Probability mass functions don't have to sum to 1?
$begingroup$
I was thinking about the discrete random variable describing the stopping time ($T$: random variable modelling the toss number where he first reaches his target) of a wealthy gambler reaching his target. It is discussed in some detail here: Gambler with infinite bankroll reaching his target and here: Probability that random walk will reach state $k$ for the first time on step $n$. I realized that when the coin is biased against the wealthy gambler, there is a finite chance he will never reach his target. So, if you calculate the summation:
$$sum_{t=0}^infty P(T=t)$$
you will only get $1$ if the coin he is using has a probability, $pgeq frac 1 2$ of heads. Otherwise, the summation above will result in a number less than $1$. Looking at the definition on Wikipedia, no where does it say that the probability mass function should sum to $1$ (emphasis: in the formal definition). However, right outside the scope of the formal definition, it does.
But this would imply that the wealthy gamblers stopping time when $p < frac 1 2$ has no PMF?
Just wanted to get the community's opinion on this.
Also, if we conclude the PMF doesn't have to sum to $1$, is there then any example of a corresponding probability density function that doesn't integrate to $1$? Perhaps the stopping time (defined as reaching a positive boundary) of a continuous time random walk with negative drift?
EDIT: saying that "never reaching the target is included in the possible outcomes" is not satisfying. We are talking about the random variable $T$. This random variable has a certain domain (which includes $infty$). Summing over the domain should give you $1$. Where in its domain should we fit "never reaching the target"? The fundamental problem remains, is $P(T=t)$ the PMF of $T$ or not? If we say it isn't because it doesn't sum to $1$ over all possible values of $T$, then does it mean $T$ doesn't have a PMF?
probability probability-theory random-variables
asked 1 hour ago
Rohit PandeyRohit Pandey
1,2281021
$endgroup$
|
show 1 more comment
$begingroup$
I was thinking about the discrete random variable describing the stopping time ($T$: random variable modelling the toss number where he first reaches his target) of a wealthy gambler reaching his target. It is discussed in some detail here: Gambler with infinite bankroll reaching his target and here: Probability that random walk will reach state $k$ for the first time on step $n$. I realized that when the coin is biased against the wealthy gambler, there is a finite chance he will never reach his target. So, if you calculate the summation:
$$sum_{t=0}^infty P(T=t)$$
you will only get $1$ if the coin he is using has a probability, $pgeq frac 1 2$ of heads. Otherwise, the summation above will result in a number less than $1$. Looking at the definition on Wikipedia, no where does it say that the probability mass function should sum to $1$ (emphasis: in the formal definition). However, right outside the scope of the formal definition, it does.
But this would imply that the wealthy gamblers stopping time when $p < frac 1 2$ has no PMF?
Just wanted to get the community's opinion on this.
Also, if we conclude the PMF doesn't have to sum to $1$, is there then any example of a corresponding probability density function that doesn't integrate to $1$? Perhaps the stopping time (defined as reaching a positive boundary) of a continuous time random walk with negative drift?
EDIT: saying that "never reaching the target is included in the possible outcomes" is not satisfying. We are talking about the random variable $T$. This random variable has a certain domain (which includes $infty$). Summing over the domain should give you $1$. Where in its domain should we fit "never reaching the target"? The fundamental problem remains, is $P(T=t)$ the PMF of $T$ or not? If we say it isn't because it doesn't sum to $1$ over all possible values of $T$, then does it mean $T$ doesn't have a PMF?
probability probability-theory random-variables
asked 1 hour ago
Rohit PandeyRohit Pandey
1,2281021
$endgroup$
$begingroup$
I mean, it says it must sum to $1$ in the wiki article you linked
$endgroup$
– Randall
1 hour ago
$begingroup$
Sorry, missed the figure on the right. So this means the stopping time of the wealthy gambler becomes a PMF for $p geq frac 1 2$ but ceases to be one otherwise? That is a weird conclusion.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
Edited the question. It didn't say the sum should be 1 in the formal definitions.
$endgroup$
– Rohit Pandey
1 hour ago
1
$begingroup$
Saying that the domain contains an arbitrarily large integer is not the same as saying it contains $infty$...
$endgroup$
– Richard Ambler
48 mins ago
$begingroup$
I am your infinitely wealthy opponent. We decide to play the biased simple random walk game, that ends when it reaches target $j$, where you win, or when it hits $0$ where I win. Given the latter case, what value does $T_j$ take? Can you compute the probability of this event? Think about this and David K’s answer ought to be more satisfying (as it is correct).
