Find the limiting probability…












1












$begingroup$


I have the following problem:



Two men are shooting at a target. We can call them $X$ and $Y$. $X$ shoots after each hit and $Y$ after each miss. Their respective probabilities of hitting the target are $alpha$ and $beta$. If $n$ is the number of shots fired, what is the limiting probability of hitting the target as $n$ approaches infinity?



What I did:



I set up the transition matrix $P$ = begin{bmatrix}alpha&1-alpha\beta&1-beta\ end{bmatrix}
Then I solve the system of equations $pi = pi P leftrightarrow (P^T-I)pi^T=0$ and $pi_1+pi_2=1$ and get $pi = left(frac{beta}{beta+1-alpha}, frac{1-alpha}{beta+1-alpha}right)$.



As we want the limiting probability that the target is hit, since $pi_1$ is the probability that $X$ shoots and $X$ only shoots after someone hits the target, I think the limiting probability should just be $pi_1 = frac{beta}{beta+1-alpha}$.



However, the textbook has the solution as $frac{beta}{1-alpha-beta}$.
I'm not sure if the textbook just has a typing error or my logic is wrong somewhere. It would be helpful if someone could corroborate my work or point out where I made a mistake.



I'm also thinking that the book might be wrong because if $alpha$ and $beta$ are both over $.5$ then the limiting probability will be negative, which makes no sense.










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$endgroup$












  • $begingroup$
    Your answer is correct. In fact $1-alpha- beta$ could even be negative!
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 23:42
















1












$begingroup$


I have the following problem:



Two men are shooting at a target. We can call them $X$ and $Y$. $X$ shoots after each hit and $Y$ after each miss. Their respective probabilities of hitting the target are $alpha$ and $beta$. If $n$ is the number of shots fired, what is the limiting probability of hitting the target as $n$ approaches infinity?



What I did:



I set up the transition matrix $P$ = begin{bmatrix}alpha&1-alpha\beta&1-beta\ end{bmatrix}
Then I solve the system of equations $pi = pi P leftrightarrow (P^T-I)pi^T=0$ and $pi_1+pi_2=1$ and get $pi = left(frac{beta}{beta+1-alpha}, frac{1-alpha}{beta+1-alpha}right)$.



As we want the limiting probability that the target is hit, since $pi_1$ is the probability that $X$ shoots and $X$ only shoots after someone hits the target, I think the limiting probability should just be $pi_1 = frac{beta}{beta+1-alpha}$.



However, the textbook has the solution as $frac{beta}{1-alpha-beta}$.
I'm not sure if the textbook just has a typing error or my logic is wrong somewhere. It would be helpful if someone could corroborate my work or point out where I made a mistake.



I'm also thinking that the book might be wrong because if $alpha$ and $beta$ are both over $.5$ then the limiting probability will be negative, which makes no sense.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Your answer is correct. In fact $1-alpha- beta$ could even be negative!
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 23:42














1












1








1





$begingroup$


I have the following problem:



Two men are shooting at a target. We can call them $X$ and $Y$. $X$ shoots after each hit and $Y$ after each miss. Their respective probabilities of hitting the target are $alpha$ and $beta$. If $n$ is the number of shots fired, what is the limiting probability of hitting the target as $n$ approaches infinity?



What I did:



I set up the transition matrix $P$ = begin{bmatrix}alpha&1-alpha\beta&1-beta\ end{bmatrix}
Then I solve the system of equations $pi = pi P leftrightarrow (P^T-I)pi^T=0$ and $pi_1+pi_2=1$ and get $pi = left(frac{beta}{beta+1-alpha}, frac{1-alpha}{beta+1-alpha}right)$.



As we want the limiting probability that the target is hit, since $pi_1$ is the probability that $X$ shoots and $X$ only shoots after someone hits the target, I think the limiting probability should just be $pi_1 = frac{beta}{beta+1-alpha}$.



However, the textbook has the solution as $frac{beta}{1-alpha-beta}$.
I'm not sure if the textbook just has a typing error or my logic is wrong somewhere. It would be helpful if someone could corroborate my work or point out where I made a mistake.



I'm also thinking that the book might be wrong because if $alpha$ and $beta$ are both over $.5$ then the limiting probability will be negative, which makes no sense.










share|cite|improve this question











$endgroup$




I have the following problem:



Two men are shooting at a target. We can call them $X$ and $Y$. $X$ shoots after each hit and $Y$ after each miss. Their respective probabilities of hitting the target are $alpha$ and $beta$. If $n$ is the number of shots fired, what is the limiting probability of hitting the target as $n$ approaches infinity?



