Citadel interview questions

For European options on the same underlying asset with the same strike $K$ and time to expiry $T$, put-call parity gives:

where $C$ is the call price, $P$ the put price, $S$ the spot price, and $r$ the risk-free rate. Gamma is the second derivative of the option price with respect to $S$:

Differentiating put-call parity twice with respect to $S$:

because $S$ is linear in $S$ (second derivative zero) and $K e^{-rT}$ is constant. Hence:

Given $\Gamma_C = 0.02$, the put gamma is also $0.02$.

Let $N$ be the number of coin flips until the first heads appears. Since the coin is fair, $N$ follows a geometric distribution with success probability $1/2$: $P(N = n) = (1/2)^n$ for $n = 1, 2, 3, \dots$ (the first $n-1$ flips are tails and the $n$th flip is heads). The payout when $N = n$ is $2^n$ dollars. The expected value of the payout is

Since the expected value is infinite, the game does not have a finite fair value. Per the problem instructions, output $-1$.

We have 15 boys and 10 girls, totaling 25 students. The students line up uniformly at random. We want the expected number of adjacent positions where one is a boy and the other is a girl (order BG or GB).

Let $X$ be the number of adjacent boy-girl pairs. For each adjacent pair of positions $(i, i+1)$ with $i = 1,\dots,24$, define an indicator $I_i$ that the two students at those positions are of opposite sexes. Then $X = \sum_{i=1}^{24} I_i$, and by linearity of expectation,

By symmetry, the probability is the same for every adjacent pair; denote it by $p$. Hence $\mathbb{E}[X] = 24 \cdot p$.

Now compute $p$. There are $25 \times 24$ equally likely ordered assignments of two distinct students to the two positions. Favorable cases are (boy, girl) and (girl, boy). There are $15 \times 10 = 150$ ways to choose a boy then a girl, and $10 \times 15 = 150$ ways to choose a girl then a boy, for a total of $300$ favorable ordered pairs. Thus

Alternatively, $\Pr(\text{boy then girl}) = \frac{15}{25} \cdot \frac{10}{24} = \frac{1}{4}$, and $\Pr(\text{girl then boy}) = \frac{10}{25} \cdot \frac{15}{24} = \frac{1}{4}$, so $p = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

Therefore,

The three calls have equally spaced strikes ($\Delta K = 10$) and prices $C_1=4$, $C_2=3.5$, $C_3=2.75$. A no-arbitrage condition for call options is that the price as a function of strike must be convex: $C_1 + C_3 \ge 2C_2$. Here $4 + 2.75 = 6.75 < 7 = 2 \times 3.5$, so convexity is violated, creating an arbitrage.

Construct a butterfly spread: buy one call at strike 1000, buy one call at strike 1020, and sell two calls at strike 1010. The net premium is

so the position generates an upfront credit of $0.25$.

The payoff at expiration $S_T$ is

Evaluating piecewise:

• $S_T \le 1000$: all options expire worthless, $V=0$.
• $1000 < S_T \le 1010$: $V = (S_T-1000) \ge 0$.
• $1010 < S_T \le 1020$: $V = (S_T-1000) - 2(S_T-1010) = 1020 - S_T \ge 0$.
• $S_T > 1020$: $V = (S_T-1000)+(S_T-1020)-2(S_T-1010)=0$.

The payoff is non-negative for all $S_T$ and strictly positive for $1000 < S_T < 1020$. Since the position was entered at a net credit of $0.25$, the guaranteed profit is $0.25$ dollars (25 cents).

Let $q_k$ be the probability that the drunkard eventually steps onto the precipice (ruin) when he is currently $k$ paces from the edge ($k\ge 0$). The precipice is at $k=0$, so $q_0=1$. We want the survival probability $1-q_1$.

For $k\ge 1$, condition on the first step:

Multiply by $3$ and rearrange to obtain the homogeneous linear recurrence

The characteristic equation is $2r^2-3r+1=0$, which factors as $(2r-1)(r-1)=0$, giving roots $r=1$ and $r=\frac12$. Hence

Boundary conditions.

• At $k=0$: $q_0 = 1 \implies A + B = 1$.
• As $k\to\infty$, the walk has a positive drift away from the edge (each step has expected change $+\frac13$). Therefore the probability of ever hitting $0$ from far away tends to $0$, i.e. $\lim_{k\to\infty}q_k = 0$. This forces $A=0$, and consequently $B=1$.

Thus $q_k = \big(\frac12\big)^k$. In particular, the ruin probability starting one pace away is $q_1 = \frac12$. The probability that he never goes over the edge is

A correlation matrix must be positive semidefinite (PSD). For three assets with pairwise correlations $\rho_{AB}=0.9$, $\rho_{BC}=0.8$, and a candidate $\rho_{AC}=0.1$, we check whether the $3\times3$ correlation matrix

is PSD. A necessary and sufficient condition for a $3\times3$ symmetric matrix with unit diagonal is that all principal minors are nonnegative. The $1\times1$ and $2\times2$ minors are clearly nonnegative, so the key condition is $\det(R) \ge 0$.

