Good Quant-Finance Interview Questions

I know there's the book by the late Mark Joshi and there is a lot of content on the internet. I thought it could be beneficial to additionally start a thread here where we could all share the most interesting interview questions in Quant finance that we have encountered (i.e. a community wiki question: each answer should include one interview question (ideally with an answer): similar to "Good quant finance jokes").

Even if there might be some duplication with other resources, perhaps the added benefit of this thread would be:

1. The thread will reflect the questions that are "currently" in fashion

2. It might add value to the quant.stackexchange website as a resource for Quants and aspiring Quants

Happy to receive constructive criticism, if others don't feel this is a good idea.

• Cool idea and thanks for the self flag Jan. I appreciate it. Commented Jul 14, 2020 at 19:11

Here is one that I got a long time ago in a quant interview:

Question: If $$x = \{ x_1, x_2, \cdots, x_n \}$$ are i.i.d. draws from a random variable $$X \sim {\mathbb U}(0,1)$$, calculate

\begin{align} {\mathbb E}[ \; \max(x) - \min(x) \; ] \end{align}

Answer: I've got two fun solutions to this problem, by CDF and by Integration:

1. CDF $$\to$$ PDF

As expectation is a linear operator, we can re-write the desired quantity as the sum of two expectations $$$$\label{minMaxUniform} {\mathbb E}[ \; \max(x) \; ] - {\mathbb E}[ \; \min(x) \; ]$$$$

Since $$X \sim {\mathbb U}(0,1)$$ is symmetical around 0.5, these must be related by $$$${\mathbb E}[ \; \max(x) \; ] = 1 - {\mathbb E}[ \; \min(x) \; ]$$$$ and we can express the desired expectation in terms of a single quantity $$$$2 \times {\mathbb E}[ \; \max(x) \; ] - 1$$$$

To calculate the expectation of the maximum of $$n$$ draws from $$X$$, let us consider $$\max(x)$$ as its own random variable, and calculate its probability distribution, $$P( \max(x) = k )$$ for $$0 \leq k \leq 1$$.

The probability that $$P( \max(x) \leq k )$$ is simply the probability that all draws $$x_i$$ are less than or equal to k, $$P( x_i \leq k \; \forall \; i \in n )$$ - and since each draw is independent, we can re-express this as a product of independent terms \begin{align} P( \max(x) \leq k ) &= P( x_i \leq k \; \forall \; i \in n )\\ &= \prod_{i=1}^n P( x_i \leq k )\\ &= k^n \end{align}

$$P( \max(x) \leq k )$$ is the cdf of $$\max(x)$$, and we can use the well-known expression to calculate its pdf \begin{align} P( \max(x) = k ) &= {\frac \partial {\partial k}} P( \max(x) \leq k )\\ &= n \cdot k^{n-1} \end{align}

Having calculated the pdf of $$\max(x)$$, we can calculate its expectation in the usual way \begin{align} {\mathbb E}[ \; \max(x) \; ] &= \int_{k=0}^{1} p( \max(x) = k ) \cdot k \cdot dk\\ &= \int_{0}^{1} n \cdot k^{n-1} \cdot k \cdot dk\\ &= \left[ {\frac n {n+1}} k^{n+1} \right]^1_0\\ &= {\frac n {n+1}} \end{align}

Putting this all together, \begin{align} {\mathbb E}[ \; \max(x) - \min(x) \; ] &= {\mathbb E}[ \; \max(x) \; ] - {\mathbb E}[ \; \min(x) \; ]\\ &= 2 \times {\mathbb E}[ \; \max(x) \; ] - 1\\ &= {\frac {2n} {n+1}} - 1\\ &= {\frac {n-1} {n+1}} \end{align} which is the answer

