An interview questions with ANZ

[xscizj]

Dear All:

I just finished my first round interview with ANZ. And one question they ask is:

suppose a stock has a process: dS(t) = sigma*dB(t), where B(t) is a standard Brownian motion, and
the current stock price is S(0). There is a barrier H>S(0). Then what is the probability that the stock
price will breach H before time T (T>0) ?

 

[vertigo]

S_t = S_0 + sigma * B_t.

Define Pi_x : = inf { t >0 such that B_t = x } to be the first hitting time of the Brownian motion on the level x at time t.

See that

inf {t > 0 such that S_t = H} = inf { t>0 such that S_0 + sigma * B_t = H } = inf { t > 0 such that B_t = y} = Pi_y,

where y = (H-S_0)/sigma.

Now, Pi_x is an Inverse Gamma distribution with parameters a = 0.5 and b = 0.5*x^2. For any random variable Z that has an Inverse Gamma distribution with parameters a_1, b_1, the distribution function is

Prob (Z < T; a_1,b_1) = = Gamma_inc(a_1,0.5*(b_1^2)*(1/T),'upper')/Gamma(a_1),

where Gamma_inc(, , 'upper') is the upper incomplete gamma function and Gamma() is the gamma function.

The answer to your question is given by Prob ( Pi_y < T;a,b).

reference: to see why Pi_x is inverse Gamma, see Klebaner's book on stochastic calculus, page 73.

 

[PepeQuant]

S(t) = S(0) + Sigma*W(t)
X(t) = S(t)-S(0) = Sigma*W(t)
M(T) = max{X(t)} for 0 < t < T
inf {t > 0 such that X(t) = H-S(0)}

P(t < T) = P(M(T) > H-S(0))
P(X(T) > H-S(0)) = P(X(T) > H-S(0) & M(T) > H-S(0)) + P(X(T) > H-S(0) & M(T) < H-S(0)), 2nd part is 0
=P(X(T) > H-S(0) & M(T) > H-S(0)) = P(X(T) > H-S(0) | M(T) > H-S(0)) * P(M(T) > H-S(0))

----> P(M(T) > H-S(0)) = P(X(T) > H-S(0)) / P(X(T) > H-S(0) | M(T) > H-S(0))
by reflection principal, P(X(T) > H-S(0) | M(T) > H-S(0)) = P(X(T) > H-S(0) | X(t) = H-S(0)) = 1/2, equallly likely for going above or below from H-S(0) if X(t) = H-S(0)

----> P(t < T) = P(M(T) > H-S(0)) = 2 * (P(X(T) > H-S(0)) = 2 * ( 1 - F((H-S(0)) / sqrt(sigma*T)) ), F~CDF of N(0, 1)

 

[stonehead]

Use reflection principle. There's a hint in this problem that the stock price is normal.