Every lecture, the number of views is going down. For everyone else doing all the problems, watching the lectures, etc. - you have my respect. I'll see you at the end!
For those confused like me regarding the fishing example: for part d): The number of fish caught can be found in 2 ways Consider 3 Random Variables: F: No. of fish caught T: Time (The values that time can take) FishTime: FishingTime (The values that fishing time takes, note that this is different from Time r.v.) 1) E[F] = E[F|02) E * (1-P(0,2)) + 1 * P(0,2) Here E will not be = lambda * tau This is because if he fishes for exactly 2 hours it means he will definitely not catch 0 fish. So we need to calculate the conditional pmf. That is P(catches k fish|catches atleast 1 fish) Then use this to find the new Expected value by multiplying with k and summing for k = 1, 2.... Then put this for E and calculate. Gives the same answer.
Regarding part d) of finding the expected number of fish to be caught, we can use the total probability theorem as Prof Tsitsiklis suggested. Define X as the rv representing the number of fish caught in the first 2 hours. Then E[# of fish]=E[X|X>0] P(X>0) + E[1|X=0] P(X=0) where E[X|X>0] P(X>0)=[1P(X=1|X>0)+2P(X=1|X>0)+...]P(X>0)=1P(X=1)+2P(X=2)+...=E[X]=1.2
<a href="#" class="seekto" data-time="976">16:16</a> I guess it should be P(0,2)*int_0^3 f_T1(t)dt (probability of catching 0 fish in 2 hours*the probability of the 1st arrival within 3 hours). Then I get the same results but somehow if you use P(0,2)*(1-P(0,3)) it's also the same result
<a href="#" class="seekto" data-time="69">1:09</a>-<a href="#" class="seekto" data-time="74">1:14</a> Notice both what he's saying and whether or not someone is walking through the door behind him.
I'm having trouble reconciling something from the fishing example (<a href="#" class="seekto" data-time="920">15:20</a>). If the rule is that Poisson stops fishing when a fish is caught (as mentioned at <a href="#" class="seekto" data-time="760">12:40</a>), then shouldn't the P of catching at least 2 fish (example (e)) be zero? At <a href="#" class="seekto" data-time="760">12:40</a>, there seems to be a contradiction: (1) "You fish for 2 hours no matter what" AND (2) "You stop fishing as soon as you catch a fish."
The probability is one instead of 1 - P(0,2) since the 0.6*2 is the unconditioned expectation. Think of it in this way, the first two hour will happen no matter what, whether you catch some fish or none. The probability of catching zero fish in the first two hours is already encoded in the expectation 0.6*2.
For the Expected number of fish ( ~ <a href="#" class="seekto" data-time="1140">19:00</a> ), shouldn't he include the possibility of catching N fish during the first two hours? Is the probability of 3 or more fish in 0 < t < 2 insignificant?
Shouldn't the answer to the fish question at <a href="#" class="seekto" data-time="946">15:46</a> be equal to p(0,2)*p(1,3) instead of p(0,2)*(1-p(0,3)) I say this because the fisherman should fish for no longer than 1 fish. In the case of 1-p(0,3), he's considering all the possibilities that he could fish 1,2,3,4,5,6,7..... any number of fishes during that interval? from hour 2....5
I think the reason why it should not be p(1,3) is that p(1,3) means just one success in time interval 3, which completely removes the possibility of getting more than 1 success in that interval.Getting more than 1 success is quite possible as it is a Poisson process. We are stopping after the first success that doesn't mean in the interval (2,5) we can't get more than 1 fish. Other way to look at it is in this example we want to have at least one fish. What's the probability of finding at least one fish? Yes, you guessed it right 1 - P(0,3)
@dhirajupadhyay01 But isn't that exactly how the problem is defined? That if the first fish is caught after the 2h period, you go home instantly. This would imply that you can't fish more than 1 fish in the period of > 2h, thus I think it is in fact P(0, 2) * P(1, 3)
He was choosing intervals at random first, which is when he got 1/2 for both types of intervals. But in the second case he is picking a point at random (not an interval), and points are distributed on the real number line and are more likely to be in intervals of size 10 rather than of size 5.
it's just a hypothetical, almost none of the processes in the world are truly memoryless. A bank having more people between 10 and 12 might mean there will be more people visiting between 12 and 2 cause it's a saturday, or it might mean less people will visit cause all of the bank's clientele visited already etc.. etc. etc.
yea, confusing example. Why isn't the life span of a light bulb normally distributed? Some last very short, most for medium length and some for very long.
The bus inter-arrival time example is a bad/counteractive example. If buses are running at the rate of 4 per hour, You go at a random time for many days and your average waiting time should be 7.5 minutes. Why would the bus company foll people by saying we are running buses whose time is determined by poisson process and not a uniform/predetermined time?
Thanks Roy, I've got the same using total expectation. Regarding f): it must be considered that it's SURE that 2 hours have already been spent fishing if no catch. I get E(T) = 2*(1-P(0,2)) + (2 + 1/0.6)*P(0,2) which reduces to 2 + 1/0.6*P(0,2) as in the lecture.
Its because you are already multiplying by the probabilities of the respective cases to get the expectation. You just have to skip the value 0 which is anyways contributing nothing to the expectation.
Actually at time 19.55 professors says that we are going to wait 2 hours no matter what whether we caught a fish at [0,2] or didn't we still wait for at least 2 hours. So probability of waiting between [0,2] is 1 ! Therefore it is actually like total expectation theorem: 1 * 0.6 * 2 + P(0,2) * 1 The first one is probability of waiting in [0,2] interval, which is 100% . Second 1 is the expected value of the fish, given that we didn't caught any in [0,2] interval.
lambda is the rate of arrival (Arrivals per unit time). When we say 1/lambda, it becomes time per arrival which is the expected value of first arrival. Not sure about the actual intuition, but analysis as above makes sense.
from the previous lecture , we have an exponential random variable from the poisson distribution in the case when k = 1 i.e arrival of the first success in a fixed time t. And expectation of an exponential random variable is 1/lambda, lambda being the arrival rate.
I think he misspoke when he said geometric. I derived the expectation of an exponential distribution, and it came out to be 1/lambda. Did we ever derive that though? Dont have that in my notes, kind of frustrating that that came out of nowhere
Looks like recitation 8 on MIT OCW, question 3, asks us to derive this in 3b. I guess thats why we were expected to know this? Dont tell me I should have been going through recitations, lectures arent enough? (crying emoji)
