Find the limiting probability…
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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
asked Dec 4 '18 at 22:59
M. Smith M. Smith
776
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add a comment |
$begingroup$
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.
probability probability-theory probability-distributions markov-chains probability-limit-theorems
asked Dec 4 '18 at 22:59
M. Smith M. Smith
776
$endgroup$
$begingroup$
Your answer is correct. In fact $1-alpha- beta$ could even be negative!
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:42
add a comment |
$begingroup$
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.
probability probability-theory probability-distributions markov-chains probability-limit-theorems
asked Dec 4 '18 at 22:59
M. Smith M. Smith
776
$endgroup$
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
probability probability-theory probability-distributions markov-chains probability-limit-theorems
asked Dec 4 '18 at 22:59
M. Smith M. Smith
776
asked Dec 4 '18 at 22:59
M. Smith M. Smith
776
edited Dec 4 '18 at 23:05
M. Smith
asked Dec 4 '18 at 22:59
M. Smith M. Smith
776
asked Dec 4 '18 at 22:59
M. Smith M. Smith
776
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 4 '18 at 23:43
Misha LavrovMisha Lavrov
44.4k555106
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add a comment |
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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.
answered Dec 4 '18 at 23:43
Misha LavrovMisha Lavrov
44.4k555106
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 23:43
Misha LavrovMisha Lavrov
44.4k555106
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 23:43
Misha LavrovMisha Lavrov
44.4k555106
$endgroup$
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.
answered Dec 4 '18 at 23:43
Misha LavrovMisha Lavrov
44.4k555106
answered Dec 4 '18 at 23:43
Misha LavrovMisha Lavrov
44.4k555106
answered Dec 4 '18 at 23:43
Misha LavrovMisha Lavrov
44.4k555106
answered Dec 4 '18 at 23:43
Misha LavrovMisha Lavrov
44.4k555106
44.4k555106
add a comment |
add a comment |
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$begingroup$
Your answer is correct. In fact $1-alpha- beta$ could even be negative!
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:42