# 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. – Bob Jansen Jul 14 '20 at 19:11

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 – Arshdeep Singh Duggal Jul 14 '20 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. – Jan Stuller Jul 14 '20 at 19:32

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}

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.)