The determinant of a $3\times3$ correlation matrix is

Substituting the given values:

Since the determinant is negative, $R$ is not positive semidefinite. Therefore $\rho_{AC}=0.1$ is not possible.

Let $M$, $T$, $W$ be the events that Monday, Tuesday, Wednesday produce a matching pair, respectively. We want

1. $P(M^c)$. Total ways to choose $2$ socks from $10$: $\binom{10}{2}=45$. Matching pairs: $5$ (one per color). So $P(M)=5/45=1/9$, hence

2. $P(T^c \mid M^c)$. After a non‑matching Monday, two socks of different colors are removed. The remaining $8$ socks consist of $3$ colors with $2$ socks each and $2$ colors with $1$ sock each. Total pairs from $8$: $\binom{8}{2}=28$. Only the three colors with $2$ socks can produce a matching pair, so $3$ matching pairs. Thus $P(T\mid M^c)=3/28$ and

3. $P(W \mid M^c \cap T^c)$. After Monday and Tuesday are both non‑matching, the composition of the remaining $6$ socks depends on how Tuesday's non‑matching pair was formed. Starting from the Monday state ($3$ colors with $2$ socks, $2$ colors with $1$ sock), Tuesday's non‑matching pair can be of three types:

• Case A: Both socks from the $2$-sock colors. Number of ways: $\binom{3}{2}\cdot2\cdot2 = 12$. Resulting state: $1$ color with $2$ socks, $4$ colors with $1$ sock. Probability of a match on Wednesday: $\frac{1}{\binom{6}{2}} = \frac{1}{15}$.
• Case B: One sock from a $2$-sock color and one from a $1$-sock color. Number of ways: $3\cdot2\cdot2\cdot1 = 12$. Resulting state: $2$ colors with $2$ socks, $2$ colors with $1$ sock, $1$ color with $0$ socks. Probability of a match on Wednesday: $\frac{2}{15}$.
• Case C: Both socks from the two $1$-sock colors. Number of ways: $1$. Resulting state: $3$ colors with $2$ socks, $2$ colors with $0$ socks. Probability of a match on Wednesday: $\frac{3}{15} = \frac15$.

Total non‑matching pairs on Tuesday: $12+12+1 = 25$, so the conditional probabilities of the cases given $T^c$ are $\frac{12}{25},\frac{12}{25},\frac{1}{25}$. Hence

4. Multiply.

Thus the desired probability is $\boxed{\frac{26}{315}}$.

(One can verify by counting all sequences: total sequences $\binom{10}{2}\binom{8}{2}\binom{6}{2}=18900$, favorable sequences $5\cdot24\cdot13=1560$, giving the same fraction.)

High-level idea

In the Black–Scholes model, the price of a European call depends only on the risk‑free rate and volatility, not on the real‑world drift. Hence two assets with the same current price, volatility, strike, maturity, and risk‑free rate have identical European call prices, regardless of their real‑world drifts.

When one asset is also subject to random downward jumps, the comparison changes — but not in the direction naive intuition suggests. Under the risk‑neutral measure the jumps must be compensated by extra drift between jumps so that the forward price is unchanged. The compensated jumps therefore act as a mean‑preserving spread of the terminal price, and the call payoff is convex, so by Jensen's inequality the call on the jump‑exposed asset is more expensive than the call on the pure‑diffusion asset.

Derivation

1. Pure diffusion (no jumps)

Under the risk‑neutral measure $\mathbb{Q}$ the asset follows

so that

The European call price is

with

The formula contains $r$ and $\sigma$ but no real‑world drift $\mu$. Therefore, if two assets share the same $S_0$, $\sigma$, $K$, $T$, and $r$, their European call prices are identical — distinct real‑world drifts are irrelevant.

2. Adding random downward jumps to one asset

Now let one asset follow a jump‑diffusion (Merton 1976). Under a suitable risk‑neutral measure its dynamics are

where $N_t$ is a Poisson process with intensity $\lambda$, $Y_t < 1$ is the downward jump multiplier, and $\kappa = \mathbb{E}^{\mathbb{Q}}[Y_t - 1]$ is the compensator that keeps the expected instantaneous return equal to $r$. For downward jumps $\kappa < 0$, so the drift between jumps is raised above $r$: the compensation exactly offsets the jumps and preserves the forward, $\mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{rT}$ for both assets.

