If $X_n$ is Gamma $(n,lambda)$ distributed then $(lambda X_n -n)/sqrt nto N(0,1)$












3












$begingroup$



Let $X_n$ be Gamma $(n,lambda)$ distributed, and $Y_n = dfrac{lambda X_n -n}{sqrt{n}}$.
Show that $Y_n rightarrow N(0,1)$.




My idea to prove this is to use Lévys theorem with the characteristic functions. If I can show that the characteristic functions $phi_{Y_n}$ of $Y_n$ converge to a function $phi$, which is continous in $0$, then there exists a random variable $Y$ such that $phi$ is the characteristic function of $Y$. My aim is to get $phi_Y (u) = e^{-u^2/2}$ since I have to show that the $Y_n$ converge to a standard normal random variable and the cF describes the distribution uniquely.



I know that the cF of a Gamma$(n,lambda)$-distributed random variable $X_n$ is given by
$$phi_{X_n}(u) =left(frac{lambda}{lambda - iu}right)^n $$
The cF rules for the linear transformation $$Y_n = frac{lambda X_n}{sqrt{n}} +frac{-n}{sqrt{n}}$$ give
$$phi_{Y_n} (u)= e^{-in frac{u}{sqrt{n}}} phi_{X_n}(frac{lambda u}{sqrt{n}}) = e^{-iu sqrt{n}} left(frac{lambda}{lambda - i frac{lambda u}{sqrt{n}}}right)^n$$
that is,
$$phi_{Y_n} (u) = e^{-iu sqrt{n}} left(frac{1}{1 - i frac{ u}{sqrt{n}}}right)^n = e^{-iu sqrt{n}} frac{1}{ (1 - i frac{u}{sqrt{n}})^n}$$
and now I don't know how to continue calculating.



Since I need the limit of this, I thought about the representation of the exponential function in terms of the limit of a sequence, e.g. $$e^x = lim_{n rightarrow infty} left( 1 + frac{x}{n}right)^n$$
Some hints or a trick would be very nice.






probability-theory central-limit-theorem characteristic-functions gamma-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    See also this question for a different approach.
    $endgroup$
    – saz
    Jan 10 at 13:43


















3












$begingroup$



Let $X_n$ be Gamma $(n,lambda)$ distributed, and $Y_n = dfrac{lambda X_n -n}{sqrt{n}}$.
Show that $Y_n rightarrow N(0,1)$.




My idea to prove this is to use Lévys theorem with the characteristic functions. If I can show that the characteristic functions $phi_{Y_n}$ of $Y_n$ converge to a function $phi$, which is continous in $0$, then there exists a random variable $Y$ such that $phi$ is the characteristic function of $Y$. My aim is to get $phi_Y (u) = e^{-u^2/2}$ since I have to show that the $Y_n$ converge to a standard normal random variable and the cF describes the distribution uniquely.



I know that the cF of a Gamma$(n,lambda)$-distributed random variable $X_n$ is given by
$$phi_{X_n}(u) =left(frac{lambda}{lambda - iu}right)^n $$
The cF rules for the linear transformation $$Y_n = frac{lambda X_n}{sqrt{n}} +frac{-n}{sqrt{n}}$$ give
$$phi_{Y_n} (u)= e^{-in frac{u}{sqrt{n}}} phi_{X_n}(frac{lambda u}{sqrt{n}}) = e^{-iu sqrt{n}} left(frac{lambda}{lambda - i frac{lambda u}{sqrt{n}}}right)^n$$
that is,
$$phi_{Y_n} (u) = e^{-iu sqrt{n}} left(frac{1}{1 - i frac{ u}{sqrt{n}}}right)^n = e^{-iu sqrt{n}} frac{1}{ (1 - i frac{u}{sqrt{n}})^n}$$
and now I don't know how to continue calculating.



Since I need the limit of this, I thought about the representation of the exponential function in terms of the limit of a sequence, e.g. $$e^x = lim_{n rightarrow infty} left( 1 + frac{x}{n}right)^n$$
Some hints or a trick would be very nice.






probability-theory central-limit-theorem characteristic-functions gamma-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    See also this question for a different approach.
    $endgroup$
    – saz
    Jan 10 at 13:43
















3












3








3


1



$begingroup$



Let $X_n$ be Gamma $(n,lambda)$ distributed, and $Y_n = dfrac{lambda X_n -n}{sqrt{n}}$.
Show that $Y_n rightarrow N(0,1)$.




