# 良い金融のインタビューの質問

（i）スレッドには、「現在」流行の質問が反映されます

（ii）Quantおよび意欲的なQuantsのリソースとして、quant.stackexchange Webサイトに付加価値を与える可能性があります

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.

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?

Answer:

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.)