Question

Poisson distirubution:

lambda=2

what is P(Y>=4|Y>=2)??

Answer 1

P(Y>=4|Y>=2) = P(Y>=2)

Answer 2

so then i could caluate that as 1-P(Y<2)??

Answer 3

really?

Answer 4

They are completely overlapping events.

Answer 5

Is it given or or?

Answer 6

yes, but \(Y>4\) is contained in \(Y>2\)

Answer 7

i could easily be wrong, but i would say it was
\[\frac{P(y>4)}{P(y>2)}\]

Answer 8

yes that is what the book has, ecxpet if >= for both num and denom

Answer 9

its*

Answer 10

but how do you know that P(>=4) goes in the numerator?

Answer 11

\[
\Pr(A \text{ if } B) =\frac{\Pr(A\text{ and }B)}{\Pr(B)}
\]

Answer 12

so its like saying A^B=A if A is a subset of B

Answer 13

yes, it is
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\] but in this case \(A\cap B=A\)

Answer 14

Oh okay, what if A was not a subset of B?

Answer 15

then it is what @wio and i both wrote above

Answer 16

okay, i believe i got it, what about this formula that i found e^((-lambda)*(t))

Answer 17

since in this case \(A\cap B=A\) it is just \(P(A|B)=\frac{P(A)}{P(B)}\)

Answer 18

\[P(y\geq 2)=1-P(y=0)-P(y=1)\]

Answer 19

is that the calculation for the bottom portion?

Answer 20

\[P(y=0)=e^{-\lambda}\]\[P(y=1)=\lambda e^{-\lambda}\]

Answer 21

yeah that is the bottom

Answer 22

alrigh

Answer 23

i think i got it from here, th

Answer 24

yw

Answer 25

:) thanks

Answer 26

you also need \[P(y=2)=\frac{\lambda^2e^{-\lambda}}{2}\] and
\[P(y=3)=\frac{\lambda^3e^{-\lambda}}{6}\]

Answer 27

general term is
\[P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!}\]

Answer 28

basically to calulate that fraction you are doing:

(1-P(y<4)/(1-P(y<2)