1. Integration

An alternative method to calculate $${\mathbb E}[ \; \max(x) \; ]$$ is to integrate over each $$x_i$$. By symmetry, the probability of any of $$n$$ variables $$x_i$$ being the maximum is $${\frac 1 n}$$, so we integrate over the region in the $$n$$-dimensional space for which $$x_1$$ is the maximum and multiply by $$n$$ \begin{align} {\mathbb E}[ \; \max(x) \; ] &= \Bigl( \int_0^1 \Bigr)^{n} \max(x) \prod_{i=1}^n dx_i\\ &= n \cdot \int_{x_1=0}^1 x_1 \Bigl( \int_0^{x_1} \Bigr)^{n-1} \prod_{i=1}^n dx_i\\ &= n \cdot \int_{x_1=0}^1 x_1 \prod_{i=1}^n \Bigl( \left[ x_i \right]^{x_1}_0 \Bigr)^{n-1} dx_1\\ &= n \cdot \int_{x_1=0}^1 x_1^n \cdot dx_1\\ &= n \cdot \left[ {\frac 1 {n+1}} x_1^{n+1}\right]_0^1\\ &= {\frac n {n+1}} \end{align}

And so using the logic from the final step of the earlier solution,

\begin{align} {\mathbb E}[ \; \max(x) - \min(x) \; ] &= 2 \times {\mathbb E}[ \; \max(x) \; ] - 1\\ &= {\frac {2n} {n+1}} - 1\\ &= {\frac {n-1} {n+1}} \end{align}

• hi was this question for trading or quantitative research/structuring etc? Commented Sep 26, 2021 at 15:21

To start the thread, let me share the most recent interview question I have been asked:

Question: Denote standard Brownian motion as $$W(t)$$. Compute the probability that:

$$\mathbb{P}(W(1)>0 \cap W(2)>0)$$

Answer: Using the independence of increments property, we have $$W(2) = W(2-1) + W(1)$$. Denote $$W(2-1)$$ as $$Y$$ and $$W(1)$$ as $$X$$. Then:

$$\mathbb{P}(W(1)>0 \cap W(2-1)+W(1)>0)=\mathbb{P}(X>0 \cap Y+X)>0)=\mathbb{P}(X>0 \cap Y>-X)$$

By definition of Brownian motion, the independent increments are jointly Normally distributed. So $$X$$ and $$Y$$ are jointly normal with density $$f_{X,Y}(u,v)$$. We can write:

$$\mathbb{P}(X>0 \cap Y>-X)=\int_{u=0}^{u=\infty}\int_{v=-u}^{v=\infty}f_{X,Y}(u,v)dv du$$

The final step is to draw the domain of the double integral: $$X>0$$ means we're interested in the right-hand side of the cartesian $$X,Y$$ plot. Then with $$Y>-X$$, this further carves out the area below the line $$Y=(-X)$$ on the right-hand side of the $$X,Y$$ plot: i.e. we cut the "bottom $$1/4$$" of the right-hand half. So we are left with $$3/4$$ of $$1/2$$ of the $$X,Y$$ domain, which is $$3/8$$. Since the jointly normal PDF is a symmetrical cone centred on $$x=0, y=0$$, the double integral is actually equal to $$3/8$$ by symmetry.

• I believe I have seen this one in Joshi's book Commented Jul 14, 2020 at 19:31
• There will inherently be some duplication here. But at least it'll be interesting to see which questions are still being asked these days compared to ten years ago. Commented Jul 14, 2020 at 19:32
• @JanStuller hi was this question for trading or quantitative research/structuring etc? Commented Sep 26, 2021 at 15:21
• @Marco: quantitative role, model validation. Commented Sep 26, 2021 at 15:51

Question: A contract pays $$P(T,T+\tau) - K$$ at $$T$$, where $$K$$ is fixed and $$P(\cdot,S)$$ is the price of a $$S$$-maturity zero-coupon bond (ZCB).

What is $$K$$ for which the contract's time $$t$$ price is null?

Replication pricing:

At time $$t$$, we go long one $$T+\tau$$-maturity ZCB and short $$P(t,T)^{-1}P(t,T+\tau)$$ $$T$$-maturity ZCB's.