My idea to prove this is to use Lévys theorem with the characteristic functions. If I can show that the characteristic functions $phi_{Y_n}$ of $Y_n$ converge to a function $phi$, which is continous in $0$, then there exists a random variable $Y$ such that $phi$ is the characteristic function of $Y$. My aim is to get $phi_Y (u) = e^{-u^2/2}$ since I have to show that the $Y_n$ converge to a standard normal random variable and the cF describes the distribution uniquely.



I know that the cF of a Gamma$(n,lambda)$-distributed random variable $X_n$ is given by
$$phi_{X_n}(u) =left(frac{lambda}{lambda - iu}right)^n $$
The cF rules for the linear transformation $$Y_n = frac{lambda X_n}{sqrt{n}} +frac{-n}{sqrt{n}}$$ give
$$phi_{Y_n} (u)= e^{-in frac{u}{sqrt{n}}} phi_{X_n}(frac{lambda u}{sqrt{n}}) = e^{-iu sqrt{n}} left(frac{lambda}{lambda - i frac{lambda u}{sqrt{n}}}right)^n$$
that is,
$$phi_{Y_n} (u) = e^{-iu sqrt{n}} left(frac{1}{1 - i frac{ u}{sqrt{n}}}right)^n = e^{-iu sqrt{n}} frac{1}{ (1 - i frac{u}{sqrt{n}})^n}$$
and now I don't know how to continue calculating.



Since I need the limit of this, I thought about the representation of the exponential function in terms of the limit of a sequence, e.g. $$e^x = lim_{n rightarrow infty} left( 1 + frac{x}{n}right)^n$$
Some hints or a trick would be very nice.






probability-theory central-limit-theorem characteristic-functions gamma-distribution







share|cite|improve this question











$endgroup$





Let $X_n$ be Gamma $(n,lambda)$ distributed, and $Y_n = dfrac{lambda X_n -n}{sqrt{n}}$.
Show that $Y_n rightarrow N(0,1)$.




My idea to prove this is to use Lévys theorem with the characteristic functions. If I can show that the characteristic functions $phi_{Y_n}$ of $Y_n$ converge to a function $phi$, which is continous in $0$, then there exists a random variable $Y$ such that $phi$ is the characteristic function of $Y$. My aim is to get $phi_Y (u) = e^{-u^2/2}$ since I have to show that the $Y_n$ converge to a standard normal random variable and the cF describes the distribution uniquely.



I know that the cF of a Gamma$(n,lambda)$-distributed random variable $X_n$ is given by
$$phi_{X_n}(u) =left(frac{lambda}{lambda - iu}right)^n $$
The cF rules for the linear transformation $$Y_n = frac{lambda X_n}{sqrt{n}} +frac{-n}{sqrt{n}}$$ give
$$phi_{Y_n} (u)= e^{-in frac{u}{sqrt{n}}} phi_{X_n}(frac{lambda u}{sqrt{n}}) = e^{-iu sqrt{n}} left(frac{lambda}{lambda - i frac{lambda u}{sqrt{n}}}right)^n$$
that is,
$$phi_{Y_n} (u) = e^{-iu sqrt{n}} left(frac{1}{1 - i frac{ u}{sqrt{n}}}right)^n = e^{-iu sqrt{n}} frac{1}{ (1 - i frac{u}{sqrt{n}})^n}$$
and now I don't know how to continue calculating.



Since I need the limit of this, I thought about the representation of the exponential function in terms of the limit of a sequence, e.g. $$e^x = lim_{n rightarrow infty} left( 1 + frac{x}{n}right)^n$$
Some hints or a trick would be very nice.






probability-theory central-limit-theorem characteristic-functions gamma-distribution




probability-theory central-limit-theorem characteristic-functions gamma-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 10:41









Did

249k23227466




249k23227466










asked Jan 9 at 19:41









Myrkuls JayKayMyrkuls JayKay

32319




32319












  • $begingroup$
    See also this question for a different approach.
    $endgroup$
    – saz
    Jan 10 at 13:43




















  • $begingroup$
    See also this question for a different approach.
    $endgroup$
    – saz
    Jan 10 at 13:43


















$begingroup$
See also this question for a different approach.
$endgroup$
– saz
Jan 10 at 13:43






$begingroup$
See also this question for a different approach.
$endgroup$
– saz
Jan 10 at 13:43