$endgroup$
– LoveTooNap29
32 mins ago
|
show 1 more comment
$begingroup$
I was thinking about the discrete random variable describing the stopping time ($T$: random variable modelling the toss number where he first reaches his target) of a wealthy gambler reaching his target. It is discussed in some detail here: Gambler with infinite bankroll reaching his target and here: Probability that random walk will reach state $k$ for the first time on step $n$. I realized that when the coin is biased against the wealthy gambler, there is a finite chance he will never reach his target. So, if you calculate the summation:
$$sum_{t=0}^infty P(T=t)$$
you will only get $1$ if the coin he is using has a probability, $pgeq frac 1 2$ of heads. Otherwise, the summation above will result in a number less than $1$. Looking at the definition on Wikipedia, no where does it say that the probability mass function should sum to $1$ (emphasis: in the formal definition). However, right outside the scope of the formal definition, it does.
But this would imply that the wealthy gamblers stopping time when $p < frac 1 2$ has no PMF?
Just wanted to get the community's opinion on this.
Also, if we conclude the PMF doesn't have to sum to $1$, is there then any example of a corresponding probability density function that doesn't integrate to $1$? Perhaps the stopping time (defined as reaching a positive boundary) of a continuous time random walk with negative drift?
EDIT: saying that "never reaching the target is included in the possible outcomes" is not satisfying. We are talking about the random variable $T$. This random variable has a certain domain (which includes $infty$). Summing over the domain should give you $1$. Where in its domain should we fit "never reaching the target"? The fundamental problem remains, is $P(T=t)$ the PMF of $T$ or not? If we say it isn't because it doesn't sum to $1$ over all possible values of $T$, then does it mean $T$ doesn't have a PMF?
probability probability-theory random-variables
asked 1 hour ago
Rohit PandeyRohit Pandey
1,2281021
$endgroup$
I was thinking about the discrete random variable describing the stopping time ($T$: random variable modelling the toss number where he first reaches his target) of a wealthy gambler reaching his target. It is discussed in some detail here: Gambler with infinite bankroll reaching his target and here: Probability that random walk will reach state $k$ for the first time on step $n$. I realized that when the coin is biased against the wealthy gambler, there is a finite chance he will never reach his target. So, if you calculate the summation:
$$sum_{t=0}^infty P(T=t)$$
you will only get $1$ if the coin he is using has a probability, $pgeq frac 1 2$ of heads. Otherwise, the summation above will result in a number less than $1$. Looking at the definition on Wikipedia, no where does it say that the probability mass function should sum to $1$ (emphasis: in the formal definition). However, right outside the scope of the formal definition, it does.
But this would imply that the wealthy gamblers stopping time when $p < frac 1 2$ has no PMF?
Just wanted to get the community's opinion on this.
Also, if we conclude the PMF doesn't have to sum to $1$, is there then any example of a corresponding probability density function that doesn't integrate to $1$? Perhaps the stopping time (defined as reaching a positive boundary) of a continuous time random walk with negative drift?
EDIT: saying that "never reaching the target is included in the possible outcomes" is not satisfying. We are talking about the random variable $T$. This random variable has a certain domain (which includes $infty$). Summing over the domain should give you $1$. Where in its domain should we fit "never reaching the target"? The fundamental problem remains, is $P(T=t)$ the PMF of $T$ or not? If we say it isn't because it doesn't sum to $1$ over all possible values of $T$, then does it mean $T$ doesn't have a PMF?
probability probability-theory random-variables
probability probability-theory random-variables
asked 1 hour ago
Rohit PandeyRohit Pandey
1,2281021
asked 1 hour ago
Rohit PandeyRohit Pandey
1,2281021
edited 1 hour ago
Rohit Pandey
asked 1 hour ago
Rohit PandeyRohit Pandey
1,2281021
asked 1 hour ago
Rohit PandeyRohit Pandey
1,2281021
asked 1 hour ago
Rohit PandeyRohit Pandey
1,2281021
1,2281021
$begingroup$
I mean, it says it must sum to $1$ in the wiki article you linked
$endgroup$
– Randall
1 hour ago
$begingroup$
Sorry, missed the figure on the right. So this means the stopping time of the wealthy gambler becomes a PMF for $p geq frac 1 2$ but ceases to be one otherwise? That is a weird conclusion.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
Edited the question. It didn't say the sum should be 1 in the formal definitions.
