Question

Let Y be the number of requests, per minute for files from a file server. Y is a random variable with Poisson distribution with parameter 6.

Let \(Y_1, Y_2, Y_3\) denote the number of requests in 3 successive minutes. Assume that \(Y_i\) have same distribution. Let for each \(i = 1, 2, 3, A_i\) be the event \(\{Y_i=0\}\). Assume also that the events \(A_1, A_2, A_3\) are mutually independent.

a) Calculate the probability that there will be no requests in entire 3 minutes period.
b) Give a full set of conditions that ensure that \(A_1,A_2,A_3\) are mutually independent.

chars limit ended

Answer 1

a) So answer should be just \(P(Y_1=0)+P(Y_2=0)+P(Y_3=0)=3\cdot e^{-6}\) ?
b) What does mutually independent means? is it mutual exclusive event or independent events?

Answer 2

What is a Poisson distribution?

Answer 3

same as binomial distribution but n approaches to infinity

Answer 4

For a, you use the same procedure as we do for all such indep probabilities: multiply the three probabilities, don't add.

Answer 5

A set of events { Pj } are mutually independent if
\[ P(P_j | P_k) = P_j \ \ \hbox{ for all } j \neq k \]

Answer 6

sorry.
i am trying to figure out what this is really asking.

\[Y_1=0\] means no calls in the first minute?

Answer 7

do I need to prove it or just enough to write \(P(Y1|Y2)=P(Y1)\) (and others) since I am asked to give full set of conditions?

Answer 8

i think yes

Answer 9

and therefore
\[P(Y_1=0)=e^{-3}\] yes?

Answer 10

@tomas.a, you can't "prove it"; you're basically writing down a careful definition of what mutually indep here means.

Answer 11

\(e^{-6}\)

Answer 12

sorry typo there, right
\[P(Y_1=0)=e^{-6}\]

Answer 13

pmf for Poisson is \(P(x)=\frac{\lambda^x e^{-\lambda}}{x!}\)

Answer 14

@jamesj correct me if i am wrong, but if this is poisson with mean 6 this means expect 6 calls per minute, and also necessarily 18 calls every 3 minutes. meaning that the probability you get no calls in 3 successive minutes must be
\[e^{-18}\]

Answer 15

yeah he said that I need to multiply, not add so \(e^{-6}\cdot e^{-6}\cdot e^{-6}=e^{-18}\)

Answer 16

which by the laws of exponents is exactly what you get if you multiply the probabilities together

Answer 17

Exactly.

Answer 18

yes you multiply because
1) they are independent and
2) it is an AND

Answer 19

statement

Answer 20

but i am confused by part 2 because if i remember correctly the fact that you have a poisson distribution means they MUST be independent

Answer 21

yeah i know that AND multiplication and OR addition but i mixed them ><