Virtu Financial Interview Questions 2026

Let C be the wholesale cost. The marked price is \(1.50C\). After the 20% sale reduction, the merchandise paid amount is \(0.80(1.50C)=1.20C\). The later refund is 25% of this amount, so the customer ultimately retains a charge of \(0.75(1.20C)=0.90C\) for the camera. The $12 delivery payment and the $12 delivery refund cancel exactly. The observed net payment therefore satisfies \(0.90C=108\), giving \(C=\$120\).

Forward-check: A $120 camera is marked at $180 and sold at $144. Adding $12 shipping produces an initial charge of $156. The shop subsequently returns $36 for merchandise and $12 for delivery, a $48 total refund. The final net payment is \(\$156-\$48=\$108\). The 25% refund applies to the discounted merchandise payment, not to the original wholesale cost or the total including shipping.

Let X and Y be the two scores. Exchangeability says \((X,Y)\) and \((Y,X)\) have the same joint distribution, even if X and Y are dependent. Therefore \(p=\Pr(X>Y)=\Pr(Y>X)\).

Exactly one of \(X>Y\), \(Y>X\), or \(X=Y\) occurs. As \(\Pr(X=Y)=0.18\), we have \(2p+0.18=1\) and \(p=\frac{1-0.18}{2}=0.41\). An independent-scores assumption would be unnecessary and potentially false; label symmetry alone establishes equality of the strict-win probabilities. A check is \(0.41+0.41+0.18=1\).

The inspector selects a time, not one of four interval labels. In an interval of length \(L\), an arrival leaves more than four minutes remaining exactly when it occurs in the first \(L-4\) minutes, provided \(L>4\). The qualifying time has length \((L-4)_+=\max(L-4,0)\).

For the four successive lengths 3, 7, 2 and 8, the qualifying portions have lengths 0, 3, 0 and 4 minutes. These portions are disjoint within the 20-minute cycle. Uniform calendar-time sampling gives

\[\Pr(\text{remaining time}>4)=\frac{0+3+0+4}{3+7+2+8}=\frac7{20}.\]

Equivalently, the inspector enters the seven-minute interval with probability \(7/20\) and then has chance \(3/7\) of seeing more than four minutes left, or enters the eight-minute interval with probability \(8/20\) and then has chance \(4/8\). Their sum is \(3/20+4/20=7/20\). Counting the long intervals as two of four without weighting by their duration would answer a different sampling question. Strict versus non-strict boundaries do not alter a continuous-time probability.

A always takes one seat. The second seat is awarded to the earliest-attending name among B,C,D,E. Thus B, when present, is always admitted and never generates a voucher.

C is turned away if both B and C attend, an event of probability \((3/4)(1/2)=\frac38\). D is turned away if D attends and at least one of B or C attends. By independence, that probability is \((1/2)[1-(1-3/4)(1-1/2)]=(1/2)(1-1/8)=\frac7{16}\). E is turned away if E attends and at least one of B, C or D attends: \((1/4)[1-(1/4)(1/2)(1/2)]=(1/4)(15/16)=\frac{15}{64}\).

Let \(I_C,I_D,I_E\) indicate these denied-boarder events. The voucher total is \(V=120I_C+80I_D+40I_E\) in pounds; simultaneous denials are allowed, so independence among these indicators is neither needed nor asserted. By linearity,

\[\mathbb E[V]=120(3/8)+80(7/16)+40(15/64)=45+35+9.375=\text{£}89.375=\text{£}\frac{715}{8}\].

This agrees with a direct enumeration of the 16 possible attendance subsets for B,C,D,E, weighting each by its independent attendance probability, admitting the earliest two and summing their denied customers' stated vouchers.

Different denied customers receive different amounts, and the priority order makes their denial probabilities unequal. The expected number denied is \(\frac38+\frac7{16}+\frac{15}{64}=\frac{67}{64}\), but multiplying that by an arbitrary unweighted 'average voucher' would not recover the expected payment. Person-specific indicators are the correct accounting units.

Let \(p\) be A's independent probability of P and \(q\) B's. A's payoff from P against B's mix is \(5(1-q)-4q=5-9q\); from D it is \(2(1-q)+q=2-q\). Indifference gives \(q=3/8\). B's payoff from P against A's mix is \(4(1-p)-5p=4-9p\); from D it is \(2(1-p)+p=2-p\). Hence \(p=1/4\). Both probabilities are interior, so this is the fully mixed Nash equilibrium. The corresponding expected scores are \(5-9(3/8)=13/8\) for A and \(4-9(1/4)=7/4\) for B.