$endgroup$
– Rohit Pandey
1 hour ago
1
$begingroup$
Saying that the domain contains an arbitrarily large integer is not the same as saying it contains $infty$...
$endgroup$
– Richard Ambler
48 mins ago
$begingroup$
I am your infinitely wealthy opponent. We decide to play the biased simple random walk game, that ends when it reaches target $j$, where you win, or when it hits $0$ where I win. Given the latter case, what value does $T_j$ take? Can you compute the probability of this event? Think about this and David K’s answer ought to be more satisfying (as it is correct).
$endgroup$
– LoveTooNap29
32 mins ago
|
show 1 more comment
$begingroup$
I mean, it says it must sum to $1$ in the wiki article you linked
$endgroup$
– Randall
1 hour ago
$begingroup$
Sorry, missed the figure on the right. So this means the stopping time of the wealthy gambler becomes a PMF for $p geq frac 1 2$ but ceases to be one otherwise? That is a weird conclusion.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
Edited the question. It didn't say the sum should be 1 in the formal definitions.
$endgroup$
– Rohit Pandey
1 hour ago
1
$begingroup$
Saying that the domain contains an arbitrarily large integer is not the same as saying it contains $infty$...
$endgroup$
– Richard Ambler
48 mins ago
$begingroup$
I am your infinitely wealthy opponent. We decide to play the biased simple random walk game, that ends when it reaches target $j$, where you win, or when it hits $0$ where I win. Given the latter case, what value does $T_j$ take? Can you compute the probability of this event? Think about this and David K’s answer ought to be more satisfying (as it is correct).
$endgroup$
– LoveTooNap29
32 mins ago
$begingroup$
I mean, it says it must sum to $1$ in the wiki article you linked
$endgroup$
– Randall
1 hour ago
$begingroup$
I mean, it says it must sum to $1$ in the wiki article you linked
$endgroup$
– Randall
1 hour ago
$begingroup$
Sorry, missed the figure on the right. So this means the stopping time of the wealthy gambler becomes a PMF for $p geq frac 1 2$ but ceases to be one otherwise? That is a weird conclusion.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
Sorry, missed the figure on the right. So this means the stopping time of the wealthy gambler becomes a PMF for $p geq frac 1 2$ but ceases to be one otherwise? That is a weird conclusion.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
Edited the question. It didn't say the sum should be 1 in the formal definitions.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
Edited the question. It didn't say the sum should be 1 in the formal definitions.
$endgroup$
– Rohit Pandey
1 hour ago
1
1
$begingroup$
Saying that the domain contains an arbitrarily large integer is not the same as saying it contains $infty$...
$endgroup$
– Richard Ambler
48 mins ago
$begingroup$
Saying that the domain contains an arbitrarily large integer is not the same as saying it contains $infty$...
$endgroup$
– Richard Ambler
48 mins ago
$begingroup$
I am your infinitely wealthy opponent. We decide to play the biased simple random walk game, that ends when it reaches target $j$, where you win, or when it hits $0$ where I win. Given the latter case, what value does $T_j$ take? Can you compute the probability of this event? Think about this and David K’s answer ought to be more satisfying (as it is correct).
$endgroup$
– LoveTooNap29
32 mins ago
$begingroup$
I am your infinitely wealthy opponent. We decide to play the biased simple random walk game, that ends when it reaches target $j$, where you win, or when it hits $0$ where I win. Given the latter case, what value does $T_j$ take? Can you compute the probability of this event? Think about this and David K’s answer ought to be more satisfying (as it is correct).
$endgroup$
– LoveTooNap29
32 mins ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If it is possible not to reach the target, then "never reaching the target" is included in the set of possible outcomes and its probability is one of the values of the PMF. When you add all the values of the PMF, this probability is included and the sum is $1.$
What you are calling the "domain" is actually the co-domain of $T.$
The domain of $T$ is whatever sample space $Omega$ you are using.
What makes $T$ a random variable is that it is a function that takes the elements of the sample space $Omega$ and maps them to outcomes.
