Let $X_n$ and $U_n$ be independent, with $P(X_n = 0, i.o.) = 1$ and $P(U_n = 0, i.o.)$, show that $P(X_n = 0,...
Any stored/leased 737s that could substitute for grounded MAXs?
As a dual citizen, my US passport will expire one day after traveling to the US. Will this work?
What does 丫 mean? 丫是什么意思？
Where did Ptolemy compare the Earth to the distance of fixed stars?
My mentor says to set image to Fine instead of RAW — how is this different from JPG?
Why not use the yoke to control yaw, as well as pitch and roll?
NIntegrate on a solution of a matrix ODE
Plotting a Maclaurin series
How to infer difference of population proportion between two groups when proportion is small?
Meaning of 境 in その日を境に
What did Turing mean when saying that "machines cannot give rise to surprises" is due to a fallacy?
Why are current probes so expensive?
How does the body cool itself in a stillsuit?
Russian equivalents of おしゃれは足元から (Every good outfit starts with the shoes)
Why are two-digit numbers in Jonathan Swift's "Gulliver's Travels" (1726) written in "German style"?
Did pre-Columbian Americans know the spherical shape of the Earth?
Proving that any solution to the differential equation of an oscillator can be written as a sum of sinusoids.
Did John Wesley plagiarize Matthew Henry...?
How to name indistinguishable henchmen in a screenplay?
3D Masyu - A Die
Short story about astronauts fertilizing soil with their own bodies
IC on Digikey is 5x more expensive than board containing same IC on Alibaba: How?
Keep at all times, the minus sign above aligned with minus sign below
draw a pulley system
Let $X_n$ and $U_n$ be independent, with $P(X_n = 0, i.o.) = 1$ and $P(U_n = 0, i.o.)$, show that $P(X_n = 0, U_{n+1} =0, i.o.) = 1$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Identify the smallest $c$ such that $P(|X_n| ge c sqrt{ln n} text{ i.o.}) = 0$ for normally distributed $X_n$$P(X_n < a$ i.o. and $X_n > b$ i.o.$) = 0$ for all $a < b$ implies that $lim_{n rightarrow infty} X_n$ exists a.e.If $P(X_nne0 text{i.o.})=1$ then $P(c_n|X_n|>1 text{i.o.})=1$ for some suitable sequence $(c_n)$Let $X_1, X_2, ..$ be independent with $E(X_n)=0, var(X_n)={sigma_{n}}^2$.Show that if $P(|X_n|>epsilon text{ i.o.} ) =0$ for every $epsilon>0$ then $X_nto 0$ almost surelyIf $E(X_n)=0$ and $E(X_n^2)=1$ then$P(X_nge n,i.o.)=0$Show $P(|X_n| geq n i.o) = 1$ for $X_i$ iid, $X_i not in L^1$.$X_n to X$ a.s. iff $mathbb{P}(|X_n-X|>epsilon_n , , text{i.o.})=0$ for some $epsilon_n to 0$Markov chain with transiency and absorptionIf $P(X_n =0 i.o.) = 1$ and $P(X_{n+1} = 1 | X_n=0) = c > 0$ for all $n$, then $P(X_n =1 i.o.) = 1$
$begingroup$
Let $X_n$ and $U_n$ be independent, with $P(X_n = 0 i.o.) = 1$ and $P(U_n = 0 i.o.)$, show that $P(X_n = 0, U_{n+1} =0, i.o.) = 1$.
Does this require additional assumptions to be true?
probability probability-theory
asked Mar 26 at 11:21
jackson5jackson5
701513
$endgroup$
add a comment |
$begingroup$
Let $X_n$ and $U_n$ be independent, with $P(X_n = 0 i.o.) = 1$ and $P(U_n = 0 i.o.)$, show that $P(X_n = 0, U_{n+1} =0, i.o.) = 1$.
