Convergence in probability: The inverse of the simple mean
up vote
1
down vote
favorite
I have a question on convergence: I have to prove that $frac{n}{U_{n}} longrightarrow 1$ in probability, where $U_{n}=sum X_{i}$, $X_{i}sim mathrm{Exp}(1)$ and because of this, $U_{n}sim mathrm{Gamma}(n,1)$.
This problem had two parts, the other part was to prove that $frac{U_{n}}{n}longrightarrow 1$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.
Thanks so much for your help! :)
probability-theory statistics convergence exponential-distribution gamma-distribution
edited Nov 29 at 9:56
Davide Giraudo
124k16150259
asked Nov 28 at 21:25
Dadadave
9118
add a comment |
up vote
1
down vote
favorite
I have a question on convergence: I have to prove that $frac{n}{U_{n}} longrightarrow 1$ in probability, where $U_{n}=sum X_{i}$, $X_{i}sim mathrm{Exp}(1)$ and because of this, $U_{n}sim mathrm{Gamma}(n,1)$.
This problem had two parts, the other part was to prove that $frac{U_{n}}{n}longrightarrow 1$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.
Thanks so much for your help! :)
probability-theory statistics convergence exponential-distribution gamma-distribution
edited Nov 29 at 9:56
Davide Giraudo
124k16150259
asked Nov 28 at 21:25
Dadadave
9118
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a question on convergence: I have to prove that $frac{n}{U_{n}} longrightarrow 1$ in probability, where $U_{n}=sum X_{i}$, $X_{i}sim mathrm{Exp}(1)$ and because of this, $U_{n}sim mathrm{Gamma}(n,1)$.
This problem had two parts, the other part was to prove that $frac{U_{n}}{n}longrightarrow 1$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.
Thanks so much for your help! :)
probability-theory statistics convergence exponential-distribution gamma-distribution
edited Nov 29 at 9:56
Davide Giraudo
124k16150259
asked Nov 28 at 21:25
Dadadave
9118
I have a question on convergence: I have to prove that $frac{n}{U_{n}} longrightarrow 1$ in probability, where $U_{n}=sum X_{i}$, $X_{i}sim mathrm{Exp}(1)$ and because of this, $U_{n}sim mathrm{Gamma}(n,1)$.
This problem had two parts, the other part was to prove that $frac{U_{n}}{n}longrightarrow 1$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.
Thanks so much for your help! :)
probability-theory statistics convergence exponential-distribution gamma-distribution
probability-theory statistics convergence exponential-distribution gamma-distribution
edited Nov 29 at 9:56
Davide Giraudo
124k16150259
asked Nov 28 at 21:25
Dadadave
9118
edited Nov 29 at 9:56
Davide Giraudo
124k16150259
asked Nov 28 at 21:25
Dadadave
9118
edited Nov 29 at 9:56
Davide Giraudo
124k16150259
edited Nov 29 at 9:56
Davide Giraudo
124k16150259
edited Nov 29 at 9:56
Davide Giraudo
124k16150259
124k16150259
asked Nov 28 at 21:25
Dadadave
9118
asked Nov 28 at 21:25
Dadadave
9118
asked Nov 28 at 21:25
Dadadave
9118
9118
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 at 21:42
angryavian
38.1k23180
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
– Dadadave
Nov 28 at 22:11
1
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
– Stockfish
Nov 29 at 10:09
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%2f3017736%2fconvergence-in-probability-the-inverse-of-the-simple-mean%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
up vote
2
down vote
accepted
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 at 21:42
angryavian
38.1k23180
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
– Dadadave
Nov 28 at 22:11
1
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
– Stockfish
Nov 29 at 10:09
add a comment |
up vote
2
down vote
accepted
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 at 21:42
angryavian
38.1k23180
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
– Dadadave
Nov 28 at 22:11
1
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
– Stockfish
Nov 29 at 10:09
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 at 21:42
angryavian
38.1k23180
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 at 21:42
angryavian
38.1k23180
answered Nov 28 at 21:42
angryavian
38.1k23180
answered Nov 28 at 21:42
angryavian
38.1k23180
answered Nov 28 at 21:42
angryavian
38.1k23180
38.1k23180
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
– Dadadave
Nov 28 at 22:11
1
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
– Stockfish
Nov 29 at 10:09
add a comment |
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
– Dadadave
Nov 28 at 22:11
1
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
– Stockfish
Nov 29 at 10:09
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
– Dadadave
Nov 28 at 22:11
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
– Dadadave
Nov 28 at 22:11
1
1
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
– Stockfish
Nov 29 at 10:09
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
– Stockfish
Nov 29 at 10:09
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3017736%2fconvergence-in-probability-the-inverse-of-the-simple-mean%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