Now consider the correlated recommendation distribution

\(\Pr(A=P,B=D)=2/5\), \(\Pr(A=D,B=P)=2/5\), \(\Pr(A=D,B=D)=1/5\), and \(\Pr(P,P)=0\).

A's expected score is \((2/5)5+(2/5)1+(1/5)2=14/5=2.8\); B's is \((2/5)1+(2/5)4+(1/5)2=12/5=2.4\). Both exceed their mixed Nash scores (1.625 and 1.75).

We must verify obedience, not just high average scores. If A is told P, B must have been told D; A earns 5 by obeying versus 2 by switching to D. If B is told P, A must have been told D; B earns 4 by obeying versus 2 by switching. If A is told D, conditional on that message B was told P with probability \((2/5)/(3/5)=2/3\), and D with probability \(1/3\). A's expected obeying score is \((2/3)1+(1/3)2=4/3\), versus \((2/3)(-4)+(1/3)5=-1\) if A deviates to P. If B is told D, A was told P with probability \(2/3\) and D with \(1/3\); B scores \(4/3\) by obeying and \((2/3)(-5)+(1/3)4=-2\) by deviating. All four inequalities favour obedience. This is a correlated equilibrium: a shared private lottery coordinates the crews in a way independent mixing cannot reproduce, without requiring any crew to observe the other's recommendation.

Let \(H\) denote the $120 payoff and \(Q\) the event that both notes say high (hence two buy requests). Conditional on H, a shared note is high with probability \(3/4\), while two independent notes are both high with probability \((3/4)^2\). With equal probabilities of sharing and independent generation,
\[ P(Q\mid H)=\tfrac12\cdot\tfrac34+\tfrac12\left(\tfrac34\right)^2 =\frac{21}{32}. \]
Under the $80 payoff, the probabilities of a false-high note are \(1/4\), so
\[ P(Q\mid L)=\tfrac12\cdot\tfrac14+\tfrac12\left(\tfrac14\right)^2 =\frac5{32}. \]
With 50–50 prior odds, Bayes' rule gives
\[ P(H\mid Q)=\frac{21/32}{21/32+5/32}=\frac{21}{26}. \]
The conditional claim value is
\[ E[V\mid Q]=120\frac{21}{26}+80\frac5{26} =\frac{2920}{26}=\boxed{\frac{1460}{13}\approx\$112.31}. \]
If the dealer incorrectly assumed independent notes with no chance of sharing, the likelihood ratio would be \((3/4)^2/(1/4)^2=9\), giving posterior \(9/10\) and value \(80+40(0.9)=\$116\). The extra $3.69 is spurious confidence from effectively counting the same inspection twice. The fair value is per claim; if both identical claims fill at a price set from this information, total expected underlying payoff is twice this per-claim value.

The combined loss \(L=A+B\) has atoms: $0 with probability 0.60, $100 with 0.20, $200 with 0.10, and $300 with 0.10. In descending loss order, the worst 25% comprises:

• 10% of weeks with $300 loss;
• 10% with $200 loss;
• 5% (not all 20%) of weeks with $100 loss.

The weighted tail loss is \(0.10(\$300)+0.10(\$200)+0.05(\$100)=\$55\) per randomly sampled week. Dividing by the tail probability 0.25 gives

\(\operatorname{ES}_{0.75}(L)=\frac{55}{0.25}=\)\(\$220\).

The 75% value-at-risk/quantile is $100: \(\Pr(L\le0)=0.60<0.75\), while \(\Pr(L\le100)=0.80\ge0.75\). The quantile lies inside a probability atom, which is why including every $100 week in the tail would overcount.

For booking A alone, loss $100 occurs in \(0.20+0.10=0.30\) of weeks. Its worst 25% consists entirely of $100 outcomes, so \(\operatorname{ES}_{0.75}(A)=\)\(\$100\).

For booking B alone, loss $200 occurs in \(0.10+0.10=0.20\) of weeks. Its worst 25% includes these 20% plus 5% of $0 weeks, giving \(\operatorname{ES}_{0.75}(B)=\frac{0.20(\$200)+0.05(\$0)}{0.25}=\)\(\$160\).

Consequently \(\operatorname{ES}_{0.75}(A)+\operatorname{ES}_{0.75}(B)=\$260>\$220=\operatorname{ES}_{0.75}(A+B)\). Measuring the bookings separately discards the diversification from their imperfectly coincident losses. All three computations use the same fractional-boundary rule.