$X_n$ ~ $Gamma(n,n)$, find the limit in Law of $X_n$












0












$begingroup$


I am investigating the following idea.
Let $X_n$ ~ $Gamma(n,n)$. I want to find the limit in Law of this random variable.



I tried using Paul Levy theorem that says the following: If I find the limit of the characteristic function equal to some function $theta(t)$ continuous in 0, then there exists a random variable X such as $X_n to X$ with $theta$ as its characteristic function.



Following this idea I tried calculating the following limit:



$$ lim Big(frac{1}{1-int}Big)^n$$
However to me this goes to 0. But that is not possible because such a characteristic function cannot exist. Is this the right approach?










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$endgroup$








  • 2




    $begingroup$
    There is another parameterization of $Gamma$ distribution for which $$ varphi_n=(1-it/n)^nto e^{-it}. $$
    $endgroup$
    – d.k.o.
    Jan 5 at 23:36






  • 1




    $begingroup$
    Indeed: scale $n$ or $1/n$?
    $endgroup$
    – Did
    Jan 6 at 0:34










  • $begingroup$
    @Did What do you mean by scale? Anyway, such a characteristic function is only for the X = 1 constant random variable, am I right?
    $endgroup$
    – qcc101
    Jan 6 at 7:19










  • $begingroup$
    "What do you mean by scale?" One of the parameters of every gamma distribution. Please se the WP page, if refreshing your memories is needed.
    $endgroup$
    – Did
    Jan 6 at 9:42










  • $begingroup$
    I see. I used the following: $ Big( 1 - frac{it}{n} Big)^{-n}$, but I get $e^{it}$
    $endgroup$
    – qcc101
    Jan 6 at 11:30
















0












$begingroup$


I am investigating the following idea.
Let $X_n$ ~ $Gamma(n,n)$. I want to find the limit in Law of this random variable.



I tried using Paul Levy theorem that says the following: If I find the limit of the characteristic function equal to some function $theta(t)$ continuous in 0, then there exists a random variable X such as $X_n to X$ with $theta$ as its characteristic function.



Following this idea I tried calculating the following limit:



$$ lim Big(frac{1}{1-int}Big)^n$$
However to me this goes to 0. But that is not possible because such a characteristic function cannot exist. Is this the right approach?










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    There is another parameterization of $Gamma$ distribution for which $$ varphi_n=(1-it/n)^nto e^{-it}. $$
    $endgroup$
    – d.k.o.
    Jan 5 at 23:36






  • 1




    $begingroup$
    Indeed: scale $n$ or $1/n$?
    $endgroup$
    – Did
    Jan 6 at 0:34










  • $begingroup$
    @Did What do you mean by scale? Anyway, such a characteristic function is only for the X = 1 constant random variable, am I right?
    $endgroup$
    – qcc101
    Jan 6 at 7:19










  • $begingroup$
    "What do you mean by scale?" One of the parameters of every gamma distribution. Please se the WP page, if refreshing your memories is needed.
    $endgroup$
    – Did
    Jan 6 at 9:42










  • $begingroup$
    I see. I used the following: $ Big( 1 - frac{it}{n} Big)^{-n}$, but I get $e^{it}$
    $endgroup$
    – qcc101
    Jan 6 at 11:30














0












0








0


1



$begingroup$


I am investigating the following idea.
Let $X_n$ ~ $Gamma(n,n)$. I want to find the limit in Law of this random variable.



I tried using Paul Levy theorem that says the following: If I find the limit of the characteristic function equal to some function $theta(t)$ continuous in 0, then there exists a random variable X such as $X_n to X$ with $theta$ as its characteristic function.



Following this idea I tried calculating the following limit:



$$ lim Big(frac{1}{1-int}Big)^n$$
However to me this goes to 0. But that is not possible because such a characteristic function cannot exist. Is this the right approach?










share|cite|improve this question









$endgroup$




I am investigating the following idea.
Let $X_n$ ~ $Gamma(n,n)$. I want to find the limit in Law of this random variable.



I tried using Paul Levy theorem that says the following: If I find the limit of the characteristic function equal to some function $theta(t)$ continuous in 0, then there exists a random variable X such as $X_n to X$ with $theta$ as its characteristic function.



