John asked in Science & MathematicsMathematics · 8 months ago

# Question about probabilities?

Hello, i`d like to know why the probabilities From the image are:

Pr[Xk+1=1 | Xk=1]=0.5

Pr[Xk+1=2 | Xk=1]=0.5

Pr[Xk+1=1 | Xk=2]=0.2

Pr[Xk+1=2 | Xk=2]=0.8

Thank you so much

### 1 Answer

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• jibz
Lv 6
8 months ago
Favourite answer

Empirically, the domain of the given Markov sequence X(1), ... , X(8) has just two elements (a.k.a. "states"): 1 and 2; so its transition matrix A will be 2×2. To construct A we inspect terms X(1), ... , X(7), noting that

2 of them are equal to 1 and

5 of them are equal to 2,

then ask: How many of them...

...is equal to 1 and is succeeded by a 1? Answer:1 term, namely X(4).

...is equal to 1 and is succeeded by a 2? Answer:1 term, namely X(5).

...is equal to 2 and is succeeded by a 2? Answer:4 terms, namely X(1), X(2), X(6), and X(7),

...is equal to 2 and is succeeded by a 1? Answer:1 term, namely X(3)

(NB our scope excluded X(8) because it doesn't have a successor.) So, to rehash using set notation where #• is used to denote the number of elements in •, we've worked out that

#{k∈{1, ... , 7} : X(k)=1} = 2.

#{k∈{1, ... , 7} : X(k)=2} = 5.

#{k∈{1, ... , 7} : X(k)=1 and X(k+1)=1} = 1.

#{k∈{1, ... , 7} : X(k)=1 and X(k+1)=2} = 1.

#{k∈{1, ... , 7} : X(k)=2 and X(k+1)=2} = 4.

#{k∈{1, ... , 7} : X(k)=2 and X(k+1)=1} = 1.

Finally, recall that given any state a and b, we have

P[X(k+1)=a | X(k)=b] = #{k∈{1, ... , 7} : X(k)=b and X(k+1)=a} / #{k∈{1, ... , 7} : X(k)=b},

and

entry A(a,b) = Pr[X(k+1) = a | X(k) = b].

• ...Show all comments
• jibz
Lv 6
8 months agoReport

oops, typo. i meant to write P[X(k+1)=1 | X(k)=2] = 1/5. you're dividing by the number of entries equal to 2.

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