Suppose a boy at a circus is given two options to catch balls in 5 minutes from a ball throwing machine, a sticky catcher and a net. The sticky catcher will always certainly catches 50 balls (p=1). The net usually catches 12 balls/per minute and follows a Poisson process. Which option should he use to catch the balls?

Utility of option 1 => 50 balls. But I am confused in describing the utility of option 2, should it be equal to => p(catching 0 balls in 5mins)*0 balls + p(1 ball)*1 + p(2)*2 + p(3)*3+........+p(huuuuge number (n) balls in 5 mins)*n?

please explain and let me know if I'm right...

  • $\begingroup$ The net follows a Poisson process, or the ball-throwing machine does? $\endgroup$ – Anna SdTC Feb 7 '17 at 23:51
  • $\begingroup$ Ah, I understand, the net catches a number of balls $X$ which is distributed like a Poisson with rate $\lambda=12$ balls/minute. Then yes, you need to compare the expected value of the number of balls to be caught with the net with the expected (sure) value of the number of balls to he caught with the sticy catcher. $\endgroup$ – Anna SdTC Feb 7 '17 at 23:57
  • $\begingroup$ The net follows the poisson process...so am i doing it correct? am i calculating it correctly? i mean calculation of expected value for the net...is that how its supposed to be done? $\endgroup$ – Muhammad Abbas Feb 8 '17 at 0:48
  • $\begingroup$ Yes, but notice that a Poisson distribution with rate parameter $\lambda$ has expected value equal to $\lambda$ itself. $\endgroup$ – Anna SdTC Feb 8 '17 at 0:58
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    $\begingroup$ Oh...yeah.... :D got it! $\endgroup$ – Muhammad Abbas Feb 8 '17 at 1:07

The Poisson process catches $12$ balls per minute on average, so over $5$ minutes that's $60$ balls on average. More exactly, the number of balls caught $N$ is Poisson-distributed with mean $60$ so we have $$ P(N=n) = e^{-60}\frac{60^n}{n!}.$$ From this you can calculate that the expected number of balls caught is $60$ and the standard deviation is $\sqrt{60}.$

So the net catches $10$ more balls on average but also experiences fluctuations. Which one has higher utility depends on how risk-averse you are (i.e. how your utility function penalizes risk). You haven't given any information as to how this is handled in your class so I can't tell if you're on the right track. You seem to be just computing the expected value which I suspect misses the point. (But if it is the case that you are just maximizing the expected number of balls, then the net is best.)

One way to quantify the risk/reward is that the net has a "Sharpe ratio" of $$\frac{10}{\sqrt{60}} \approx 1.3$$ relative to the risk-free sticky catcher.


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