Options & Derivatives Interview Questions

Sample Answer

No, the stock's expected return $\mu$ does not appear in the Black-Scholes price. The derivation constructs a riskless hedge that eliminates exposure to the stock's drift, so the option price depends only on $\sigma$, $r$, $S$, $K$, and $T$. Under risk-neutral pricing, you replace $\mu$ with $r$ and take expectations under the risk-neutral measure $\mathbb{Q}$, which means investors' risk preferences and the real-world drift are irrelevant to the price. This is one of the most counterintuitive and important results in the framework.

Sample Answer

You could use Black-Scholes with a single constant $\sigma$ estimated from historical data, or you could use the market's implied volatility for that specific strike and expiry. Using implied volatility wins here because it is a model-free extraction of the market's consensus and naturally incorporates the skew and kurtosis that heavy tails produce, giving you a price consistent with traded options. Constant historical vol fails when the return distribution is fat-tailed because Black-Scholes assumes lognormal returns, systematically underpricing OTM puts and deep OTM calls. However, implied vol itself breaks down as a pricing tool when you try to use a single number across strikes, since the volatility smile or skew means each strike has its own implied vol, revealing that Black-Scholes is being used as a quoting convention rather than a true model.

Sample Answer

Start by noting that under $\mathbb{Q}$, $\ln S_T \sim \mathcal{N}(\ln S_0 + (r - \sigma^2/2)T,\, \sigma^2 T)$. You want $\mathbb{Q}(S_T > K)$, so you standardize: $\mathbb{Q}\left(\frac{\ln S_T - \mathbb{E}[\ln S_T]}{\sigma\sqrt{T}} > \frac{\ln K - \mathbb{E}[\ln S_T]}{\sigma\sqrt{T}}\right)$, and flipping the inequality gives you $N(d_2)$ where $d_2 = \frac{\ln(S/K) + (r - \sigma^2/2)T}{\sigma\sqrt{T}}$. Now $N(d_1)$ is subtler: it arises from $\mathbb{E}^{\mathbb{Q}}[S_T \mathbf{1}_{S_T > K}]$, which after a change of numeraire to the stock measure $\mathbb{Q}^S$ becomes $S_0 e^{rT} N(d_1)$. So $N(d_1)$ is the probability that $S_T > K$ under the stock-price measure, or equivalently, it equals the option's delta.

Sample Answer

You could sell short-dated ATM straddles to reduce vega, or you could sell longer-dated options instead. Selling short-dated straddles is problematic because short-dated ATM options have very high gamma relative to their vega, so you would destroy your gamma neutrality while barely denting your vega exposure. The better approach is to sell longer-dated options, which carry high vega per unit of gamma, allowing you to reduce vega without significantly disturbing your gamma-neutral position. You can then fine-tune any residual gamma with short-dated instruments.

Sample Answer

Start by thinking about what drives the value of a deep OTM call: its entire value is time value, so it depends heavily on the probability of the underlying reaching the strike before expiry. Vega is typically the dominant sensitivity here because any increase in implied volatility dramatically raises the probability of finishing in the money, which is a large percentage change relative to the small premium. Theta comes next: the option bleeds time value daily, and since the option is all time value, theta as a fraction of price is meaningful, though the absolute dollar amount is moderate. Rho is the smallest because interest rate sensitivity scales with the present value of the strike weighted by exercise probability, which for a deep OTM call is very low. So the ranking is vega, then theta, then rho in terms of $|\text{Greek} \times \text{unit change}| / \text{option price}$.

Sample Answer

Start by thinking about supply and demand: portfolio managers systematically buy OTM puts for downside protection, which bids up their prices and therefore their implied vols. Layer on the empirical fact that equity returns exhibit negative skewness and excess kurtosis, meaning large down moves are more frequent than Black-Scholes assumes, so higher put implied vols reflect genuine tail risk. The leverage effect reinforces this: when spot drops, firm leverage rises, which increases equity volatility, creating a negative correlation between returns and vol. Now consider steepening: if there is a sudden market crash or a spike in credit spreads, demand for downside protection surges while realized negative skewness increases, pushing OTM put vols sharply higher relative to ATM and OTM call vols.