What I did:



I set up the transition matrix $P$ = begin{bmatrix}alpha&1-alpha\beta&1-beta\ end{bmatrix}
Then I solve the system of equations $pi = pi P leftrightarrow (P^T-I)pi^T=0$ and $pi_1+pi_2=1$ and get $pi = left(frac{beta}{beta+1-alpha}, frac{1-alpha}{beta+1-alpha}right)$.



As we want the limiting probability that the target is hit, since $pi_1$ is the probability that $X$ shoots and $X$ only shoots after someone hits the target, I think the limiting probability should just be $pi_1 = frac{beta}{beta+1-alpha}$.



However, the textbook has the solution as $frac{beta}{1-alpha-beta}$.
I'm not sure if the textbook just has a typing error or my logic is wrong somewhere. It would be helpful if someone could corroborate my work or point out where I made a mistake.



I'm also thinking that the book might be wrong because if $alpha$ and $beta$ are both over $.5$ then the limiting probability will be negative, which makes no sense.







probability probability-theory probability-distributions markov-chains probability-limit-theorems






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edited Dec 4 '18 at 23:05







M. Smith

















asked Dec 4 '18 at 22:59









M. Smith M. Smith

776




776












  • $begingroup$
    Your answer is correct. In fact $1-alpha- beta$ could even be negative!
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 23:42


















  • $begingroup$
    Your answer is correct. In fact $1-alpha- beta$ could even be negative!
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 23:42
















$begingroup$
Your answer is correct. In fact $1-alpha- beta$ could even be negative!
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:42




$begingroup$
Your answer is correct. In fact $1-alpha- beta$ could even be negative!
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:42










1 Answer
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$begingroup$

You are correct and the textbook's answer is wrong.



A quick way of getting the limiting probability is that in the long run, we switch from $X$ to $Y$ a $pi_X (1-alpha)$ fraction of the time, and from $Y$ to $X$ a $pi_Ybeta$ fraction of the time. These must be equal from which we deduce that $$pi_X : pi_Y = beta : 1-alpha.$$ Normalizing, $pi_X = frac{beta}{(1-alpha) + beta}$ and $pi_Y = frac{1-alpha}{(1-alpha) + beta}$.



Your method, of course, works equally well.






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    $begingroup$

    You are correct and the textbook's answer is wrong.



    A quick way of getting the limiting probability is that in the long run, we switch from $X$ to $Y$ a $pi_X (1-alpha)$ fraction of the time, and from $Y$ to $X$ a $pi_Ybeta$ fraction of the time. These must be equal from which we deduce that $$pi_X : pi_Y = beta : 1-alpha.$$ Normalizing, $pi_X = frac{beta}{(1-alpha) + beta}$ and $pi_Y = frac{1-alpha}{(1-alpha) + beta}$.



    Your method, of course, works equally well.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You are correct and the textbook's answer is wrong.



      A quick way of getting the limiting probability is that in the long run, we switch from $X$ to $Y$ a $pi_X (1-alpha)$ fraction of the time, and from $Y$ to $X$ a $pi_Ybeta$ fraction of the time. These must be equal from which we deduce that $$pi_X : pi_Y = beta : 1-alpha.$$ Normalizing, $pi_X = frac{beta}{(1-alpha) + beta}$ and $pi_Y = frac{1-alpha}{(1-alpha) + beta}$.



      Your method, of course, works equally well.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You are correct and the textbook's answer is wrong.



        A quick way of getting the limiting probability is that in the long run, we switch from $X$ to $Y$ a $pi_X (1-alpha)$ fraction of the time, and from $Y$ to $X$ a $pi_Ybeta$ fraction of the time. These must be equal from which we deduce that $$pi_X : pi_Y = beta : 1-alpha.$$ Normalizing, $pi_X = frac{beta}{(1-alpha) + beta}$ and $pi_Y = frac{1-alpha}{(1-alpha) + beta}$.



        Your method, of course, works equally well.






        share|cite|improve this answer









        $endgroup$



        You are correct and the textbook's answer is wrong.



        A quick way of getting the limiting probability is that in the long run, we switch from $X$ to $Y$ a $pi_X (1-alpha)$ fraction of the time, and from $Y$ to $X$ a $pi_Ybeta$ fraction of the time. These must be equal from which we deduce that $$pi_X : pi_Y = beta : 1-alpha.$$ Normalizing, $pi_X = frac{beta}{(1-alpha) + beta}$ and $pi_Y = frac{1-alpha}{(1-alpha) + beta}$.



        Your method, of course, works equally well.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 23:43









        Misha LavrovMisha Lavrov

        47k657107




        47k657107






























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