Solving the SDE, the terminal price factorizes as

where $S_T^{\text{diff}}$ is the pure‑diffusion terminal price and $J$ is an independent jump factor with $\mathbb{E}^{\mathbb{Q}}[J] = 1$. Conditioning on $S_T^{\text{diff}}$ and applying Jensen's inequality to the convex payoff $x \mapsto (x - K)^{+}$,

Taking expectations and discounting shows the jump‑exposed call is worth at least as much for every strike — and strictly more whenever the jumps are genuinely random. Equivalently: at the same forward, the jumps add total risk‑neutral variance, and extra dispersion always benefits a convex payoff (a mean‑preserving spread raises the value of a convex function's expectation).

The tempting argument that "downward jumps create negative skew and therefore depress the call" is wrong because it ignores the compensator: between jumps the asset drifts upward faster than $r$, and through the convexity of the payoff this more than makes up for the heavier left tail.

As a numerical check, with $S_0 = K = 100$, $r = 0.02$, $\sigma = 0.2$, $T = 1$, $\lambda = 1$, and jump multiplier $Y = 0.8$, Monte Carlo gives a jump‑diffusion call price of about $12.45$ versus a Black–Scholes price of about $8.92$.

Consequently, the call on the asset with random downward jumps is more expensive than the call on the pure‑diffusion asset.

Final answer

Without jumps the two European calls have the same price. When one asset is also subject to random downward jumps (compensated under the risk‑neutral measure so the forward is unchanged), its call becomes more expensive than the call on the pure‑diffusion asset: the jumps act as a mean‑preserving spread of the terminal price, and the call payoff is convex.

We start from the Ornstein–Uhlenbeck SDE

Step 1. Solve the SDE exactly. Define $Y_t = e^{\theta t} X_t$. By Itô’s lemma,

Substituting $dX_t$,

The terms containing $X_t$ cancel. Integrate from $0$ to $t$,

with $Y_0 = X_0$. Since $\int_0^t \theta e^{\theta s}\, ds = e^{\theta t} - 1$, we obtain

Step 2. Expected value. The Itô integral in (1) is a martingale starting at zero; its expectation is zero. Hence

Step 3. Variance. Only the stochastic term in (1) contributes to the variance. Using Itô isometry,

Equations (2) and (3) are the required expressions.

Idea

Track the game state as the longest suffix of the flip sequence that is a prefix of either target pattern. This yields a small Markov chain whose first-step equations determine the win probability exactly.

States and Transitions

Target patterns: Player 1 wins on $HTH$; Player 2 wins on $HHT$.

The transient states are $\{\varepsilon,\, H,\, HH,\, HT\}$, where $\varepsilon$ denotes no useful suffix (start, or after a progress-resetting tail).

Remark on $HH \xrightarrow{H} HH$: after any run of heads, the longest suffix that prefixes a target is still $HH$ (the length-2 prefix of $HHT$).

System of Equations

Let $p_s$ denote the probability that Player 1 wins from state $s$.

Equation (3) reflects that $HH$ is a trap: every additional $H$ keeps the game in $HH$, and the inevitable first $T$ completes $HHT$, so Player 1 cannot win from $HH$.

Solution

Substituting $(3)$ into $(2)$:

Combined with $(1)$, we have $p_{\varepsilon} = p_H = \tfrac{1}{2}p_{HT}$. Substituting into $(4)$:

The game begins in state $\varepsilon$, so Player 1 wins with probability $\dfrac{1}{3}$.

A virtual function call in C++ adds overhead from an indirect branch, loss of inlining and subsequent compiler optimisations, and the per‑object vptr.

Function pointer alternative – Replacing the virtual method with a stored function pointer removes the vtable indirection but still requires an indirect call; the call cost itself is unchanged. Inlining remains impossible unless the compiler can prove the pointer value at compile time. For truly dynamic dispatch, performance is on par with virtual functions, while the code loses type safety and polymorphic extensibility.

Templates (CRTP / static polymorphism) – By resolving the target type at compile time, templates enable direct calls that are fully inlineable. This eliminates all dispatch overhead and can yield large speedups for small, frequently‑called operations. The trade‑off is the loss of runtime polymorphism: all concrete types must be known at compile time. Template instantiation may also increase binary size and compilation time.

Performance verdict – In hot inner loops where the type is statically known, templates consistently outperform virtual functions. Function pointers rarely provide a meaningful advantage; they replace one indirection with another while sacrificing design clarity. In most real‑world code, virtual function overhead is dwarfed by other costs (memory latency, I/O, algorithmic complexity) and modern branch predictors hide much of the remaining penalty.

Conclusion – Virtual functions remain the right default for runtime polymorphism. They should be replaced with templates only when profiling identifies a hot virtual call site and compile‑time dispatch is feasible. The flexibility and maintainability they offer are well worth their overhead in the vast majority of code.