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







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$endgroup$












  • $begingroup$
    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

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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












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1 Answer
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1 Answer
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active

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active

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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


















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