Does this require additional assumptions to be true?
probability probability-theory
asked Mar 26 at 11:21
jackson5jackson5
701513
$endgroup$
add a comment |
$begingroup$
Let $X_n$ and $U_n$ be independent, with $P(X_n = 0 i.o.) = 1$ and $P(U_n = 0 i.o.)$, show that $P(X_n = 0, U_{n+1} =0, i.o.) = 1$.
Does this require additional assumptions to be true?
probability probability-theory
asked Mar 26 at 11:21
jackson5jackson5
701513
$endgroup$
Let $X_n$ and $U_n$ be independent, with $P(X_n = 0 i.o.) = 1$ and $P(U_n = 0 i.o.)$, show that $P(X_n = 0, U_{n+1} =0, i.o.) = 1$.
Does this require additional assumptions to be true?
probability probability-theory
probability probability-theory
asked Mar 26 at 11:21
jackson5jackson5
701513
asked Mar 26 at 11:21
jackson5jackson5
701513
edited Mar 26 at 12:06
jackson5
asked Mar 26 at 11:21
jackson5jackson5
701513
asked Mar 26 at 11:21
jackson5jackson5
701513
asked Mar 26 at 11:21
jackson5jackson5
701513
701513
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Answer to the question before it was modified:
Let $X_n=1$ for $n$ odd, $0$ for $n$ even and $U_n=X_n$ for all $n$. Then $X_n=0$ implies $U_{n+1} neq 0$ so the conclusion is false.
answered Mar 26 at 11:50
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
But $U_n$ and $X_n$ are independent here, how can we have $U_n = X_n$??
$endgroup$
– jackson5
Mar 26 at 11:53
$begingroup$
All the random variables I have defined are constant random variables. Any constant random variable is independent of itself!. @jackson5
$endgroup$
– Kavi Rama Murthy
Mar 26 at 11:54
$begingroup$
Oh yes, of course! What are some minimal conditions I can put on X_n, U_n to make this true? What if U_n is Bernoulli(1/2)?
$endgroup$
– jackson5
Mar 26 at 11:59
$begingroup$
Ok, I will ask a new question
$endgroup$
– jackson5
Mar 26 at 12:05
add a comment |
Your Answer
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%2f3163044%2flet-x-n-and-u-n-be-independent-with-px-n-0-i-o-1-and-pu-n-0%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
$begingroup$
Answer to the question before it was modified:
Let $X_n=1$ for $n$ odd, $0$ for $n$ even and $U_n=X_n$ for all $n$. Then $X_n=0$ implies $U_{n+1} neq 0$ so the conclusion is false.
answered Mar 26 at 11:50
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
But $U_n$ and $X_n$ are independent here, how can we have $U_n = X_n$??
$endgroup$
– jackson5
Mar 26 at 11:53
$begingroup$
All the random variables I have defined are constant random variables. Any constant random variable is independent of itself!. @jackson5
$endgroup$
– Kavi Rama Murthy
Mar 26 at 11:54
$begingroup$
Oh yes, of course! What are some minimal conditions I can put on X_n, U_n to make this true? What if U_n is Bernoulli(1/2)?
$endgroup$
– jackson5
Mar 26 at 11:59
$begingroup$
Ok, I will ask a new question
$endgroup$
– jackson5
Mar 26 at 12:05
add a comment |
$begingroup$
Answer to the question before it was modified:
Let $X_n=1$ for $n$ odd, $0$ for $n$ even and $U_n=X_n$ for all $n$. Then $X_n=0$ implies $U_{n+1} neq 0$ so the conclusion is false.
answered Mar 26 at 11:50
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
But $U_n$ and $X_n$ are independent here, how can we have $U_n = X_n$??
$endgroup$
– jackson5
Mar 26 at 11:53
$begingroup$
All the random variables I have defined are constant random variables. Any constant random variable is independent of itself!. @jackson5
$endgroup$
– Kavi Rama Murthy
Mar 26 at 11:54
$begingroup$
Oh yes, of course! What are some minimal conditions I can put on X_n, U_n to make this true? What if U_n is Bernoulli(1/2)?