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This question is checking whether you can identify event-driven vol dynamics and reason about term structure mechanics. A sharply inverted term structure almost always signals a known near-term event, most likely an earnings announcement or FDA decision, that is expected to cause a large discrete move. The short-dated options embed this jump risk, inflating their implied vol, while longer-dated options dilute the event's impact over a larger time horizon. Once the event passes and the jump is realized (or not), you should expect the front-end vol to collapse rapidly as the event premium evaporates, and the term structure to revert toward a flatter or normally upward-sloping shape, with the 3-month vol adjusting modestly based on the magnitude of the realized move.

Sample Answer

This question is checking whether you can distinguish between European put-call parity (an equality) and the American version (inequalities), and whether you understand early exercise. For American options on non-dividend-paying stocks, the relationship is $S - K \leq C - P \leq S - K e^{-rT}$. A deep ITM American put can be worth more than $K - S$ because it carries the right to early exercise and capture interest on the strike proceeds, so $P > K - S$ is not itself a violation. You need to check whether the observed prices actually breach the American bounds before claiming arbitrage. Your colleague is likely confusing the European parity equality with the looser American inequalities.

Sample Answer

The standard move is to use the dividend-adjusted put-call parity: $C - P = S - PV(D) - K e^{-rT}$. But here, the discrete dividend matters because you must discount it to the present. You compute $PV(D) = 2 e^{-0.05 \times 0.25} \approx 1.975$, and $K e^{-rT} = 100 e^{-0.05 \times 0.5} \approx 97.53$. The right side is $105 - 1.975 - 97.53 = 5.495$, while the left side is $9 - 5 = 4$. Since $C - P$ is too low, you buy the call, sell the put, short the stock (collecting the dividend obligation), and lend $PV(D) + K e^{-rT}$ at the risk-free rate, netting roughly $1.50 in arbitrage profit.

Sample Answer

The standard move is to check whether the underlying distribution admits a closed-form solution. But here, the arithmetic average of lognormal variables is not lognormal, so Black-Scholes does not directly apply. You have two practical paths: either use a moment-matching approximation (Turnbull-Wakeman or Levy) where you fit a lognormal to the first two moments of the arithmetic average, giving you an adjusted vol $\sigma_A$ and forward $F_A$ to plug into a Black-Scholes-like formula, or you run Monte Carlo simulation on the daily price path and average the terminal payoffs. The moment-matching approach is fast and usually accurate enough for trading, while Monte Carlo gives you flexibility to handle discrete sampling, term structure effects, and skew.

Sample Answer

Get this wrong in production and you end up with massive P&L swings from trying to hedge an effectively infinite delta near the strike at expiry. The digital's delta behaves like the derivative of a step function, approximated by $\Delta \approx \frac{e^{-rT}}{\sigma\sqrt{T}} \phi(d_2)$, which blows up as $T \to 0$ when $S \approx K$. In practice, you cannot delta hedge this cleanly, so you instead replicate the digital as a tight call spread: long a call at $K - \epsilon$, short a call at $K + \epsilon$, scaled by $1/(2\epsilon)$. This call spread super-replication caps your delta exposure at the cost of a small overhedge, and you choose $\epsilon$ based on your transaction costs and the tick size of the underlying. The residual risk is pure pin risk, which you manage through position limits and accepting the binary outcome near expiry.

Sample Answer

Get this wrong in production and you face margin calls and forced liquidation at the worst possible moment. What happened is that your individual deltas were hedged assuming independent moves, but as correlations spiked toward 1, all your short puts moved in the money together, creating a massive net short delta at the portfolio level that your name-by-name hedges did not capture. The right call is to immediately buy index futures or SPX calls as a macro hedge to stop the bleeding, because hedging 50 names individually during a selloff is too slow and too expensive given widened spreads. Going forward, you need to stress test your book under correlation regimes, not just individual Greeks, and maintain a standing tail hedge via index puts or a correlation swap overlay.

Sample Answer

Selling straddles sounds reasonable but gives you path-dependent P&L because you are taking on gamma risk alongside the vega hedge, meaning you have not isolated the volatility exposure cleanly. VIX futures do not work well here because VIX tracks SPX implied vol, which can diverge significantly from single-stock implied vol due to dispersion and idiosyncratic moves. That leaves variance swaps on the individual names (or a basket approximation) as the cleanest hedge, since they provide pure exposure to realized versus implied variance without the gamma rebalancing noise of straddles. You should be aware that single-stock variance swaps can be illiquid, so in practice you may use a combination: variance swaps on the most liquid names and index variance swaps plus dispersion trades to approximate the rest.