Q1-4

Quant Interview Questions

Acknowledgement: Questions are from 150-Most-Frequently-Asked-Questions-on-Quant-Interview

Q1

Put options with strikes 30 and 20 on the same underlying asset and with the same maturity are trading for $6 and $4, respectively. Can you find an arbitrage?

Answer:- CFA I, key words:option, portfolio,strike

Q2

The number \(2^{29}\) has 9 digits, all different. Without computing \(2^{29}\), find the missing digit.

Answer:- Key words: Combinations theory, number theory, recall:

\[ 9 \mid n- D(n) \] which sum of all its digits is divisible by 9, thus we have:

\[ 9 \mid 2^{29}- D(2^{29}) \]

\[ 9 \mid \sum_{j=0}^9 j -D(2^{29}) \]

Q3 Tower Rule

Exception of \(\Phi(X+a)\), assume another \(Y\sim N(0,1)\), such that \(\Phi(X+a)= P(Y<X+a)\) , take its expectation:

\[ E_x(\Phi(X+a)) = E_x(P(Y<X+a)|x) \]

Since X,Y are independent, we can apply the tower rule:

\[ \begin{align} =P(Y<X+a)=P(Y-X<a)\\ =P(\frac{Y-X}{\sqrt 2} < \frac{a}{\sqrt 2} ) \\ = \Phi(\frac{a}{\sqrt{2}}) \end{align} \]

Q4 Throw Coins, dices

Throw coins can be converted to a markov chain question, Expected step to recurrent states

draw a graph solving this: Expected number of throwing HHH sequence:

\[ \begin{align} u_0=0.5(1+u_0)+0.5(1+u_1)=1+0.5u_0+0.5u_1 \\ u_1 =0.5(1+u_0)+0.5(1+u_2)\\ u_2 = 0.5(1+u_0+0.5) \end{align} \]

Note, u_0 are steps from that state to the recurrent state.

To understand this, think about u_0 first, 0.5(1+u_0) means, there’s 0.5 chance the first throw is not H, hence waste this step, and plus itself step u_0, and 0.5(1+u_1) is assuming the first toss is H already, we are remaining u_1 step, due to 0.5+0.5 =1 has to be, there’s no terms for u_2.

Dices

Expected sum for throwing 6 side dices, until the first 1 is appear.

thoughts: Dice is a discrete uniform distribution. Using Law of total expectation could find the expected tosses of getting a 1:

\[ E = \frac{1}{6}*0+ \frac{5}{6}(1+E) = 5 \]

5/6 chance to throw not a 1, 1/6 chance to throw 1 at the first time.

\[ E(sum) = 5 * \frac{2+3+4+5+6}{5}+1=21 \]

first term indicate the expected sum on the mean 5 tosses not getting a 1 ,plus 1 indicate the 1 toss that get a 1.