Time $$t$$ cost of this position is $$0$$ as:

$$(-1)\cdot P(t,T+\tau) + P(t,T)^{-1}P(t,T+\tau)\cdot P(t,T) = 0.$$

At time $$T$$, as the shorted bond matures, we have a flow of $$- P(t,T)^{-1}P(t,T+\tau).$$

But we are also expecting $$1$$ dollar flow at $$T+\tau$$, whose price at time $$T$$ is:

$$P(T,T+\tau).$$

Hence, the $$t$$ price of payout (at time $$T$$)

$$P(T,T+\tau) - P(t,T)^{-1}P(t,T+\tau)$$

is $$0$$. This is of course exactly our contract with

$$K = P(t,T)^{-1}P(t,T+\tau).$$

Pricing under $$T$$-forward measure:

$$V_t = P(t,T)\mathbf{E}^{T}_t[P(T,T+\tau) - K]$$

Setting $$V_t$$ to $$0$$ implies:

$$K = \mathbf{E}^{T}_t[P(T,T+\tau)]$$

As $$P(t,T+\tau)$$ is a traded asset, under $$T$$-forward measure, process $$\left(P(t,T)^{-1} P(t,T+\tau)\right)_{t\geq 0}$$ is a martingale, which leads to: $$\mathbf{E}^{T}_t[P(T,T)^{-1} P(T,T+\tau)] = P(t,T)^{-1} P(t,T+\tau).$$ Due to $$P(T,T)=1$$, we have:

$$K = \mathbf{E}^{T}_t[P(T,T+\tau)] = P(t,T)^{-1}P(t,T+\tau)$$

Pricing under money market account measure:

$$V_t = \beta_t\mathbf{E}_t[\beta_T^{-1} (P(T,T+\tau) - K)]$$

Setting $$V_t$$ to $$0$$ implies:

$$K = \mathbf{E}_t[\beta_T^{-1}]^{-1}\mathbf{E}_t[\beta_T^{-1} P(T,T+\tau)]$$

$$= P(t,T)\mathbf{E}_t\left[\beta_T^{-1} \mathbf{E}_T[\beta_T \beta_{T+\tau}^{-1} ] \right]$$

$$= P(t,T)^{-1}\mathbf{E}_t\left[ \mathbf{E}_T[ \beta_{T+\tau}^{-1} ] \right]$$

$$= P(t,T)^{-1}\mathbf{E}_t\left[ \beta_{T+\tau}^{-1} \right]$$

$$= P(t,T)^{-1}P(t,T+\tau),$$

using tower property of conditional expectations in the penultimate equality.

(Note: not necessarily a recent question, but expected to be asked - I flunked the replication pricing part that the interviewer was obviously enamored with; this is covered by both Brigo/Mercurio's book, in the context of FRA pricing, and by Andersen/Piterbarg's book, forward bond price.)

• hi was this question for trading or quantitative research/structuring etc? Commented Sep 26, 2021 at 15:21
• @Mining quantitative analyst
– ir7
Commented Nov 8, 2021 at 21:24

Question
Let $$\mathbf{C}$$ be a $$n\times n$$ covariance matrix such that all diagonal elements are equal to 1, and the non-diagonal ones to $$\rho$$ with $$-1\leq\rho\leq1$$. Which range of values is admissible for $$\rho$$?

Solution 1
Let $$X_1,\dots,X_n$$ be a sequence of independent random variables with unit variance and pairwise correlation $$\rho$$ for any $$i\not= j$$. Let $$Y:=\sum_iX_i$$ then: \begin{align} \notag V\left(Y\right) &=\sum_{i=1}^nV\left(X_i\right)+\sum_{i\not=j}Cov(X_i,X_j) \\ &=n+n(n-1)\rho \end{align} The variance of $$Y$$ is positive, therefore: \begin{align} n+n(n-1)\rho\geq0 \quad\Leftrightarrow\quad \boxed{\rho\geq\frac{1}{1-n}} \end{align}

Solution 2
The covariance matrix $$C$$ cen be written as

$$C=(1-\rho)\mathbf{I}+\mathbf{u}\mathbf{u}^T$$ where $$\mathbf{I}$$ is the identity matrix and $$\mathbf{u}$$ is a vector composed of $$\sqrt{\rho}$$. A covariance matrix must be positive semidefinite , hence its smallest eigenvalue $$\lambda_0$$, must be $$\lambda_0\geq 0$$.

