Levy Khinchine formula for Gamma distribution
$begingroup$
I am searching for the Levy-Khinchin formula for the Gamma distribution $Gamma(lambda, beta)$. My attempt is following:
For the characteristic function I get
$$varphi_{Gamma(lambda,beta)}(t)=Big(frac{beta}{beta -it}Big)^lambda$$
Therefore
$$[varphi_{Gamma(frac{lambda}{n},beta)}(t)]^n=Big[Big(frac{beta}{beta -it}Big)^frac{lambda}{n}Big]^n=varphi_{Gamma(lambda,beta)}(t)$$
and the gamma distribution in infinitely divisible. Now the idea is to find a triplet $(b,sigma,nu)$ for the Levy-Khinchine formula. I have a hint to use $nu$ which has a density regarding to the lebesgue measure as follows
$$frac{lambda}{x}e^{-beta x}1_{[0,infty)}(x)$$
So the Levy Kinchin formula is the following:
$$psi(t)=itb-frac{t^2sigma^2}{2}+int_{mathbb{R}}Big(e^{itx}-1-itxcdot1_{{|x|<1}}(x)Big)frac{lambda}{x}e^{-beta x}1_{[0,infty)}(x) dx$$
Now I take derivates of $logcircvarphi_{Gamma(lambda,beta)}$ and $psi$ to find $b$:
$$psi'(t)=ib-tsigma^2+frac{lambda i}{beta-it}+lambda iBig(frac{1}{beta}e^{-beta}-frac{1}{beta}Big)$$
$$(logcircvarphi_{Gamma(lambda,beta)})'(t)=frac{lambda i}{beta-it}$$
So I take $sigma=0$ and $b=-lambda Big(frac{1}{beta}e^{-beta}-frac{1}{beta}Big)$ to get $psi'(t)=(logcircvarphi_{Gamma(lambda,beta)})'(t)$. We also have $psi(0)=varphi_{Gamma(lambda,beta)}(0)=0.$
Now I get
begin{align} varphi_{Gamma(lambda,beta)}(t)&=exp(log(varphi_{Gamma(lambda,beta)}(t)))\
&=exp(log(varphi_{Gamma(lambda,beta)}(t)-log(varphi_{Gamma(lambda,beta)}(0)))\
&=exp(int_0^t psi'(s)ds)text{, what we just calculated}\
&=exp(psi(t)-psi(0))\
&=exp(psi(t))
end{align}
Is this the right way?
probability-theory stochastic-processes stochastic-calculus
asked Nov 29 '18 at 15:55
user408858user408858
470110
$endgroup$
add a comment |
$begingroup$
I am searching for the Levy-Khinchin formula for the Gamma distribution $Gamma(lambda, beta)$. My attempt is following:
For the characteristic function I get
$$varphi_{Gamma(lambda,beta)}(t)=Big(frac{beta}{beta -it}Big)^lambda$$
Therefore
$$[varphi_{Gamma(frac{lambda}{n},beta)}(t)]^n=Big[Big(frac{beta}{beta -it}Big)^frac{lambda}{n}Big]^n=varphi_{Gamma(lambda,beta)}(t)$$
and the gamma distribution in infinitely divisible. Now the idea is to find a triplet $(b,sigma,nu)$ for the Levy-Khinchine formula. I have a hint to use $nu$ which has a density regarding to the lebesgue measure as follows
$$frac{lambda}{x}e^{-beta x}1_{[0,infty)}(x)$$
So the Levy Kinchin formula is the following:
$$psi(t)=itb-frac{t^2sigma^2}{2}+int_{mathbb{R}}Big(e^{itx}-1-itxcdot1_{{|x|<1}}(x)Big)frac{lambda}{x}e^{-beta x}1_{[0,infty)}(x) dx$$
Now I take derivates of $logcircvarphi_{Gamma(lambda,beta)}$ and $psi$ to find $b$:
$$psi'(t)=ib-tsigma^2+frac{lambda i}{beta-it}+lambda iBig(frac{1}{beta}e^{-beta}-frac{1}{beta}Big)$$
$$(logcircvarphi_{Gamma(lambda,beta)})'(t)=frac{lambda i}{beta-it}$$
So I take $sigma=0$ and $b=-lambda Big(frac{1}{beta}e^{-beta}-frac{1}{beta}Big)$ to get $psi'(t)=(logcircvarphi_{Gamma(lambda,beta)})'(t)$. We also have $psi(0)=varphi_{Gamma(lambda,beta)}(0)=0.$
Now I get
begin{align} varphi_{Gamma(lambda,beta)}(t)&=exp(log(varphi_{Gamma(lambda,beta)}(t)))\
&=exp(log(varphi_{Gamma(lambda,beta)}(t)-log(varphi_{Gamma(lambda,beta)}(0)))\
&=exp(int_0^t psi'(s)ds)text{, what we just calculated}\
&=exp(psi(t)-psi(0))\
&=exp(psi(t))
end{align}
Is this the right way?