The probability of an outcome is the measure of the subset of the sample space whose elements map to that outcome,
and the measure of the entire sample space is $1.$
Consider what it means if the you add up the probabilities of all possible outcomes produced by $T$ and the sum is not $1.$ That implies that there is some part of the sample space (in fact, a part of the sample space with positive measure) that $T$ fails to map to any outcome.
In that case, not only do you not have a PMF that sums to less than $1$;
not only does $T$ not have a PMF;
$T$ is not even a random variable,
because it fails to meet the necessary requirements to be a function
over the sample space.
So if there is a positive chance that the gambler never reaches the goal,
and you want to have a random variable that returns $n$ if and only if the goal is reached at time $n,$ then your random variable $T$ must return something
in the case where the goal is not reached.
You can call that outcome what you like, but you have to include it in the range of $T.$
If you really do not want to do that, an alternative is to define a different random variable that returns the time at which the goal is reached,
conditioned on the event that the goal is reached.
Since you conditioned that variable on the goal being reached,
it only needs to take values that are finite integers
(since those are the possible stopping times).
It is still the case, however, that if you correctly define this conditional random variable, the sum of the probabilities of its outcomes will be $1.$
answered 1 hour ago
David KDavid K
53.2k341115
$endgroup$
$begingroup$
But if you sum $P(T)$ over all it possible values of $T$, it doesn't sum to $1$.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
$T$ is a random variable. It has a certain domain. Where in the domain should we include "never reaching the target"?
$endgroup$
– Rohit Pandey
1 hour ago
2
$begingroup$
Typically, for (almost surely) finite stopping times their domain is $[0,infty)$. For random variables like the one here, we simply include $infty$ so the domain is $[0,infty]$. Then $T=infty$ corresponds to never happening and gets a non-zero mass.
$endgroup$
– apsad
11 mins ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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votes
$begingroup$
If it is possible not to reach the target, then "never reaching the target" is included in the set of possible outcomes and its probability is one of the values of the PMF. When you add all the values of the PMF, this probability is included and the sum is $1.$
What you are calling the "domain" is actually the co-domain of $T.$
The domain of $T$ is whatever sample space $Omega$ you are using.
What makes $T$ a random variable is that it is a function that takes the elements of the sample space $Omega$ and maps them to outcomes.
The probability of an outcome is the measure of the subset of the sample space whose elements map to that outcome,
and the measure of the entire sample space is $1.$
Consider what it means if the you add up the probabilities of all possible outcomes produced by $T$ and the sum is not $1.$ That implies that there is some part of the sample space (in fact, a part of the sample space with positive measure) that $T$ fails to map to any outcome.
In that case, not only do you not have a PMF that sums to less than $1$;
not only does $T$ not have a PMF;
$T$ is not even a random variable,
because it fails to meet the necessary requirements to be a function
over the sample space.
So if there is a positive chance that the gambler never reaches the goal,
and you want to have a random variable that returns $n$ if and only if the goal is reached at time $n,$ then your random variable $T$ must return something
in the case where the goal is not reached.
You can call that outcome what you like, but you have to include it in the range of $T.$
If you really do not want to do that, an alternative is to define a different random variable that returns the time at which the goal is reached,
conditioned on the event that the goal is reached.
Since you conditioned that variable on the goal being reached,
it only needs to take values that are finite integers
(since those are the possible stopping times).
It is still the case, however, that if you correctly define this conditional random variable, the sum of the probabilities of its outcomes will be $1.$
answered 1 hour ago
David KDavid K
53.2k341115
$endgroup$
$begingroup$
But if you sum $P(T)$ over all it possible values of $T$, it doesn't sum to $1$.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
$T$ is a random variable. It has a certain domain. Where in the domain should we include "never reaching the target"?
$endgroup$
– Rohit Pandey
1 hour ago
2
$begingroup$
Typically, for (almost surely) finite stopping times their domain is $[0,infty)$. For random variables like the one here, we simply include $infty$ so the domain is $[0,infty]$. Then $T=infty$ corresponds to never happening and gets a non-zero mass.
$endgroup$
– apsad
11 mins ago
add a comment |
$begingroup$
If it is possible not to reach the target, then "never reaching the target" is included in the set of possible outcomes and its probability is one of the values of the PMF. When you add all the values of the PMF, this probability is included and the sum is $1.$
What you are calling the "domain" is actually the co-domain of $T.$
The domain of $T$ is whatever sample space $Omega$ you are using.
