (Cleo–Alice–Ben), and $58 (Cleo–Ben–Alice). Their average is exactly $39. No independence of customers' wait times is needed: the indicators for different ordered pairs are dependent, but expectation is additive.

Assumptions

All three customers arrive simultaneously and each preparation time is deterministic.
Every order is equally likely and there is no idle time or abandonment.
Compensation is paid in cash from arrival until service begins, excluding each customer’s own preparation time.

Final Answer

$\$39$ expected total compensation.

Expectation and expected value

↗

A night guard chooses two of three storage rooms to inspect. The guard commits to a possibly randomized route before a thief independently chooses one room to target, and neither sees the other's choice. A checked room cannot be robbed. If an unchecked room is targeted, the theft loss is $6 from room 1, $8 from room 2 or $12 from room 3.

The guard can choose routes {1,2}, {1,3}, or {2,3}. Every chosen room is checked except on route {1,3}: the guard checks room 1 first, then independently gets delayed with probability 1/2 and misses room 3 before the theft. No other route has delays. The thief always attempts exactly one room and seeks to maximize expected loss; the guard seeks to minimize it.

Find an optimal randomization over the three routes and the minimax expected loss. Exhibit a randomized thief strategy that proves the guard cannot do better.

Get hint

Hint 1

Let x,y,z be probabilities of routes {1,2},{1,3},{2,3}. Compute the loss separately if each room is attacked.

Hint 2

The three conditional attack losses are $6z$, $8y$ and $12x+6y$. Find a mixture that equalizes them, then use an attacker mixture to certify the lower bound.

Show solution

Let $x,y,z\ge0$ be guard probabilities for routes {1,2}, {1,3}, {2,3}, summing to 1. Against an attack on room 1, loss occurs only on route {2,3}, hence $L_{1}=6z$. Against room 2, only {1,3} fails to cover it, so $L_{2}=8y$. Against room 3, route {1,2} always misses it and route {1,3} misses it with probability $\frac{1}{2}$, giving $L_{3}=12x+12(\frac{1}{2})y=12x+6y$. The guard minimizes $\max(L_{1},L_{2},L_{3})$.

Set all three losses equal: $6z=8y=12x+6y$. The equalities imply $z=(\frac{4}{3})y$ and $x=\frac{y}{6}$. With $x+y+z=1$, we obtain $y=\frac{2}{5}$, $x=\frac{1}{15}$, $z=\frac{8}{15}$. Each targeted room then has expected loss $\frac{16}{5}$ dollars, so this patrol guarantees loss no greater than $3.20 for any attack or mixture of attacks.

To show no other guard schedule can do better, let the thief independently target rooms (1,2,3) with probabilities $(\frac{8}{15},\frac{1}{5},\frac{4}{15})$. If the guard always uses {1,2}, expected loss is $12(\frac{4}{15})=\frac{16}{5}$. For {1,3} it is $8(\frac{1}{5})+12(\frac{1}{2})(\frac{4}{15})=\frac{16}{5}$. For {2,3} it is $6(\frac{8}{15})=\frac{16}{5}$. Any random mix of those routes must also suffer expected loss $\frac{16}{5}$ against this thief mix, by linearity. Therefore the guard cannot guarantee less and the thief cannot force more against the stated patrol. The matching upper and lower bounds establish the minimax value and both optimal mixed strategies, including the delay risk.

Assumptions

The thief picks exactly one room before observing the guard route or any delay.
A successful inspection prevents the theft at that room with certainty; no loss occurs at unchosen rooms.
The probability-1/2 delay is independent of both players' randomizations.
The route {1,3} delay affects room 3 only; room 1 is always inspected on that route.

Final Answer

Choose routes {1,2}, {1,3}, {2,3} with probabilities $\frac{1}{15}$, $\frac{2}{5}$, $\frac{8}{15}$. The minimax expected theft loss is $$\frac{16}{5}$=$3.20. A matching thief strategy targets rooms 1,2,3 with probabilities $\frac{8}{15}$, $\frac{1}{5}$, $\frac{4}{15}$.

Game theory and strategic decisions

↗

A broker needs 100 identical spare parts, each priced at $10 before fees. Two suppliers can deliver all 100 parts for the same $1,000 base price; the broker is choosing only between their billing arrangements.

Supplier A fills the order in either one invoice for 100 parts, with probability 1/2, or four separate invoices for 25 parts each, with probability 1/2. Its processing charge is the greater of $12 or 2% of that invoice's value, calculated independently for every invoice. Supplier B guarantees one invoice and charges a flat 3% of the $1,000 total, with no minimum.

Which supplier has the lower expected all-in cost? If the chance of A using four invoices were p instead of 1/2, at what p would the broker be indifferent?

Get hint

Hint 1

Apply Supplier A’s minimum separately to each invoice. Four invoices for $250 each do not incur the same fee as one invoice for $1,000.

Hint 2

A’s total charge is $20 in the one-invoice outcome and $48 in the four-invoice outcome. Compare 20(1−p)+48p with B’s fixed $30.

Show solution

Both suppliers charge $1,000 for the parts, so compare their processing fees, adding the common base at the end.

For A's one-invoice outcome, invoice value is $1,000 and 2% is $20; its fee is $\max(12,20)$=$20. For A's four-invoice outcome, each invoice is $250; 2% is $5, below the $12 minimum. Each of four invoices therefore incurs $12, making the total fee $48. With the stated 50/50 distribution,

$E[\text{fee A}]=(\frac12)\cdot20+(\frac12)\cdot48$=$34,

so A's expected all-in cost is $1,034.

