# Proof of Markov Chain: Bunny rabbit has three dens A, B and C. It likes A better than

#### coconi

##### New member
Hi all,

I have the following question in my assignment but I do not know how to proof:

Bunny rabbit has three dens A, B and C. It likes A better than B and C. If it's in B or C on
any night, it will always take chance 0.9 to go to A and chance 0.1 to go to the other den for the
following night. Once it reaches A, it will stay there for two nights and the third night will be in
B or C with equal chance 0.5. Let Xn be the stats Bunny is in for night n.

Question:
If the state space of the {Xn} is given by {First night at den A; Second night at den A; den B; den C}
then is {Xn} a MC?

Thanks.

#### Subhotosh Khan

##### Super Moderator
Staff member
Hi all,

I have the following question in my assignment but I do not know how to proof:

Bunny rabbit has three dens A, B and C. It likes A better than B and C. If it's in B or C on
any night, it will always take chance 0.9 to go to A and chance 0.1 to go to the other den for the
following night. Once it reaches A, it will stay there for two nights and the third night will be in
B or C with equal chance 0.5. Let Xn be the stats Bunny is in for night n.

Question:
If the state space of the {Xn} is given by {First night at den A; Second night at den A; den B; den C}
then is {Xn} a MC?

Thanks.

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

http://www.freemathhelp.com/forum/announcement.php?f=33

#### coconi

##### New member

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

http://www.freemathhelp.com/forum/announcement.php?f=33

Hi Subhotosh,

Thanks. I would read the rules first.

Also, here is my idea:

To prove it it a MC, I may need to list all the possible conditional probability likes:

P(Xn+1 = "A" | Xn = "A", Xn-1 = "A") = 0 (Impossible to stay 3 nights)
P(Xn+1 = "A" | Xn = "A", Xn-1 = "B") = 1 (must stay 2 nights at A)
P(Xn+1 = "A" | Xn = "A", Xn-1 = "C") = 1 (must stay 2 nights at A)

So it is not a MC?