probability-theory stochastic-processes stochastic-calculus
asked Nov 29 '18 at 15:55
user408858user408858
470110
$endgroup$
add a comment |
$begingroup$
I am searching for the Levy-Khinchin formula for the Gamma distribution $Gamma(lambda, beta)$. My attempt is following:
For the characteristic function I get
$$varphi_{Gamma(lambda,beta)}(t)=Big(frac{beta}{beta -it}Big)^lambda$$
Therefore
$$[varphi_{Gamma(frac{lambda}{n},beta)}(t)]^n=Big[Big(frac{beta}{beta -it}Big)^frac{lambda}{n}Big]^n=varphi_{Gamma(lambda,beta)}(t)$$
and the gamma distribution in infinitely divisible. Now the idea is to find a triplet $(b,sigma,nu)$ for the Levy-Khinchine formula. I have a hint to use $nu$ which has a density regarding to the lebesgue measure as follows
$$frac{lambda}{x}e^{-beta x}1_{[0,infty)}(x)$$
So the Levy Kinchin formula is the following:
$$psi(t)=itb-frac{t^2sigma^2}{2}+int_{mathbb{R}}Big(e^{itx}-1-itxcdot1_{{|x|<1}}(x)Big)frac{lambda}{x}e^{-beta x}1_{[0,infty)}(x) dx$$
Now I take derivates of $logcircvarphi_{Gamma(lambda,beta)}$ and $psi$ to find $b$:
$$psi'(t)=ib-tsigma^2+frac{lambda i}{beta-it}+lambda iBig(frac{1}{beta}e^{-beta}-frac{1}{beta}Big)$$
$$(logcircvarphi_{Gamma(lambda,beta)})'(t)=frac{lambda i}{beta-it}$$
So I take $sigma=0$ and $b=-lambda Big(frac{1}{beta}e^{-beta}-frac{1}{beta}Big)$ to get $psi'(t)=(logcircvarphi_{Gamma(lambda,beta)})'(t)$. We also have $psi(0)=varphi_{Gamma(lambda,beta)}(0)=0.$
Now I get
begin{align} varphi_{Gamma(lambda,beta)}(t)&=exp(log(varphi_{Gamma(lambda,beta)}(t)))\
&=exp(log(varphi_{Gamma(lambda,beta)}(t)-log(varphi_{Gamma(lambda,beta)}(0)))\
&=exp(int_0^t psi'(s)ds)text{, what we just calculated}\
&=exp(psi(t)-psi(0))\
&=exp(psi(t))
end{align}
Is this the right way?
probability-theory stochastic-processes stochastic-calculus
asked Nov 29 '18 at 15:55
user408858user408858
470110
$endgroup$
I am searching for the Levy-Khinchin formula for the Gamma distribution $Gamma(lambda, beta)$. My attempt is following:
For the characteristic function I get
$$varphi_{Gamma(lambda,beta)}(t)=Big(frac{beta}{beta -it}Big)^lambda$$
Therefore
$$[varphi_{Gamma(frac{lambda}{n},beta)}(t)]^n=Big[Big(frac{beta}{beta -it}Big)^frac{lambda}{n}Big]^n=varphi_{Gamma(lambda,beta)}(t)$$
and the gamma distribution in infinitely divisible. Now the idea is to find a triplet $(b,sigma,nu)$ for the Levy-Khinchine formula. I have a hint to use $nu$ which has a density regarding to the lebesgue measure as follows
$$frac{lambda}{x}e^{-beta x}1_{[0,infty)}(x)$$
So the Levy Kinchin formula is the following:
$$psi(t)=itb-frac{t^2sigma^2}{2}+int_{mathbb{R}}Big(e^{itx}-1-itxcdot1_{{|x|<1}}(x)Big)frac{lambda}{x}e^{-beta x}1_{[0,infty)}(x) dx$$
Now I take derivates of $logcircvarphi_{Gamma(lambda,beta)}$ and $psi$ to find $b$:
$$psi'(t)=ib-tsigma^2+frac{lambda i}{beta-it}+lambda iBig(frac{1}{beta}e^{-beta}-frac{1}{beta}Big)$$
$$(logcircvarphi_{Gamma(lambda,beta)})'(t)=frac{lambda i}{beta-it}$$
So I take $sigma=0$ and $b=-lambda Big(frac{1}{beta}e^{-beta}-frac{1}{beta}Big)$ to get $psi'(t)=(logcircvarphi_{Gamma(lambda,beta)})'(t)$. We also have $psi(0)=varphi_{Gamma(lambda,beta)}(0)=0.$
Now I get
begin{align} varphi_{Gamma(lambda,beta)}(t)&=exp(log(varphi_{Gamma(lambda,beta)}(t)))\
&=exp(log(varphi_{Gamma(lambda,beta)}(t)-log(varphi_{Gamma(lambda,beta)}(0)))\
&=exp(int_0^t psi'(s)ds)text{, what we just calculated}\
&=exp(psi(t)-psi(0))\
&=exp(psi(t))
end{align}
Is this the right way?
probability-theory stochastic-processes stochastic-calculus
probability-theory stochastic-processes stochastic-calculus
asked Nov 29 '18 at 15:55
user408858user408858
470110
asked Nov 29 '18 at 15:55
user408858user408858
470110
asked Nov 29 '18 at 15:55
user408858user408858
470110
asked Nov 29 '18 at 15:55
user408858user408858
470110
asked Nov 29 '18 at 15:55
user408858user408858
470110
470110
add a comment |
add a comment |
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