Belvedere Trading Interview Questions 2026

Two small sacks contain \(2\times3=6\) kg and cost \(2\times\$12=\$24\). The large sack contains 8 kg and costs $24. Altogether, the caterer buys \(6+8=14\) usable kilograms and spends $48. Therefore
\[ \text{combined price per kg}=\frac{\$48}{14\text{ kg}} =\boxed{\frac{24}{7}\approx\$3.4286\text{ per kg}}. \]

The individual rates are indeed \(\$12/3=\$4/\mathrm{kg}\) and \(\$24/8=\$3/\mathrm{kg}\), but the caterer did not buy equal kilogram quantities at those rates. The correct quantity-weighted rate is
\[ \frac{6}{14}(\$4/\text{kg})+\frac{8}{14}(\$3/\text{kg}) =\frac{24+24}{14}=\$\frac{24}{7}/\text{kg}. \]

The arithmetic average \((4+3)/2=3.50\) treats the two sack types as equally represented by quantity. Even averaging across the three individual sacks would be wrong without kilogram weights: the large sack supplies 8 kg, much more than either small sack. Because all purchased flour remains usable, there is no need to remove discarded kilograms from the denominator.

Let the letters also denote the events that their corresponding inspections pass. The original rule approves on \(O=A\cap B\), whereas the revised rule approves on \(N=A\cap B\cap C\). Every outcome in \(N\) is also in \(O\), so \(N\subseteq O\), which proves

\[ P(N)\le P(O) \]

with no independence assumption. More precisely, \(O\) is the disjoint union of \(N\) and \(A\cap B\cap C^c\). Therefore

\[ P(O)-P(N)=P(A\cap B\cap C^c). \]

It follows that equality holds if and only if the extra test never rejects a batch that would otherwise be approved, except on a probability-zero set: \(P(A\cap B\cap C^c)=0\). A strict decrease occurs exactly when that event has positive probability.

For example, toss a fair coin to determine A; let B always pass, and define C to pass exactly when A does. Then \(P(C\text{ fails})=1/2\), but both old and new rules approve precisely on heads, with probability \(1/2\). The result does not depend on whether individual test failures are common or rare; it follows from event inclusion alone.

Let A be the event that delivery is on time, C the observed coarse bulletin “amber”, and \(M_{1},\ldots,M_{r}\) the mutually exclusive, collectively exhaustive detailed messages that can occur given C. Set \(w_{j}=P(M_{j}\mid C)>0\). Conditional total probability gives the coherence identity

\(P(A\mid C)=\sum_{j=1}^{r}w_jP(A\mid C,M_j)\), with \(\sum_{j}w_{j}=1\).

If all detailed posteriors were strictly larger than the original \(p=P(A\mid C)\), their weighted average would also be strictly greater than p. That would contradict the identity. More generally, if every posterior were at least p and one were strictly greater, the average would exceed p; therefore a genuine upward revision on some positive-probability branches must be offset by downward revisions on others (unless no branch changes probability).

In the proposed 20-scenario refinement, North has 8 cases with 6 successes, so \(P(A\mid C,\text{North})=6/8=\)\(3/4\). South has 12 cases with 4 successes, so \(P(A\mid C,\text{South})=4/12=\)\(1/3\). Their conditional message probabilities are \(P(\text{North}\mid C)=8/20=2/5\) and \(P(\text{South}\mid C)=12/20=3/5\). The weighted posterior average is

\((2/5)(3/4)+(3/5)(1/3)=3/10+1/5=\)\(1/2\).

The unconditional-on-detail amber probability is also \((6+4)/20=10/20=1/2\), as required. More information can raise the probability after one particular message, but it cannot guarantee a rise after every possible positive-probability message.

Let \(D\) be the sum of paid pledges. There are \(2^3=8\) subsets of participating donors; independence lets us multiply the corresponding payment and nonpayment probabilities. Every subset with at most one donor totals at most \(100\), so its match is zero. The four positive cases are:

Consequently

\(\mathbb E[M]=10(1/8)+40(1/12)+70(1/24)+80(1/24)\)
\(=30/24+80/24+70/24+80/24=260/24=\boxed{65/6\text{ dollars}}\).

A separate check is that the probabilities of the four remaining zero-payment cases (nobody, only A, only B, only C) are \(1/4\), \(1/4\), \(1/8\) and \(1/12\). Those total \(17/24\), while the four positive case probabilities total \(7/24\); altogether the eight disjoint probabilities sum to one. The $100 donation by C alone produces zero matching payment because the offer is strictly above \(100\), and the $210 total yields only \(80\), not \(110\), because of the cap.

Let demands be \((d_1,d_2,d_3)=(1,2,3)\) and S the set of farms using the pump. Their combined pump demand is \(D=\sum_{i\in S}d_i\). A farm i in S pays \(d_i(1+D)\); otherwise it pays \(5d_i\). A unilateral move into the pump changes \(D\) to \(D+d_i\), a point that must be incorporated in the deviation calculation.

