Show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant $2p$
$begingroup$
Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.
I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.
My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$
I can also use the following criterion:
Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$
To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?
probability convergence
$endgroup$
add a comment |
$begingroup$
Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.
I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.
My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$
I can also use the following criterion:
Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$
To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?
probability convergence
$endgroup$
$begingroup$
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
$endgroup$
– dem0nakos
Jan 4 at 20:38
$begingroup$
Can you use the law of large numbers?
$endgroup$
– Lundborg
Jan 4 at 20:54
$begingroup$
@dem0nakos I didn't know that inequality, super useful.
$endgroup$
– qcc101
Jan 4 at 20:58
add a comment |
$begingroup$
Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.
I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.
My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$
I can also use the following criterion:
Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$
To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?
probability convergence
$endgroup$
Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.
I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.
My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$
I can also use the following criterion:
Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$
To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?
probability convergence
probability convergence
edited Jan 4 at 21:57
Davide Giraudo
127k16153268
127k16153268
asked Jan 4 at 20:30
qcc101qcc101
627213
627213
$begingroup$
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
$endgroup$
– dem0nakos
Jan 4 at 20:38
$begingroup$
Can you use the law of large numbers?
$endgroup$
– Lundborg
Jan 4 at 20:54
$begingroup$
@dem0nakos I didn't know that inequality, super useful.
$endgroup$
– qcc101
Jan 4 at 20:58
add a comment |
$begingroup$
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
$endgroup$
– dem0nakos
Jan 4 at 20:38
$begingroup$
Can you use the law of large numbers?
$endgroup$
– Lundborg
Jan 4 at 20:54
$begingroup$
@dem0nakos I didn't know that inequality, super useful.
$endgroup$
– qcc101
Jan 4 at 20:58
$begingroup$
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
$endgroup$
– dem0nakos
Jan 4 at 20:38
$begingroup$
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
$endgroup$
– dem0nakos
Jan 4 at 20:38
$begingroup$
Can you use the law of large numbers?
$endgroup$
– Lundborg
Jan 4 at 20:54
$begingroup$
Can you use the law of large numbers?
$endgroup$
– Lundborg
Jan 4 at 20:54
$begingroup$
@dem0nakos I didn't know that inequality, super useful.
$endgroup$
– qcc101
Jan 4 at 20:58
$begingroup$
@dem0nakos I didn't know that inequality, super useful.
$endgroup$
– qcc101
Jan 4 at 20:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
$endgroup$
add a comment |
$begingroup$
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
$endgroup$
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%2f3062073%2fshow-that-the-sequence-t-n-n-geq-1-converges-in-probability-to-the-constan%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
$endgroup$
add a comment |
$begingroup$
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
$endgroup$
add a comment |
$begingroup$
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
$endgroup$
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
answered Jan 4 at 20:39
Will M.Will M.
2,835315
2,835315
add a comment |
add a comment |
$begingroup$
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
$endgroup$
add a comment |
$begingroup$
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
$endgroup$
add a comment |
$begingroup$
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
$endgroup$
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
answered Jan 4 at 20:58
LundborgLundborg
892516
892516
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%2f3062073%2fshow-that-the-sequence-t-n-n-geq-1-converges-in-probability-to-the-constan%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
$begingroup$
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
$endgroup$
– dem0nakos
Jan 4 at 20:38
$begingroup$
Can you use the law of large numbers?
$endgroup$
– Lundborg
Jan 4 at 20:54
$begingroup$
@dem0nakos I didn't know that inequality, super useful.
$endgroup$
– qcc101
Jan 4 at 20:58