1 Answer
1






active

oldest

votes


















4












$begingroup$

To sum up, at this moment, you know that $phi_{Y_n} (u)=psi(v_n)^{-n}$ where $$psi(v)=e^{v}(1-v)qquad v_n=frac{iu}{sqrt n}$$
Now, by the expansion of the exponential, when $vto0$, $$psi(v)=left(1+v+frac12v^2+o(v^2)right)(1-v)=1-frac12v^2+o(v^2)$$
Thus, for every fixed $u$, when $ntoinfty$, $$psi(v_n)=1+frac12frac{u^2}n+oleft(frac1nright)$$ hence $$psi(v_n)^nto e^{u^2/2}$$
That is, $Y_n$ converges in distribution to the standard normal distribution, as desired.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Myrkuls JayKay
    Jan 9 at 21:20












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%2f3067872%2fif-x-n-is-gamma-n-lambda-distributed-then-lambda-x-n-n-sqrt-n-to-n%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









4












$begingroup$

To sum up, at this moment, you know that $phi_{Y_n} (u)=psi(v_n)^{-n}$ where $$psi(v)=e^{v}(1-v)qquad v_n=frac{iu}{sqrt n}$$
Now, by the expansion of the exponential, when $vto0$, $$psi(v)=left(1+v+frac12v^2+o(v^2)right)(1-v)=1-frac12v^2+o(v^2)$$
Thus, for every fixed $u$, when $ntoinfty$, $$psi(v_n)=1+frac12frac{u^2}n+oleft(frac1nright)$$ hence $$psi(v_n)^nto e^{u^2/2}$$
That is, $Y_n$ converges in distribution to the standard normal distribution, as desired.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Myrkuls JayKay
    Jan 9 at 21:20
















4












$begingroup$

To sum up, at this moment, you know that $phi_{Y_n} (u)=psi(v_n)^{-n}$ where $$psi(v)=e^{v}(1-v)qquad v_n=frac{iu}{sqrt n}$$
Now, by the expansion of the exponential, when $vto0$, $$psi(v)=left(1+v+frac12v^2+o(v^2)right)(1-v)=1-frac12v^2+o(v^2)$$
Thus, for every fixed $u$, when $ntoinfty$, $$psi(v_n)=1+frac12frac{u^2}n+oleft(frac1nright)$$ hence $$psi(v_n)^nto e^{u^2/2}$$
That is, $Y_n$ converges in distribution to the standard normal distribution, as desired.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Myrkuls JayKay
    Jan 9 at 21:20














4












4








4





$begingroup$

To sum up, at this moment, you know that $phi_{Y_n} (u)=psi(v_n)^{-n}$ where $$psi(v)=e^{v}(1-v)qquad v_n=frac{iu}{sqrt n}$$
Now, by the expansion of the exponential, when $vto0$, $$psi(v)=left(1+v+frac12v^2+o(v^2)right)(1-v)=1-frac12v^2+o(v^2)$$
Thus, for every fixed $u$, when $ntoinfty$, $$psi(v_n)=1+frac12frac{u^2}n+oleft(frac1nright)$$ hence $$psi(v_n)^nto e^{u^2/2}$$
That is, $Y_n$ converges in distribution to the standard normal distribution, as desired.






share|cite|improve this answer











$endgroup$



To sum up, at this moment, you know that $phi_{Y_n} (u)=psi(v_n)^{-n}$ where $$psi(v)=e^{v}(1-v)qquad v_n=frac{iu}{sqrt n}$$
Now, by the expansion of the exponential, when $vto0$, $$psi(v)=left(1+v+frac12v^2+o(v^2)right)(1-v)=1-frac12v^2+o(v^2)$$
Thus, for every fixed $u$, when $ntoinfty$, $$psi(v_n)=1+frac12frac{u^2}n+oleft(frac1nright)$$ hence $$psi(v_n)^nto e^{u^2/2}$$
That is, $Y_n$ converges in distribution to the standard normal distribution, as desired.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 21:00

























answered Jan 9 at 20:51









DidDid

249k23227466




249k23227466












  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Myrkuls JayKay
    Jan 9 at 21:20


















  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Myrkuls JayKay
    Jan 9 at 21:20
















$begingroup$
Great answer, thank you!
$endgroup$
– Myrkuls JayKay
Jan 9 at 21:20




$begingroup$
Great answer, thank you!
$endgroup$
– Myrkuls JayKay
Jan 9 at 21:20


















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%2f3067872%2fif-x-n-is-gamma-n-lambda-distributed-then-lambda-x-n-n-sqrt-n-to-n%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

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more