Probability

Question

A company entered into a new market (sells product X) and started operating. In the first month of operation, it is found that  25% of customers are using the product X and while 75% of people still using other substitutes of X.

The company has it's own records as follows: (from other cities or etc.)

• If somebody uses product X in current month, for him, there is 80% chance of keep using X in the next month, and 20% chance of switching to other substitute.
• If somebody not using product X in current month, for him, there is 30% chance of using X in the next month, and 70% chance of not changing to X.

Find the % of customers using the product X after a long time. 

Comments

Yeah

---

In long run if we assume 

• x = % of people using product X
• y = % of people not using product X , then, 

Solution fo the following equation gives the value of x and y.

$$\begin{align*} \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 0.8 & 0.2\\ 0.3 & 0.7 \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix} \end{align*}$$ 

x = 0.6

diagrammatically,

Assuming steady state exists,

Total probabilities along the red paths must be equal to x (No change) $=>x.0.8+y.0.3 = x$

Total probabilities along the green path must be equal to y (No change) $=>x.0.2 + y.0.7 = y$

Special case Regular Makarov chain

---

Can u pls give some more added information in this matrix like solution . I feel this is easy to understand .

Answer 1

P.S-: I am editing my Answer as Previous answer had some error.

$\Rightarrow$ Let $U_{n}$ be the Event that a Person is using the product in the $nth$ month.

Given$P\left ( U_{1} \right )=0.25$

$\Rightarrow$ Probablity that if person is using the product in the $nth$ month,he will use it in next month$\Rightarrow$

$P \left ( \left( U_{n+1} \right )|P\left ( U_{n} \right ) \right )=$0.80

$\Rightarrow$ Probablity that if person is not using the product in the $nth$ month,he will use it in next month$\Rightarrow$

$P \left ( \left( U_{n+1} \right )|P\left ( U_{n} \right ){}' \right )=$0.30

Let us first find the probablity that the person uses the product in next month 

$P\left ( U_{n+1} \right )=P\left (U_{n+1} | U_{n} \right )*P\left ( U_{n} \right )+P\left (U_{n+1} | U_{n}{}' \right )*P\left ( U_{n}{}' \right )$

$P\left ( U_{n+1} \right )=0.80*P\left ( U_{n} \right )+0.30*P\left ( U_{n}{}' \right )$

$P\left ( U_{n+1} \right )=0.80*P\left ( U_{n} \right )+0.30*\left ( 1- P\left ( U_{n} \right )\right )$

$P\left ( U_{n+1} \right )=0.5* P\left ( U_{n} \right )+0.3$

Solving the recurrence relation

---

Let $P_{n+1}=P\left ( U_{n+1} \right )$, with $P\left ( U_{1} \right )=0.25$

So our reccurence will be simplified to

$P_{n+1}=0.5*P_{n}+0.3$

our roots are $r-0.5=0\Rightarrow 0.5$

Solving for $P_{n} $,

$P_{n}=\alpha * root^{n}+\beta \Rightarrow P_{n}=\alpha * 2^{-n}+\beta$

$P_{n+1}=\alpha * 2^{-\left ( n+1 \right )}+\beta$$\cdots \cdots \cdots \cdots eqtn 1$

Also ,$P_{n+1}=0.5*P_{n}+0.3$ $

so 

$P_{n+1}=0.5 *\alpha * 2^{-n}+\beta+0.3$

             =$1/2 \left ( \alpha * 2^{-n}+\beta \right )+0.3$

             =$\left ( \alpha * 2^{-\left ( n+1 \right )}+\beta /2 \right )+0.3$$\cdots \cdots \cdots \cdots eqtn 2$

From eqtn 1 and 2,

$\left ( \alpha * 2^{-\left ( n+1 \right )}+\beta /2 \right )+0.3=\alpha * 2^{-\left ( n+1 \right )}+\beta$

$\Rightarrow \beta=6/10$

Given that $P\left ( U_{1}\right )=25$

We have to find $\alpha$

we have $P_{n}=P\left ( U_{n}\right )=\alpha * 2^{-n}+\beta$

take n=1$\Rightarrow P_{1}=P\left ( U_{1}\right )=\alpha * 2^{-1}+0.6=25$

            $\Rightarrow \alpha * 0.5^{1}+0.6=25 \Rightarrow \alpha \approx 24$

we have 

       $P_{n+1}=\alpha * 2^{-\left ( n+1 \right )}+\beta$

       $P_{n+1}=24* 2^{-\left ( n+1 \right )}+0.6$

Taking $n$ as $\infty$(As we are not given a limited month tenure for which it will last ,it is asking for "long time ")

$P_{n+1}=0.6$

Thus Answer =60%

Comments

Actually few weeks back .I asked that head tail pair question in math.stack-exchange. Almost similar question in gate 2016, There someone solved it using this kind of matrix equations. Then I did not get his method at all. Later i viewed his reference and found few, good questions (may be possible inside gate scope I don't know)  

---

@Debasish how r u putting P_n is constant k?

The question is very interesting

it seems whatever we take last month we must use product X,

So, probability of X and Other(O) will be P(X),P(OX),P(XX),P(OOX),P(OXX),P(XOX),P(XXX)..........................

---

It the method to find particular solution of a recurrence relation after getting the homogeneous solution from the roots for nonhomogeneous recurrence relations.

http://web.cs.wpi.edu/~cs504/s00m/notes/recurrence/solve/step2/step2.html

Answer 2

One using X product P(X)=0.25

One using Not X "   P(NX)=0.75

One using X product , again use X product in next month=P(XX)=0.25$\times$0.8=0.2

One using Not X product , use X product in next month=P(NXX)=0.75X0.3=0.225

So, If we put it in a recurrence relation

0.25a_n=0.2a_n+1+0.225a_n+1

So, Solving this recurrence we get value as r=58.8%