Mixing
![What is $mathbbE[W(tau)^4]$ for a Wiener process $W$ and a stopping time $tau$?](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Wald’s equation for a lower bound on expected stopping time
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I saw it mentioned here that Wald's Equation can still be used when the number of random variables being summed is not independent of the random variables themselves.
Consider finding the expected value of the stopping time $N$ defined below, where $U_i$ are iid and uniform in (0, 1).
$$N = min Big n: sum_i = 1^n U_i > 1 Big$$
As mentioned here the expected value is $e$. Let $S_n = sum_i = 1^n U_i$. Since, $S_n geq 1$ due to the stopping rule, would it be valid to use Wald's equation for a crude lower bound as follows
$$ mathbbES_n = mathbbEU_i mathbbEN geq 1\
mathbbEN geq frac1mathbbEU_i = 2$$
probability stochastic-processes
asked Jul 26 at 0:07
renewalquestion
61
add a comment |Â
up vote
1
down vote
favorite
I saw it mentioned here that Wald's Equation can still be used when the number of random variables being summed is not independent of the random variables themselves.
Consider finding the expected value of the stopping time $N$ defined below, where $U_i$ are iid and uniform in (0, 1).
$$N = min Big n: sum_i = 1^n U_i > 1 Big$$
As mentioned here the expected value is $e$. Let $S_n = sum_i = 1^n U_i$. Since, $S_n geq 1$ due to the stopping rule, would it be valid to use Wald's equation for a crude lower bound as follows
$$ mathbbES_n = mathbbEU_i mathbbEN geq 1\
mathbbEN geq frac1mathbbEU_i = 2$$
probability stochastic-processes
asked Jul 26 at 0:07
renewalquestion
61
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I saw it mentioned here that Wald's Equation can still be used when the number of random variables being summed is not independent of the random variables themselves.
Consider finding the expected value of the stopping time $N$ defined below, where $U_i$ are iid and uniform in (0, 1).
$$N = min Big n: sum_i = 1^n U_i > 1 Big$$
As mentioned here the expected value is $e$. Let $S_n = sum_i = 1^n U_i$. Since, $S_n geq 1$ due to the stopping rule, would it be valid to use Wald's equation for a crude lower bound as follows
$$ mathbbES_n = mathbbEU_i mathbbEN geq 1\
mathbbEN geq frac1mathbbEU_i = 2$$
probability stochastic-processes
asked Jul 26 at 0:07
renewalquestion
61
I saw it mentioned here that Wald's Equation can still be used when the number of random variables being summed is not independent of the random variables themselves.
Consider finding the expected value of the stopping time $N$ defined below, where $U_i$ are iid and uniform in (0, 1).
$$N = min Big n: sum_i = 1^n U_i > 1 Big$$
As mentioned here the expected value is $e$. Let $S_n = sum_i = 1^n U_i$. Since, $S_n geq 1$ due to the stopping rule, would it be valid to use Wald's equation for a crude lower bound as follows
$$ mathbbES_n = mathbbEU_i mathbbEN geq 1\
mathbbEN geq frac1mathbbEU_i = 2$$
probability stochastic-processes
asked Jul 26 at 0:07
renewalquestion
61
asked Jul 26 at 0:07
renewalquestion
61
asked Jul 26 at 0:07
renewalquestion
61
asked Jul 26 at 0:07
renewalquestion
61
61
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
There seems to be some notational confusion – if I understand correctly, where you write $S_n$, except for the first time, you mean $S_N$.
Interpreted in this way, your derivation is correct. Wikipedia states the independence assumption under which Wald's equation holds:
$$
E[X_n1_Nge n]=E[X_n]P(Nge n);.
$$
This is fulfilled if $N$ is a stopping time for the sequence $X_n$.
Your result also follows intuitively if you imagine starting a new “pile†to add the $X_n$ to as soon as the previous pile is $ge1$. Then you use all the $X_n$ in these piles, and at any point in the process the sum of all the piles is the sum of all the $X_n$ up to that point, which couldn't work if you'd have less than $2$ of them per pile on average.
answered Jul 26 at 3:54
joriki
164k10180328
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
There seems to be some notational confusion – if I understand correctly, where you write $S_n$, except for the first time, you mean $S_N$.
Interpreted in this way, your derivation is correct. Wikipedia states the independence assumption under which Wald's equation holds:
$$
E[X_n1_Nge n]=E[X_n]P(Nge n);.
$$
This is fulfilled if $N$ is a stopping time for the sequence $X_n$.
Your result also follows intuitively if you imagine starting a new “pile†to add the $X_n$ to as soon as the previous pile is $ge1$. Then you use all the $X_n$ in these piles, and at any point in the process the sum of all the piles is the sum of all the $X_n$ up to that point, which couldn't work if you'd have less than $2$ of them per pile on average.
answered Jul 26 at 3:54
joriki
164k10180328
add a comment |Â
up vote
0
down vote
There seems to be some notational confusion – if I understand correctly, where you write $S_n$, except for the first time, you mean $S_N$.
Interpreted in this way, your derivation is correct. Wikipedia states the independence assumption under which Wald's equation holds:
$$
E[X_n1_Nge n]=E[X_n]P(Nge n);.
$$
This is fulfilled if $N$ is a stopping time for the sequence $X_n$.
Your result also follows intuitively if you imagine starting a new “pile†to add the $X_n$ to as soon as the previous pile is $ge1$. Then you use all the $X_n$ in these piles, and at any point in the process the sum of all the piles is the sum of all the $X_n$ up to that point, which couldn't work if you'd have less than $2$ of them per pile on average.
answered Jul 26 at 3:54
joriki
164k10180328
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There seems to be some notational confusion – if I understand correctly, where you write $S_n$, except for the first time, you mean $S_N$.
Interpreted in this way, your derivation is correct. Wikipedia states the independence assumption under which Wald's equation holds:
$$
E[X_n1_Nge n]=E[X_n]P(Nge n);.
$$
This is fulfilled if $N$ is a stopping time for the sequence $X_n$.
Your result also follows intuitively if you imagine starting a new “pile†to add the $X_n$ to as soon as the previous pile is $ge1$. Then you use all the $X_n$ in these piles, and at any point in the process the sum of all the piles is the sum of all the $X_n$ up to that point, which couldn't work if you'd have less than $2$ of them per pile on average.
answered Jul 26 at 3:54
joriki
164k10180328
There seems to be some notational confusion – if I understand correctly, where you write $S_n$, except for the first time, you mean $S_N$.
Interpreted in this way, your derivation is correct. Wikipedia states the independence assumption under which Wald's equation holds:
$$
E[X_n1_Nge n]=E[X_n]P(Nge n);.
$$
This is fulfilled if $N$ is a stopping time for the sequence $X_n$.
Your result also follows intuitively if you imagine starting a new “pile†to add the $X_n$ to as soon as the previous pile is $ge1$. Then you use all the $X_n$ in these piles, and at any point in the process the sum of all the piles is the sum of all the $X_n$ up to that point, which couldn't work if you'd have less than $2$ of them per pile on average.
answered Jul 26 at 3:54
joriki
164k10180328
answered Jul 26 at 3:54
joriki
164k10180328
answered Jul 26 at 3:54
joriki
164k10180328
answered Jul 26 at 3:54
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862952%2fwald-s-equation-for-a-lower-bound-on-expected-stopping-time%23new-answer', 'question_page');
);
Post as a guest
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
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
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