Find the limiting probability…
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
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
$endgroup$
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
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
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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026304%2ffind-the-limiting-probability%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
47k657107
47k657107
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026304%2ffind-the-limiting-probability%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Your answer is correct. In fact $1-alpha- beta$ could even be negative!
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:42