Question

Consider a sequence of independent trials. At each trial, the possible outcomes are “S” and “F”. On trial n, the probability of getting “S” is equal to Pn = (1/4)^n, n = 1, 2, . . . Let T be the waiting time until the first occurrence of “S”. What is E(T)? (expectation of T)

Answer 1

i will guess 4

Answer 2

oh maybe i am misinterpreting the problem let me think before i speak. it looks the probability of getting a success on the nth trial is not \(\frac{1}{4}\) but rather \(\frac{1}{4^n}\)

Answer 3

Yeah.. someone was telling me though expectation infinity, I'm not sure how they got it.

Answer 4

is infinity*

Answer 5

me neither. i cannot figure out how to add this up

Answer 6

when T = 1 (success on the first trial), p = 1/4

Answer 7

when T = 2 (success on the 2nd trial), P = (3/4)(1/16)
i think the problem iE(T) s the sum of all probabilities.

Answer 8

probabilities are
\[\frac{1}{4}\]\[\frac{3}{4}\times \frac{1}{4^2}\]\[\frac{3}{4}\times \frac{4^2-1}{4^2}\times \frac{1}{4^3}\]\[\frac{3}{4}\times \frac{4^2-1}{4^2}\times \frac{4^3-1}{4^3}\times \frac{1}{4^4}\]
etc

Answer 9

then you have to multiply these by 1,2,3,... and the add
my addition is not that good

Answer 10

would this be some kind of geometric distribution?

Answer 11

if it was i could add it, but it does not look like one to me as there is not common ratio

Answer 12

the method used for the expected waiting time for bernoulli trials comes from summing the derivative of the geometric series, but this isn't that one either

Answer 13

I'm not too sure what the underlying pmf would be... Can we say though that P(T < infinity) < 1?

Answer 14

in fact i cant even figure out how to write this one as a formula

Answer 15

\[P(T=n)=\frac{(4-1)(4^2-1)...(4^{n-1}-1)}{4^{\frac{n(n+1)}{2}}}\] or something like that

Answer 16

then you have to multiply each term by \(n\) and then you have to sum it up
yikes

Answer 17

o.o oh goodness..

Answer 18

this question seemed fairly simple but it is giving me such a headache

Answer 19

i may be wrong, but i think that is what you have to do
on the other hand it is also possible that this series doesn't converge

Answer 20

no it is not simple
lets think for a second
maybe the terms don't even go to zero

Answer 21

well maybe they do
sorry i cannot be of more help, maybe re-post and you will get a better answer

Answer 22

Alright I appreciate it the help anyways :)

Answer 23

the*