I have a general question about a homework assignment that deals with a uniform distribution of a continuous variable. Here is the question (and the parts of the question):
Suppose parking rules are enforced from 8AM to 12AM (midnight), say from $t = 8$ to $t = 24$. It is also known that the parking rules are enforced once a day. It is also known that a ticket can be issued anytime during the 16-hour period (ie. any time point is equally likely). Let $T$ denote the time of enforcement.
a. Suppose you do not have the right to park your vehicle in the parking lot, but you parked there from 10AM ($t = 10$) to 1PM ($t = 13$). What is the probability of receiving a ticket?
b. If a ticket is 160 dollars, what is the expected loss from the illegal parking?
c. The parking rate is 2 dollars per hour. What is an appropriate action based on the expected loss. Justify your answer.
My answer for a and b:
a. First of all $S_t = (8,24)$, and since T is continuous, $f(t) = \frac{1}{24 - 8} = \frac{1}{16}, 8 < t < 24$. To find the probability of getting a ticket between $t = 10$ and $t = 13$, I did this:
$P(10 < t < 13) = \int_{10}^{13}\frac{1}{16}dt = \frac{3}{16} = 0.1875$
b. This is where I get a little confused. So far, I made a new variable called $L$ which is loss. And the support is $S_L = (0,160)$, where 0 is no ticket and 160 is the price of getting a ticket.
For the expected loss, since I think I'm trying to find $E(160)$, then this would just $= 160$. Is it really that easy?
Finally, I'm confused for part c. If the parking rate is 2 dollars per hour, then the expected loss if you pay for parking for 3 hours would be $E(6) = 6$. This would mean that in order to be worth it (I think), $E(pay) > E(160)$. But since this never happens in a 16 hour period, then the appropriate action would be to just purchase parking. Is this correct, or am I not taking something into account?