What makes $T$ a random variable is that it is a function that takes the elements of the sample space $Omega$ and maps them to outcomes.
The probability of an outcome is the measure of the subset of the sample space whose elements map to that outcome,
and the measure of the entire sample space is $1.$
Consider what it means if the you add up the probabilities of all possible outcomes produced by $T$ and the sum is not $1.$ That implies that there is some part of the sample space (in fact, a part of the sample space with positive measure) that $T$ fails to map to any outcome.
In that case, not only do you not have a PMF that sums to less than $1$;
not only does $T$ not have a PMF;
$T$ is not even a random variable,
because it fails to meet the necessary requirements to be a function
over the sample space.
So if there is a positive chance that the gambler never reaches the goal,
and you want to have a random variable that returns $n$ if and only if the goal is reached at time $n,$ then your random variable $T$ must return something
in the case where the goal is not reached.
You can call that outcome what you like, but you have to include it in the range of $T.$
If you really do not want to do that, an alternative is to define a different random variable that returns the time at which the goal is reached,
conditioned on the event that the goal is reached.
Since you conditioned that variable on the goal being reached,
it only needs to take values that are finite integers
(since those are the possible stopping times).
It is still the case, however, that if you correctly define this conditional random variable, the sum of the probabilities of its outcomes will be $1.$
answered 1 hour ago
David KDavid K
53.2k341115
$endgroup$
$begingroup$
But if you sum $P(T)$ over all it possible values of $T$, it doesn't sum to $1$.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
$T$ is a random variable. It has a certain domain. Where in the domain should we include "never reaching the target"?
$endgroup$
– Rohit Pandey
1 hour ago
2
$begingroup$
Typically, for (almost surely) finite stopping times their domain is $[0,infty)$. For random variables like the one here, we simply include $infty$ so the domain is $[0,infty]$. Then $T=infty$ corresponds to never happening and gets a non-zero mass.
$endgroup$
– apsad
11 mins ago
add a comment |
$begingroup$
If it is possible not to reach the target, then "never reaching the target" is included in the set of possible outcomes and its probability is one of the values of the PMF. When you add all the values of the PMF, this probability is included and the sum is $1.$
What you are calling the "domain" is actually the co-domain of $T.$
The domain of $T$ is whatever sample space $Omega$ you are using.
What makes $T$ a random variable is that it is a function that takes the elements of the sample space $Omega$ and maps them to outcomes.
The probability of an outcome is the measure of the subset of the sample space whose elements map to that outcome,
and the measure of the entire sample space is $1.$
Consider what it means if the you add up the probabilities of all possible outcomes produced by $T$ and the sum is not $1.$ That implies that there is some part of the sample space (in fact, a part of the sample space with positive measure) that $T$ fails to map to any outcome.
In that case, not only do you not have a PMF that sums to less than $1$;
not only does $T$ not have a PMF;
$T$ is not even a random variable,
because it fails to meet the necessary requirements to be a function
over the sample space.
So if there is a positive chance that the gambler never reaches the goal,
and you want to have a random variable that returns $n$ if and only if the goal is reached at time $n,$ then your random variable $T$ must return something
in the case where the goal is not reached.
You can call that outcome what you like, but you have to include it in the range of $T.$
If you really do not want to do that, an alternative is to define a different random variable that returns the time at which the goal is reached,
conditioned on the event that the goal is reached.
Since you conditioned that variable on the goal being reached,
it only needs to take values that are finite integers
(since those are the possible stopping times).
It is still the case, however, that if you correctly define this conditional random variable, the sum of the probabilities of its outcomes will be $1.$
answered 1 hour ago
David KDavid K
53.2k341115
$endgroup$
If it is possible not to reach the target, then "never reaching the target" is included in the set of possible outcomes and its probability is one of the values of the PMF. When you add all the values of the PMF, this probability is included and the sum is $1.$
What you are calling the "domain" is actually the co-domain of $T.$
The domain of $T$ is whatever sample space $Omega$ you are using.
What makes $T$ a random variable is that it is a function that takes the elements of the sample space $Omega$ and maps them to outcomes.
The probability of an outcome is the measure of the subset of the sample space whose elements map to that outcome,
and the measure of the entire sample space is $1.$
Consider what it means if the you add up the probabilities of all possible outcomes produced by $T$ and the sum is not $1.$ That implies that there is some part of the sample space (in fact, a part of the sample space with positive measure) that $T$ fails to map to any outcome.