$endgroup$
– jackson5
Mar 26 at 11:59
$begingroup$
Ok, I will ask a new question
$endgroup$
– jackson5
Mar 26 at 12:05
add a comment |
$begingroup$
Answer to the question before it was modified:
Let $X_n=1$ for $n$ odd, $0$ for $n$ even and $U_n=X_n$ for all $n$. Then $X_n=0$ implies $U_{n+1} neq 0$ so the conclusion is false.
answered Mar 26 at 11:50
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
Answer to the question before it was modified:
Let $X_n=1$ for $n$ odd, $0$ for $n$ even and $U_n=X_n$ for all $n$. Then $X_n=0$ implies $U_{n+1} neq 0$ so the conclusion is false.
answered Mar 26 at 11:50
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
edited Mar 26 at 12:01
answered Mar 26 at 11:50
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Mar 26 at 11:50
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Mar 26 at 11:50
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
76.6k53471
$begingroup$
But $U_n$ and $X_n$ are independent here, how can we have $U_n = X_n$??
$endgroup$
– jackson5
Mar 26 at 11:53
$begingroup$
All the random variables I have defined are constant random variables. Any constant random variable is independent of itself!. @jackson5
$endgroup$
– Kavi Rama Murthy
Mar 26 at 11:54
$begingroup$
Oh yes, of course! What are some minimal conditions I can put on X_n, U_n to make this true? What if U_n is Bernoulli(1/2)?
$endgroup$
– jackson5
Mar 26 at 11:59
$begingroup$
Ok, I will ask a new question
$endgroup$
– jackson5
Mar 26 at 12:05
add a comment |
$begingroup$
But $U_n$ and $X_n$ are independent here, how can we have $U_n = X_n$??
$endgroup$
– jackson5
Mar 26 at 11:53
$begingroup$
All the random variables I have defined are constant random variables. Any constant random variable is independent of itself!. @jackson5
$endgroup$
– Kavi Rama Murthy
Mar 26 at 11:54
$begingroup$
Oh yes, of course! What are some minimal conditions I can put on X_n, U_n to make this true? What if U_n is Bernoulli(1/2)?
$endgroup$
– jackson5
Mar 26 at 11:59
$begingroup$
Ok, I will ask a new question
$endgroup$
– jackson5
Mar 26 at 12:05
$begingroup$
But $U_n$ and $X_n$ are independent here, how can we have $U_n = X_n$??
$endgroup$
– jackson5
Mar 26 at 11:53
$begingroup$
But $U_n$ and $X_n$ are independent here, how can we have $U_n = X_n$??
$endgroup$
– jackson5
Mar 26 at 11:53
$begingroup$
All the random variables I have defined are constant random variables. Any constant random variable is independent of itself!. @jackson5
$endgroup$
– Kavi Rama Murthy
Mar 26 at 11:54
$begingroup$
All the random variables I have defined are constant random variables. Any constant random variable is independent of itself!. @jackson5
$endgroup$
– Kavi Rama Murthy
Mar 26 at 11:54
$begingroup$
Oh yes, of course! What are some minimal conditions I can put on X_n, U_n to make this true? What if U_n is Bernoulli(1/2)?
$endgroup$
– jackson5
Mar 26 at 11:59
$begingroup$
Oh yes, of course! What are some minimal conditions I can put on X_n, U_n to make this true? What if U_n is Bernoulli(1/2)?
$endgroup$
– jackson5
Mar 26 at 11:59
$begingroup$
Ok, I will ask a new question
$endgroup$
– jackson5
Mar 26 at 12:05
$begingroup$
Ok, I will ask a new question
$endgroup$
– jackson5
Mar 26 at 12:05
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%2f3163044%2flet-x-n-and-u-n-be-independent-with-px-n-0-i-o-1-and-pu-n-0%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
