A thief robs a house every night. His profit each night is independent of others, and is a random variable with $Exp(1/\lambda)$ distribution. Every night, there is a probability $0<q<1$ that he is caught and has to return his total profit. What is an optimal strategy for when to stop in order to maximise his total expected profit?
If anyone can help me out with this then that is very good!!
On
His expected gain from the next burglary is $\lambda (1-q)$ while if he already has $V$ (matching kaine's comment) then his expected loss is $Vq$. So, assuming he takes nothing else into account, he should stop when $$ \lambda (1-q) \le Vq$$ which could be written as $\lambda \le V\frac{q}{1-q}$ or $V \ge \lambda \frac{1-q}{q}$ or $q \ge \frac{\lambda}{\lambda + V}$ or as kaine's comment has it $\frac{V}{\lambda} \ge \frac{1-q}{q}$ or $\frac{\lambda}{V} \ge \frac{q}{1-q}$ . The last expression can be interpreted as a statement about comparing the odds of being caught with the expected proportional gain of the next attempt.
Thief wakes up and takes a look at his finances:
He has robbed enough houses to earn V dollars. If he robs another house he expects to win $\lambda$ dollars. He has, however a probability q of getting caught and losing all V dollars he already won.
Therefore: he expects his money to change by $E = \lambda*(1-q)-Vq$.
If $V/\lambda>\frac{1-q}{q}$ then his potential earning (E) from robbing a house will actually be a negative reducing his maximum potential winnings so he shouldn't do it. His predicted winnings, however, haven't been determined like this and he will earn less than $\lambda\frac{1-q}{q}$ on average as he has a chance to lose before reaching that point.