Stationary Distribution for this conditional probability table

Question

What is the stationary distribution for the following conditional probability table?

Answer 1

Hey there, 

Here, as it is seen, from sunny weather, the probability of the weather to be sunny is 0.6, to be cloudy is 0.4, and to be rainy is 0.1. So, as main features, we look at the column names, if we want to choose the weather sunny (row sun) and find the probabilites of the transition of weather to go from sunny to sunny, cloudy, rainy, we look at sun column. Based on these, you can see the stationary distribution:

Psun​=0.6Psun​+0.4Pcloudy​+0.1Prain ​
\pi_{cloudy} = 0.3 \pi_{sun} + 0.2 \pi_{cloudy} + 0.4 \pi_{rain}Pcloudy​=0.3Psun​+0.2Pcloudy​+0.4Prain​
\pi_{rain} = 0.1 \pi_{sun} + 0.4 \pi_{cloudy} + 0.5 \pi_{rain}Prain​=0.1Psun​+0.4Pcloudy​+0.5Prain      and their sum = 1

After simpifying them:

1. Psun − 0.6 Psun = 0.4 Pcloudy + 0.1 Prain
0.4 Psun = 0.4 Pcloudy + 0.1 Prain -> p_sun = Pcloudy + 0.25 Prain 

2. Pcloudy − 0.2 Pcloudy = 0.3 Psun + 0.4 Prain

0.8 Pcloudy = 0.3 Psun + 0.4 Prain -> Pcloudy = 0.375 Psun + 0.5 Prain

3.  We put 2 to 1 :

Pcloudy = 0.375(Pcloudy + 0.25 Prain) + 0.5 Prain​ ->   Pcloudy = 0.95 Prain

Let's find Psun ->   Psun = Pcloudy + 0.25 Prain = 0.95 Prain + 0.25 Prain 

Psun = 1.2 Prain

Psun + Pcloudy + Prain = 1

1.2 Prain + 0.95 Prain + Prain = 1

3.15 Prain = 1

Prain ≈ 0.317 others :

Pcloudy = 0.95 × 0.317 = 0.301

Psun = 1.2 × 0.317 = 0.381

Answer 2

S = Sun

C = Cloudy

R = Rain

We find π = (a, b, c) such that πP = π and a + b + c = 1

Translation matrix:

       0.6  0.3. 0.1

P =  0.4  0.2. 0.4

       0.1  0.4  0.5

Solving it:

a = 0.6a + 0.4b + 0.1c

b = 0.3a + 0.2b + 0.4c

a + b + c = 1

So, this gives us:

4a - 4b - c = 0

-3a + 8b - 4c = 0

b = 19/24a

c = 5/6a

Using a + b + c = 1

a = 8/21, b = 19/63, c = 20/63

So Final stationary distribution is:

π = (8/21, 19/63, 20/63)

So, in the long run the system spends about 38.1% Sun, 30.2% Cloudy and 31.7% Rain

Answer 3

Hello professor, 

Let the stationary distribution be:

• s = probability of Sun

• c = probability of Cloudy

• r = probability of Rain

Then:

s = 0.6s + 0.4c + 0.1r

c = 0.3s + 0.2c + 0.4r

r = 0.1s + 0.4c + 0.5r

Also:

s + c + r = 1

From the first equation:

s - 0.6s = 0.4c + 0.1r

0.4s = 0.4c + 0.1r

4s = 4c + r

r = 4s - 4c

From the second equation:

c - 0.2c = 0.3s + 0.4r

0.8c = 0.3s + 0.4r

Substitute r = 4s - 4c:

0.8c = 0.3s + 0.4(4s - 4c)

0.8c = 0.3s + 1.6s - 1.6c

0.8c + 1.6c = 1.9s

2.4c = 1.9s

c = 19s / 24

Then:

r = 4s - 4c
= 4s - 4(19s / 24)
= 4s - 19s / 6
= 5s / 6

Now use:

s + c + r = 1

s + 19s / 24 + 5s / 6 = 1

24s / 24 + 19s / 24 + 20s / 24 = 1

63s / 24 = 1

s = 8 / 21

Then:

c = 19 / 63

r = 5 / 18

Using decimal values:

• Sun = 0.381

• Cloudy = 0.302

• Rain = 0.278

So the stationary distribution is approximately:

(0.381, 0.302, 0.278)

Answer 4

Salam!
 

Let the stationary distribution be:

pi = (s, c, r)

A stationary distribution means the probabilities stay the same after one step, so:

pi = piP

That gives:

s = 0.6s + 0.4c + 0.1r

c = 0.3s + 0.2c + 0.4r

r = 0.1s + 0.4c + 0.5r

Also:

s + c + r = 1

Solving these gives:

s = 8/21

c = 19/63

r = 20/63

So the stationary distribution is:

pi = (8/21, 19/63, 20/63)

In decimal form:

pi ≈ (0.3810, 0.3016, 0.3175)

Final answer:

Sun = 8/21 ≈ 0.3810

Cloudy = 19/63 ≈ 0.3016

Rain = 20/63 ≈ 0.3175

Answer 5

Let pi_S, pi_C, and pi_R represent the stationary probabilities for Sun, Cloudy, and Rain. Based on the transition matrix, they must satisfy the stationarity condition and sum to 1:

