Question

Please help explain this problem for Stat midterm review:

A population starts with one member at time t=1. It either divides in two with probability p or dies with probability 1-p. If it divides, then both of its children behave independently with the same two alternatives at time t=2. Let X혵, t=1,2,... be the number of members of the population at the time t.

1. Find the probability distribution of the random variable X₂ (this means describe the possible values of X₂ and their probabilities. Note that X₁ takes the value 1 with the probability 1).

Answer 1

2. Find the probability distribution of the random variable X₃.

Answer 2

And I'm not sure why there's an odd symbol at "Let Xt, t = 1,2,..."

Answer 3

if i a reading this right, Xt means population at time t
so X1 means populatoin at time t = 1
X2 means population at time t = 2
and in general Xt means population at time t

Answer 4

correct

Answer 5

so X1 = 1 because you start at time t = 1 with one

Answer 6

X2 can take on two values, either X2 = 0 or X2 = 2 (dies or splits)

Answer 7

it is 2 with probabity p and 0 with probability 1-p

Answer 8

so you can write
\[P(X2=2)=p,P(X2=0)=1-p\] and that is the complete distribution because those are the only possible values for X2

Answer 9

X3 can have several values:

0 meaning died at first step or they both die
3 meaning you had two, one died and one split
4 meaning both split
and i think that is it. does that seem reasonable?

Answer 10

Aren't possible outcomes at t=3: 0, 2, 4?

Answer 11

maybe
at step 2 you either have 0 or 2
now if both split you have 4
if one dies and one splits you have 3
if both die you have 0
no i don't think you can have 2

Answer 12

Why wouldn't they both either die or split? That makes the possible outcomes either 2 or 4. I'm confused :/

Answer 13

you have two separate whatever these are

they could both die at the next step
they could both split at the next step

but one could die and one could split
they don't have to both die or both split

Answer 14

if they both die you don't have any , so that would be 0 not 2
if they both spit you would have 4
and if one dies and one splits you would have 3
i think those are the only possibilities

still have to find the probability of each

Answer 15

ok, so considering that those are my possible outcomes, how would I go about answering part 2? Would I start with P(0) = (1-p) + p(1-p)(1-p)?

Answer 16

yes exactly what i was going to say!! i think you got the rest yes?

Answer 17

that represents
1) died at first step
2) split at first step and then both died
so i believe that is correct

Answer 18

\[P(X3=4)=p^3\] i fiam not mistaken
split on first step and then both split at second step


\[P(X3=3)\] is probably the annoying one

Answer 19

well not really
split split no split times 2 because you have to ways for one to split and on not
so i am guessing it should be
\[P(X3=3)=2p^2(1-p)\] but don't take my word for it

Answer 20

How did you get 2p²(1-p)? p [p･(1-p)･p] ?

Answer 21

And should I leave my answer as just that?

Answer 22

I think the outcomes at t=3: 0, 2, 4?
For some reason Sat was thinking died, split --> 3 But died->0 and split->2 for a total of 2

Answer 23

That's what I was saying. But even so, I still get my distribution to be 2p²(1-p) for p(2).
Am I right?

Answer 24

Let me use q= (1-p) to save typing