In that case, not only do you not have a PMF that sums to less than $1$;
not only does $T$ not have a PMF;
$T$ is not even a random variable,
because it fails to meet the necessary requirements to be a function
over the sample space.
So if there is a positive chance that the gambler never reaches the goal,
and you want to have a random variable that returns $n$ if and only if the goal is reached at time $n,$ then your random variable $T$ must return something
in the case where the goal is not reached.
You can call that outcome what you like, but you have to include it in the range of $T.$
If you really do not want to do that, an alternative is to define a different random variable that returns the time at which the goal is reached,
conditioned on the event that the goal is reached.
Since you conditioned that variable on the goal being reached,
it only needs to take values that are finite integers
(since those are the possible stopping times).
It is still the case, however, that if you correctly define this conditional random variable, the sum of the probabilities of its outcomes will be $1.$
answered 1 hour ago
David KDavid K
53.2k341115
edited 1 min ago
answered 1 hour ago
David KDavid K
53.2k341115
answered 1 hour ago
David KDavid K
53.2k341115
answered 1 hour ago
David KDavid K
53.2k341115
53.2k341115
$begingroup$
But if you sum $P(T)$ over all it possible values of $T$, it doesn't sum to $1$.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
$T$ is a random variable. It has a certain domain. Where in the domain should we include "never reaching the target"?
$endgroup$
– Rohit Pandey
1 hour ago
2
$begingroup$
Typically, for (almost surely) finite stopping times their domain is $[0,infty)$. For random variables like the one here, we simply include $infty$ so the domain is $[0,infty]$. Then $T=infty$ corresponds to never happening and gets a non-zero mass.
$endgroup$
– apsad
11 mins ago
add a comment |
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But if you sum $P(T)$ over all it possible values of $T$, it doesn't sum to $1$.
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– Rohit Pandey
1 hour ago
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$T$ is a random variable. It has a certain domain. Where in the domain should we include "never reaching the target"?
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– Rohit Pandey
1 hour ago
2
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Typically, for (almost surely) finite stopping times their domain is $[0,infty)$. For random variables like the one here, we simply include $infty$ so the domain is $[0,infty]$. Then $T=infty$ corresponds to never happening and gets a non-zero mass.
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– apsad
11 mins ago
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But if you sum $P(T)$ over all it possible values of $T$, it doesn't sum to $1$.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
But if you sum $P(T)$ over all it possible values of $T$, it doesn't sum to $1$.
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
$T$ is a random variable. It has a certain domain. Where in the domain should we include "never reaching the target"?
$endgroup$
– Rohit Pandey
1 hour ago
$begingroup$
$T$ is a random variable. It has a certain domain. Where in the domain should we include "never reaching the target"?
$endgroup$
– Rohit Pandey
1 hour ago
2
2
$begingroup$
Typically, for (almost surely) finite stopping times their domain is $[0,infty)$. For random variables like the one here, we simply include $infty$ so the domain is $[0,infty]$. Then $T=infty$ corresponds to never happening and gets a non-zero mass.
$endgroup$
– apsad
11 mins ago
$begingroup$
Typically, for (almost surely) finite stopping times their domain is $[0,infty)$. For random variables like the one here, we simply include $infty$ so the domain is $[0,infty]$. Then $T=infty$ corresponds to never happening and gets a non-zero mass.
$endgroup$
– apsad
11 mins ago
add a comment |
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I mean, it says it must sum to $1$ in the wiki article you linked
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– Randall
1 hour ago
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Sorry, missed the figure on the right. So this means the stopping time of the wealthy gambler becomes a PMF for $p geq frac 1 2$ but ceases to be one otherwise? That is a weird conclusion.
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– Rohit Pandey
1 hour ago
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Edited the question. It didn't say the sum should be 1 in the formal definitions.
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– Rohit Pandey
1 hour ago
1
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Saying that the domain contains an arbitrarily large integer is not the same as saying it contains $infty$...
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– Richard Ambler
48 mins ago
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I am your infinitely wealthy opponent. We decide to play the biased simple random walk game, that ends when it reaches target $j$, where you win, or when it hits $0$ where I win. Given the latter case, what value does $T_j$ take? Can you compute the probability of this event? Think about this and David K’s answer ought to be more satisfying (as it is correct).
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– LoveTooNap29
32 mins ago