The eigenvalues are found from the roots of the determinant equation:

$$\mathrm{det}(\Sigma-\lambda\mathbf{I})=\mathrm{det}((1-\rho-\lambda)\mathbf{I}+\mathbf{u}\mathbf{u}^T)=0$$

By the matrix determinant lemma, the determinant is found as

\begin{align} \mathrm{det}(\Sigma-\lambda\mathbf{I})&=\left(\mathbf{u}^T\left((1-\rho-\lambda)\mathbf{I}\right)^{-1}\mathbf{u}\right)\mathbf{det}\left((1-\rho-\lambda)\mathbf{I}\right)\\ &=\left(1-\lambda+(n-1)\rho\right)\left(1-\rho-\lambda\right)^{n-1} \end{align}

The first eigenvalue is $$\lambda_1=1-\rho$$, with multiplicity $$n-1$$. The second eigenvalue is $$\lambda_2=1+(n-1)\rho$$. The smallest admissible eigenvalue, zero, is reached at either $$\rho=1$$ or $$\rho=-\frac{1}{n-1}$$. Hence,

$$\boxed{-\frac{1}{n-1}\leq\rho\leq1}$$

• How do we know this inequality is best possible ? Thx
– dm63
Commented Oct 30, 2022 at 11:46
• Thinking about my own question: you can generate an inequality like this using any linear combination of the $X_i$. Indeed it does seem that the ‘best’ inequality occurs when all the coefficients are 1.
– dm63
Commented Oct 30, 2022 at 13:47
• @dm63 I don't have a clear proof for the general case so I have restricted the problem to the case where the associated variance matrix is the unit matrix. Commented Nov 1, 2022 at 21:57

Background
Consider the affine Linear Gauss Markov (LGM) model for Interest Rates, characterized by a single-factor state variable $$x_t$$ with normal dynamics... \begin{align} \text{d}x_t&=\sigma(t)\text{d}W_t \end{align} ... specified in a measure under which the price process $$N_t$$: \begin{align} N_t&:=\frac{1}{P(0,t)}e^{H(t)x_t+a(t)}, \end{align} is a valid numéraire, where $$P$$ is the price of a zero-coupon bond, while $$H(t)$$ and $$a(t)$$ are two deterministic functions. Note this model is also known as the Hagan and Woodward parameterization of Hull-White, see this answer.

Question
Determine the structure of the function $$a(t)$$ to ensure the LGM model is arbitrage-free.

We know a model is arbitrage-free if and only if there exists an equivalent martingale measure (EMM), namely a probability measure such that the price of a traded asset is equal to the conditional expectation of its discounted cash flows. The basic asset in any rate model is the zero-coupon bond, which pays $$\\\1$$ at expiry. Hence our LGM model must satisfy: $$P(0,t)=E\left(\frac{1}{N_t}\right)$$ Per the definition of $$N_t$$, the equivalent condition is: $$E\left(e^{-H(t)x_t-a(t)}\right)=1\tag{1}$$ The state variable $$x_t$$ is normally distributed, with zero mean and total variance up to $$t$$ equal to: $$\Sigma(t):=\int_0^t\sigma^2(u)\text{d}u$$ Expectation $$(1)$$ can be explicitly calculated, for example by invoking the Laplace transform of a normal variable, and we get: $$\boxed{a(t) = \frac{1}{2}H^2(t)\Sigma(t)}$$ Interestingly, we note that compared to the Hull-White parameterisation, where the calibrated parameter needs to be updated whenever the curve changes to remain arbitrage-free (e.g. see the specification of the function $$\theta(t)$$ in this answer), the LGM model is arbitrage-free by design provided we set the function $$a(t)$$ to be equal to the expression above.