$X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$












1












$begingroup$


Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$, everything independent. I want to find the convergence in law of:
$$ Z_n = frac{1}{n}X_n + beta Y_n $$



With $beta in mathbb{R}$.



Now, my plan is to use the characteristic function of $Z_n$.



This is what I have done:
$$phi_{Z_n} = phi_{frac{1}{n}X_n} *phi_{beta Y_n} $$



$$phi_{Z_n} = frac{ e^{-frac{1}{n}} e^{ibeta t}}{1-(1-e^{-frac{1}{n}})e^{ibeta t} } *e^{n(e^{it} - 1)}$$



I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?






probability weak-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is your fraction just 1? Is there an exponent missing?
    $endgroup$
    – Paul
    Jan 22 at 14:50










  • $begingroup$
    @Paul Edited, thank you for catching the mistake.
    $endgroup$
    – qcc101
    Jan 22 at 15:06
















1












$begingroup$


Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$, everything independent. I want to find the convergence in law of:
$$ Z_n = frac{1}{n}X_n + beta Y_n $$



With $beta in mathbb{R}$.



Now, my plan is to use the characteristic function of $Z_n$.



This is what I have done:
$$phi_{Z_n} = phi_{frac{1}{n}X_n} *phi_{beta Y_n} $$



$$phi_{Z_n} = frac{ e^{-frac{1}{n}} e^{ibeta t}}{1-(1-e^{-frac{1}{n}})e^{ibeta t} } *e^{n(e^{it} - 1)}$$



I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?






probability weak-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is your fraction just 1? Is there an exponent missing?
    $endgroup$
    – Paul
    Jan 22 at 14:50










  • $begingroup$
    @Paul Edited, thank you for catching the mistake.
    $endgroup$
    – qcc101
    Jan 22 at 15:06














1












1








1





$begingroup$


Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$, everything independent. I want to find the convergence in law of:
$$ Z_n = frac{1}{n}X_n + beta Y_n $$



With $beta in mathbb{R}$.



Now, my plan is to use the characteristic function of $Z_n$.



This is what I have done:
$$phi_{Z_n} = phi_{frac{1}{n}X_n} *phi_{beta Y_n} $$



$$phi_{Z_n} = frac{ e^{-frac{1}{n}} e^{ibeta t}}{1-(1-e^{-frac{1}{n}})e^{ibeta t} } *e^{n(e^{it} - 1)}$$



I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?






probability weak-convergence







share|cite|improve this question











$endgroup$




Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$, everything independent. I want to find the convergence in law of:
$$ Z_n = frac{1}{n}X_n + beta Y_n $$



With $beta in mathbb{R}$.



Now, my plan is to use the characteristic function of $Z_n$.



This is what I have done:
$$phi_{Z_n} = phi_{frac{1}{n}X_n} *phi_{beta Y_n} $$



$$phi_{Z_n} = frac{ e^{-frac{1}{n}} e^{ibeta t}}{1-(1-e^{-frac{1}{n}})e^{ibeta t} } *e^{n(e^{it} - 1)}$$



I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?






probability weak-convergence




probability weak-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 15:06







qcc101

















asked Jan 22 at 14:46









qcc101qcc101

627213




627213












  • $begingroup$
    Is your fraction just 1? Is there an exponent missing?
    $endgroup$
    – Paul
    Jan 22 at 14:50










  • $begingroup$
    @Paul Edited, thank you for catching the mistake.
    $endgroup$
    – qcc101
    Jan 22 at 15:06


















  • $begingroup$
    Is your fraction just 1? Is there an exponent missing?
    $endgroup$
    – Paul
    Jan 22 at 14:50










  • $begingroup$
    @Paul Edited, thank you for catching the mistake.
    $endgroup$
    – qcc101
    Jan 22 at 15:06
















$begingroup$
Is your fraction just 1? Is there an exponent missing?
$endgroup$
– Paul
Jan 22 at 14:50




$begingroup$
Is your fraction just 1? Is there an exponent missing?
$endgroup$
– Paul
Jan 22 at 14:50












$begingroup$
@Paul Edited, thank you for catching the mistake.
$endgroup$
– qcc101
Jan 22 at 15:06




$begingroup$
@Paul Edited, thank you for catching the mistake.
$endgroup$
– qcc101
Jan 22 at 15:06










1 Answer
1






active

oldest

votes


















1












$begingroup$

Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get



$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$



and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).



Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$



$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$



and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)



Convergence in probability is stable under summation and implies convergence in distribution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, super clear. What if I wanted to use the characteristic function though?
    $endgroup$
    – qcc101
    Jan 22 at 15:42










  • $begingroup$
    @qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
    $endgroup$
    – Lundborg
    Jan 22 at 16:25













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083258%2fx-n-poissonn-y-n-geometrice-frac1n%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









1












$begingroup$

Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get



$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$



and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).



Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$



$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$



and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)



Convergence in probability is stable under summation and implies convergence in distribution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, super clear. What if I wanted to use the characteristic function though?
    $endgroup$
    – qcc101
    Jan 22 at 15:42










  • $begingroup$
    @qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
    $endgroup$
    – Lundborg
    Jan 22 at 16:25


















1












$begingroup$

Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get



$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$



and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).



Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$



$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$



and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)



Convergence in probability is stable under summation and implies convergence in distribution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, super clear. What if I wanted to use the characteristic function though?
    $endgroup$
    – qcc101
    Jan 22 at 15:42










  • $begingroup$
    @qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
    $endgroup$
    – Lundborg
    Jan 22 at 16:25
















1












1








1





$begingroup$

Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get



$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$



and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).



Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$



$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$



and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)



Convergence in probability is stable under summation and implies convergence in distribution.






share|cite|improve this answer









$endgroup$



Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get



$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$



and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).



Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$



$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$



and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)



Convergence in probability is stable under summation and implies convergence in distribution.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 22 at 15:33









LundborgLundborg

880416




880416












  • $begingroup$
    Thank you, super clear. What if I wanted to use the characteristic function though?
    $endgroup$
    – qcc101
    Jan 22 at 15:42










  • $begingroup$
    @qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
    $endgroup$
    – Lundborg
    Jan 22 at 16:25




















  • $begingroup$
    Thank you, super clear. What if I wanted to use the characteristic function though?
    $endgroup$
    – qcc101
    Jan 22 at 15:42










  • $begingroup$
    @qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
    $endgroup$
    – Lundborg
    Jan 22 at 16:25


















$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42




$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42












$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25






$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25




















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083258%2fx-n-poissonn-y-n-geometrice-frac1n%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

File:Tiny Toon Adventures Wacky Sports JP Title.png

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! File:Tiny Toon Adventures Wacky Sports JP Title.png From The Cutting Room Floor Jump to: navigation, search File File history File usage No higher resolution available. Tiny_Toon_Adventures_Wacky_Sports_JP_Title.png ‎ (160 × 144 pixels, file size: 3 KB, MIME type: image/png ) This image is a screenshot of a(n) Game Boy title screen. File history Click on a date/time to view the file as it appeared at that time. Date/Time Thumbnail Dimensions User Comment current 21:31, 20 August 2016 160 × 144 (3 KB) TheBluegeek (Talk | contribs) 17:25, 29 August 2013 160 × 144 (3 KB) Luigi-San (Talk | contribs) {{titlescreen|Game Boy}} You cannot overwrite this file. File usage The following page links to this file: Tiny Toon Advent
Read more