Enumerating all eight subsets gives (pump set; \(D\); total bill; Nash?): empty;0;$30;no. {1};1;$27;no, because farm 2 would join and pay $8 rather than $10. {2};2;$26;no, because farm 1 would join and pay $4 rather than $5. {3};3;$27;yes. {1,2};3;$27;yes. {1,3};4;$30;yes. {2,3};5;$35;no, because existing pump users prefer the tanker. {1,2,3};6;$42;no, because all would pay less using a tanker alone.

To verify the equilibria, at \(S=\{3\}\), farm 3 pays \(3(1+3)=\text{\$}12<\text{\$}15\) tanker; farm 1 joining would pay \(1(1+4)=\text{\$}5\), exactly its tanker price; farm 2 joining would pay \(2(1+5)=\text{\$}12>\text{\$}10\). At \(S=\{1,2\}\), the pump price per unit is $4, cheaper for both users than the $5 tanker; farm 3 joining would face $7 per unit, so stays on its tanker. At \(S=\{1,3\}\), the pump unit price is $5, making both users indifferent to leaving; farm 2 joining would face $7 per unit, so stays outside. Thus no unilateral strict improvements occur in exactly those three profiles.

Total cost as a function of \(D\) is \(C(D)=D(1+D)+(6-D)5=D^{2}-4D+30=(D-2)^{2}+26\). Feasible \(D\) values are 0 through 6 because subset sums of 1,2,3 cover every integer in that range. The unique minimum occurs at \(D=2\): only farm 2 uses the pump, with $6 pump cost plus $20 tanker cost = $26. It is not Nash-stable because farm 1 would join. The best equilibrium cost is $27, a $1 gap; the worst is $30, a $4 gap (price of anarchy \(\frac{30}{26}=\frac{15}{13}\)). Unequal, indivisible farm demands mean that setting two continuous-route costs equal is not a valid substitute for checking discrete deviations.

There are two possible states: the contracted customer arrives with probability 0.30, or does not arrive with probability 0.70. Selling now to the dealer pays $200 immediately, but leaves no unit to fulfil the commitment. In the arrival state the kiosk must pay $240 compensation, so its final cash is \(200-240=-\$40\). If no customer arrives, it keeps the $200. Expected cash from selling is
\[ E[C_{\rm sell}]=0.30(-40)+0.70(200)=\boxed{\$128}. \]
Keeping the unit yields $120 in the arrival state, because the kiosk completes the promised rental, and $50 in the no-arrival state, because it disposes of the unused pack. Its expected cash is
\[ E[C_{\rm keep}]=0.30(120)+0.70(50)=\boxed{\$71}. \]
Without the risk rule, the seller would accept the dealer's offer: $128 is $57 higher than $71. But the constraint requires \(C\ge-20\) in every state. Selling fails in the customer-arrival state (\(-40<-20\)); keeping produces at least $50, and so passes both states. The only feasible choice is therefore \(\boxed{\text{keep the pack}}\).

The conclusion is not based on average customer demand or on the average cancellation cost. A hard terminal shortfall limit vetoes an action as soon as one permitted state breaches it, even if that state is relatively unlikely and the action has higher expected cash.

Both policies start with $50 in each asset. After period one, A is worth \(50\times2=\$100\) and B is worth \(50\times(1/2)=\$25\), so midpoint wealth is $125. The actual buy-and-hold weights have drifted to \(100/125=80\%\) in A and 20% in B; a 50:50 policy requires a trade.

Rebalancing sells $37.50 of A and uses that exact amount to buy more B. It is self-financing and leaves $62.50 in each asset. On the reversal path (\(A\times1/2,\ B\times2\)), terminal wealth is \(62.50(1/2)+62.50(2)=31.25+125=\$156.25\). On the continuing-trend path (\(A\times2,\ B\times1/2\)), it is \(62.50(2)+62.50(1/2)=125+31.25=\$156.25\). Its expected terminal wealth is thus $156.25.

Buying and holding preserves the midpoint values $100 in A and $25 in B. On reversal, final wealth is \(100(1/2)+25(2)=\$100\). On continuing trend it is \(100(2)+25(1/2)=\$212.50\). Weighting the same two paths as above gives

\[E[W_{\mathrm{hold}}]=\tfrac23(100)+\tfrac13(212.50)=\$137.50.\]

The expected advantage of rebalancing is \(156.25-137.50=\boxed{\$18.75}\). Buy-and-hold wins on the continuing-trend path, but rebalancing wins on reversal. The comparison is not based on assuming that constant weights and unchanged share quantities mean the same thing: the midpoint trade changes the exposure to the second-period returns. Both policies use identical exogenous return paths and the rebalance requires no injected capital.