• 0.6(pi_S) + 0.4(pi_C) + 0.1(pi_R) = pi_S

• 0.3(pi_S) + 0.2(pi_C) + 0.4(pi_R) = pi_C

• 0.1(pi_S) + 0.4(pi_C) + 0.5(pi_R) = pi_R

• pi_S + pi_C + pi_R = 1

By substituting pi_R = 1 - pi_S - pi_C into the first two equations and solving the system, you get the exact fractions:

• pi_S = 8/21

• pi_C = 19/63

• pi_R = 20/63

Final Distribution Vector: [8/21, 19/63, 20/63]

Answer 6

p_s = 0.6*p_s + 0.4*p_c + 0.1*p_r => 0.4*p_s = 0.4*p_c + 0.1*p_r => p_s = p_c + 0.25*p_r

p_c = 0.3*p_s + 0.2*p_c + 0.4*p_r => 0.8*p_c = 0.2*p_s + 0.4*p_r 

p_r = 0.1*p_s + 0.4*p_c + 0.5*p_r => 0.5*p_r = 0.1*p_s + 0.4*p_c => p_r = 0.2*p_s + 0.8*p_c

p_r = 0.2*p_c + 0.2*0.25*p_r + 0.8*p_c

0.95*p_r = p_c

p_s = 0.95*p_r + 0.25*p_r = 1.2*p_r 

p_c = 0.3*1.2*p_r + 0.2*p_c + 0.4*p_r

0.8*p_c = 0.76*p_r => p_c = 0.76*p_r/0.8

p_s + p_c + p_r = 1 => 1.2*p_r + 0.76*p_r/0.8 + p_r = 1 => p_r = 1/(1.2+0.76/0.8+1) = 8/25.2 = 2/6.3

p_s = 0.8/2.1

p_c = 0.19/6.3 

p_r = 2/6.3

Answer 7

Hello,

Here is my solution according to the given:

Transition table:

From Sun -> (Sun 0.6, Cloudy 0.3, Rain 0.1)

From Cloudy -> (Sun 0.4, Cloudy 0.2, Rain 0.4)

From Rainy -> (Sun 0.1, Cloudy 0.4, Rain 0.5)

Let the stationary distribution be:

pi = (s, c, r)

where

s = P(Sun)

c = P(Cloudy)

r = P(Rainy)

For stationary distribution:

pi = pi P

and also:

s + c + r = 1

So we write the equations:

For Sun:

s = 0.6s + 0.4c + 0.1r

s - 0.6s = 0.4c + 0.1r

0.4s = 0.4c + 0.1r

4s = 4c + r

r = 4s - 4c

For Cloudy:

c = 0.3s + 0.2c + 0.4r

c - 0.2c = 0.3s + 0.4r

0.8c = 0.3s + 0.4r

For Rainy:

r = 0.1s + 0.4c + 0.5r

r - 0.5r = 0.1s + 0.4c

0.5r = 0.1s + 0.4c

5r = s + 4c

r = (s + 4c) / 5

Now use the two expressions for r:

4s - 4c = (s + 4c) / 5

5(4s - 4c) = s + 4c

20s - 20c = s + 4c

19s = 24c

c = 19s / 24

Now find r:

r = 4s - 4c

r = 4s - 4(19s/24)

r = 4s - 19s/6

r = 24s/6 - 19s/6

r = 5s/6

Now apply:

s + c + r = 1

s + 19s/24 + 5s/6 = 1

s + 19s/24 + 20s/24 = 1

24s/24 + 19s/24 + 20s/24 = 1

63s/24 = 1

21s/8 = 1

s = 8/21

Then:

c = 19s/24 = 19/63

r = 5s/6 = 20/63

Final stationary distribution:

pi = (8/21, 19/63, 20/63)

Approximately:

P(Sun) = 0.381

P(Cloudy) = 0.302

P(Rainy) = 0.317

So the stationary distribution is:

Sun = 8/21 ~= 0.381

Cloudy = 19/63 ~= 0.302

Rainy = 20/63 ~= 0.317

Answer 8

Hello,

Let the stationary distribution be π = (s, c, r) for (Sun, Cloudy, Rain).

Solve πP = π with s + c + r = 1:

From equations:

c = (19/24)s, r = (5/6)s

Normalize:

s + c + r = s + (19/24)s + (5/6)s = 1

-> s × (63/24) = 1

-> s = 8/21

Then:

c = 19/63

r = 20/63

Final answer:

π = (8/21, 19/63, 20/63)

(≈ Sun: 0.381, Cloudy: 0.302, Rain: 0.317)

Answer 9

Stationary distribution: πP=π  and probabilities sum to 1.

From the table

$$s=0.6s+0.4c+0.1r$$
$$c=0.3s+0.2c+0.4r$$
$$r=0.1s+0.4c+0.5r$$

$$s+c+r=1$$

$$After\; solving\; final\; answers\; are\; approximately:$$Sun: 0.381, Cloudy: 0.302, Rain: 0.317

Answer 10

Hello,

Stationary distribution π=(x,y,z):

πP=π, x+y+z=1

πP=π:

x=0.6x+0.4y+0.1z

y=0.3x+0.2y+0.4z

z=0.1x+0.4y+0.5z

x=8/21​

y=19/63​

z=20​/63

π=(​8/21,19/63​,20/63​)​=(24/63,19/63,20/63)