Question

On a given day, a particular raccoon will eat the trash from one of three different houses. If he eats from the trash of a particular house, he has a 50% chance to eat from the same house the next day, and a 25% chance each to eat from one of the other two houses. What is the stochastic matrix for this scenario

Answer 1

Answer:

Step-by-step explanation:

On a given day , a particular raccoon will eat the trash from one of three different houses.

Let assume $Y_n$ be a random variable that illustrating  the house raccoon will eat on an unknown given nth day.

If he eats from the trash of a particular house, he has a 50% chance to eat from the same house the next day, and a 25% chance each to eat from one of the other two houses.

There are three states given in the above statement.

So, we can have state 1, state 2 and state 3

Assuming that:

state 1 = house 1

state 2 = house 2

state 3 = house 2

If he eats from the trash of a particular house,

For state 1 : he has a 50% chance to eat from the same house the next day

i.e state 1 = 0.50

and a 25% chance each to eat from one of the other two houses.

For state 2 and state 3: = 0.25

i.e state 2 = 0.25

state 3 = 0.25

NOW:

$\mathtt{P[Y_{n+1 }= 0 \ | \ Y_n = 0] = 0.5}$

$\mathtt{P[Y_{n+1 }= 1 \ | \ Y_n = 0] = 0.25}$

$\mathtt{P[Y_{n+2}= 2 \ | \ Y_n = 0] = 0.25}$

$\mathtt{P[Y_{n+1 }= 0 \ | \ Y_n = 1] = 0.25}$

$\mathtt{P[Y_{n+1 }= 1 \ | \ Y_n = 1] = 0.5}$

$\mathtt{P[Y_{n+1 }= 2 \ | \ Y_n = 1] = 0.25}$

$\mathtt{P[Y_{n+1 }= 0 \ | \ Y_n = 2] = 0.25}$

$\mathtt{P[Y_{n+1 }= 1 \ | \ Y_n = 2] = 0.25}$

$\mathtt{P[Y_{n+1 }= 2 \ | \ Y_n = 2] = 0.5}$

The stochastic matrix for this scenario can be computed as:

                  0          1       2

$P = \left\begin{array}{c}0\\1\\2\end{array}\right \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]$

$\mathbf{ P =\left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right] }$