Supplier B charges exactly $3\%\cdot\$1{,}000$=$30, for a fixed all-in cost of $1,030. Choose B, cheaper in expectation by $4.

For a variable four-invoice probability p, A's expected fee is

$\[20(1-p)+48p=20+28p.\]$

Indifference with B requires $20+28p=30$, giving $p=\frac{10}{28}=\frac5{14}$. For $p<\frac5{14}$ A is cheaper; for $p>\frac5{14}$ B is cheaper; at equality the expected costs coincide. The order always contains the same 100 parts: the expected-cost difference comes entirely from random fragmentation into minimum-fee invoices, not from uncertain purchased quantity.

Assumptions

Both suppliers deliver exactly 100 parts at a fixed $10 base price each.
A produces either one 100-part invoice or four distinct 25-part invoices at the given probabilities.
Minimum processing charges apply per invoice, not per order; B has no minimum fee.
All monetary amounts and outcomes are evaluated under risk-neutral expected cost.

Final Answer

Choose B at the stated probabilities: A costs $1,034 in expectation, B costs $1,030. The fee break-even probability for A's four-invoice outcome is $p=\frac5{14}\approx35.714\%$; A is cheaper below this and B above it.

Market making and microstructure

↗

A collector starts with £100 of equity and borrows against it to buy units of an inventory portfolio worth £100 per unit today. Fractional units are allowed. Let $L\ge1$ be the initial leverage: the purchase costs £$100L$ in total, financed by £100 of equity and a loan of £$100(L-1)$. The loan bears no interest, and the lender takes no additional security.

After day one, each item's price is £70 with probability 0.10, £90 with probability 0.30, or £110 with probability 0.60. The lender checks immediately: if equity divided by the then-current collateral value is strictly below 25%, the entire position is liquidated and cannot recover. Otherwise it survives. On day two, every non-liquidated item's price becomes £120, regardless of the day-one move.

The collector insists that the probability of liquidation be at most 10%. What is the largest allowed L? State what happens exactly at a margin threshold, and explain why positive terminal equity on every hypothetical two-day path would not settle this question.

Get hint

Hint 1

Write interim collateral value as Lp and outstanding debt as $100(L-1)$, where p is the day-one item price.

Hint 2

Solve $\frac{Lp-100(L-1)}{Lp}\ge\frac{1}{4}$ for L at $p=70,90,110$. Compare the probabilities of failing the successive thresholds, including the fact that equality does not trigger liquidation.

Show solution

The initial purchase buys L items in the stylized divisible-unit model. At interim price p the collateral is worth $Lp$ pounds, the debt is £$100(L-1)$, and equity is $E_1=Lp-100(L-1)$. The lender's survival condition is

$\frac{E_1}{Lp}=1-\frac{100(L-1)}{Lp}\ge\frac{1}{4}$ .

Equivalently, $100(L-1)\le\frac{3}{4}Lp$, so for these prices $L\le\frac{100}{100-\frac{3}{4}p}$. This yields the three survival ceilings:

$p$=£70: $L\le\frac{100}{47.5}=\frac{40}{19}\approx2.105263$;

$p$=£90: $L\le\frac{100}{32.5}=\frac{40}{13}\approx3.076923$;

$p$=£110: $L\le\frac{100}{17.5}=\frac{40}{7}\approx5.714286$.

As leverage increases from 1, none of the paths triggers a call until $L$ exceeds $\frac{40}{19}$. For $\frac{40}{19}<L\le\frac{40}{13}$, only the £70 state fails, so liquidation probability is 0.10. Once $L$ exceeds $\frac{40}{13}$, both the £70 and £90 states fail, giving probability at least $0.10+0.30=0.40$; above $\frac{40}{7}$ all three states fail. Therefore the largest $L$ compatible with a 10% cap is $\frac{40}{13}$. At $L=\frac{40}{13}$, the £90 state has collateral £$\left(90\times\frac{40}{13}\right)$ and debt £$\left[100\times\left(\frac{27}{13}\right)\right]$; the difference is exactly one quarter of the collateral, so it is not liquidated under the 'strictly below' rule. The £70 state is below margin, giving the allowed 10% failure probability.

If a position could survive to day two, its hypothetical final equity would be £$\left[120L-100(L-1)\right]$=£$(100+20L)$>$0$ for every $L\ge1$. But that terminal recovery is unavailable on any path liquidated at the day-one check. The constraint is a probability of interim survival, not terminal solvency. Enumerating the three interim price states, rather than their common terminal price, is essential.

Assumptions

The items are divisible for this model, allowing any real initial leverage $L\ge1$; all purchased units have identical price paths.
Debt is constant over the two days with no fees or interest; the lender checks exactly once at day one's price.
Liquidation occurs only if equity/collateral is strictly less than 25%; equality survives.
All surviving units are worth £120 each on day two, and a liquidated position cannot be reopened.

Final Answer

Maximum initial leverage $L=\frac{40}{13}\approx3.076923$. At that leverage liquidation occurs only on the £70 interim path, with probability 0.10; the £90 path exactly meets the 25% requirement and survives.

Risk, covariance and diversification

At the checkout you have a grocery basket with a pre-discount total of $x$ dollars, where $x\ge0$. Only one of two coupons can be applied. Coupon A…

Unlock the rest

Keep practising with 93 more questions, available hints and full worked solutions.

Explore Premium