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.

  • 2
    $\begingroup$ Cool idea and thanks for the self flag Jan. I appreciate it. $\endgroup$
    – Bob Jansen
    Commented Jul 14, 2020 at 19:11

5 Answers 5


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

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

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}

  • $\begingroup$ hi was this question for trading or quantitative research/structuring etc? $\endgroup$
    – Mining
    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.

  • $\begingroup$ I believe I have seen this one in Joshi's book $\endgroup$
    – Arshdeep
    Commented Jul 14, 2020 at 19:31
  • $\begingroup$ 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. $\endgroup$ Commented Jul 14, 2020 at 19:32
  • $\begingroup$ @JanStuller hi was this question for trading or quantitative research/structuring etc? $\endgroup$
    – Mining
    Commented Sep 26, 2021 at 15:21
  • $\begingroup$ @Marco: quantitative role, model validation. $\endgroup$ 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.)

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

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,

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

  • $\begingroup$ How do we know this inequality is best possible ? Thx $\endgroup$
    – dm63
    Commented Oct 30, 2022 at 11:46
  • $\begingroup$ 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. $\endgroup$
    – dm63
    Commented Oct 30, 2022 at 13:47
  • $\begingroup$ @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. $\endgroup$ Commented Nov 1, 2022 at 21:57

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.

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.

  • $\begingroup$ hi was this question for trading or quantitative research/structuring etc? $\endgroup$
    – Mining
    Commented Sep 26, 2021 at 15:22
  • $\begingroup$ @Marco quant research. $\endgroup$ Commented Sep 26, 2021 at 15:25

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