Following this idea I tried calculating the following limit:



$$ lim Big(frac{1}{1-int}Big)^n$$
However to me this goes to 0. But that is not possible because such a characteristic function cannot exist. Is this the right approach?







probability limits characteristic-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 22:34









qcc101qcc101

548113




548113








  • 2




    $begingroup$
    There is another parameterization of $Gamma$ distribution for which $$ varphi_n=(1-it/n)^nto e^{-it}. $$
    $endgroup$
    – d.k.o.
    Jan 5 at 23:36






  • 1




    $begingroup$
    Indeed: scale $n$ or $1/n$?
    $endgroup$
    – Did
    Jan 6 at 0:34










  • $begingroup$
    @Did What do you mean by scale? Anyway, such a characteristic function is only for the X = 1 constant random variable, am I right?
    $endgroup$
    – qcc101
    Jan 6 at 7:19










  • $begingroup$
    "What do you mean by scale?" One of the parameters of every gamma distribution. Please se the WP page, if refreshing your memories is needed.
    $endgroup$
    – Did
    Jan 6 at 9:42










  • $begingroup$
    I see. I used the following: $ Big( 1 - frac{it}{n} Big)^{-n}$, but I get $e^{it}$
    $endgroup$
    – qcc101
    Jan 6 at 11:30














  • 2




    $begingroup$
    There is another parameterization of $Gamma$ distribution for which $$ varphi_n=(1-it/n)^nto e^{-it}. $$
    $endgroup$
    – d.k.o.
    Jan 5 at 23:36






  • 1




    $begingroup$
    Indeed: scale $n$ or $1/n$?
    $endgroup$
    – Did
    Jan 6 at 0:34










  • $begingroup$
    @Did What do you mean by scale? Anyway, such a characteristic function is only for the X = 1 constant random variable, am I right?
    $endgroup$
    – qcc101
    Jan 6 at 7:19










  • $begingroup$
    "What do you mean by scale?" One of the parameters of every gamma distribution. Please se the WP page, if refreshing your memories is needed.
    $endgroup$
    – Did
    Jan 6 at 9:42










  • $begingroup$
    I see. I used the following: $ Big( 1 - frac{it}{n} Big)^{-n}$, but I get $e^{it}$
    $endgroup$
    – qcc101
    Jan 6 at 11:30








2




2




$begingroup$
There is another parameterization of $Gamma$ distribution for which $$ varphi_n=(1-it/n)^nto e^{-it}. $$
$endgroup$
– d.k.o.
Jan 5 at 23:36




$begingroup$
There is another parameterization of $Gamma$ distribution for which $$ varphi_n=(1-it/n)^nto e^{-it}. $$
$endgroup$
– d.k.o.
Jan 5 at 23:36




1




1




$begingroup$
Indeed: scale $n$ or $1/n$?
$endgroup$
– Did
Jan 6 at 0:34




$begingroup$
Indeed: scale $n$ or $1/n$?
$endgroup$
– Did
Jan 6 at 0:34












$begingroup$
@Did What do you mean by scale? Anyway, such a characteristic function is only for the X = 1 constant random variable, am I right?
$endgroup$
– qcc101
Jan 6 at 7:19




$begingroup$
@Did What do you mean by scale? Anyway, such a characteristic function is only for the X = 1 constant random variable, am I right?
$endgroup$
– qcc101
Jan 6 at 7:19












$begingroup$
"What do you mean by scale?" One of the parameters of every gamma distribution. Please se the WP page, if refreshing your memories is needed.
$endgroup$
– Did
Jan 6 at 9:42




$begingroup$
"What do you mean by scale?" One of the parameters of every gamma distribution. Please se the WP page, if refreshing your memories is needed.
$endgroup$
– Did
Jan 6 at 9:42












$begingroup$
I see. I used the following: $ Big( 1 - frac{it}{n} Big)^{-n}$, but I get $e^{it}$
$endgroup$
– qcc101
Jan 6 at 11:30




$begingroup$
I see. I used the following: $ Big( 1 - frac{it}{n} Big)^{-n}$, but I get $e^{it}$
$endgroup$
– qcc101
Jan 6 at 11:30










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