Quant job interview question: martingale

Hello!

One more question to warm up to your next job interview.

A standing martingale (what?)

1) consider the process:

$X(t)=\int_0^t W(\tau) d\tau$

where $W$ is a Wiener process under a probability measure $P$. Is $X$ a martingale under $P$?

2) What about this other random variable?

$X(t)=\int_0^t y(\tau) d\tau$

with $dy=\sigma dW$.

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Answer:

1) let's remember that a Martingale read

$X(t)=E[X(T)|F(t)]$

for all $t<T$. The filtration $F(t)$ denotes the availability of information at $t$. Therefore, a Martingale process is one that the best estimate of the conditional expectation value in a future time is simply the current value of the process. So in order to answer this question we need to prove such condition.

Hence, piece of cake:

$E[X(T)|F(t)]=E \left[\int_0^T W(\tau)d\tau|F(t) \right]$

$=E\left[\int_0^tW(\tau)d\tau + \int_t^TW(\tau)d\tau|F(t) \right]$

$=X(t)+E\left[\int_t^TW(\tau)d\tau|F(t) \right]$

$=X(t)+\int_t^TE\left[W(\tau)|F(t) \right]d\tau$

Using the conditional expectation, one finally gets

$=X(t)+\int_t^TW(t)d\tau$

$=X(t)+W(t) \left(T-t \right)$

Therefore, the process $X(t)$ is not a Martingale.

2) Similarly to the solution above, let's consider the expectation of the process

$E \left[X(T)|F(t)\right] = E\left[\int_0^T y(\tau) d\tau|F(t) \right]$

$= E\left[\int_0^t y(\tau) d\tau + \int_t^T y(\tau) d\tau|F(t) \right]$

$= X(t) + E\left[\int_t^T y(\tau) d\tau|F(t) \right]$

$= X(t) + \int_t^T E\left[y(\tau)|F(t) \right] d\tau$

By definition, the random variable $y(t)$ is a Martingale, which can be easily verified by the fact the differential stochastic equation does not have a term going with $dt$. Hence

$= X(t) + \int_t^T y(t) d\tau$

And finally

$= X(t) + y(t)\left( T-t \right)$

Therefore, the process $X(t)$ is also not a Martingale.


Good luck!

May the Force be with you.

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Diogo de Moura Pedroso

LinkedIn: www.linkedin.com/in/diogomourapedroso