equivalence of definitions
Why are both definitions equivalent?
Random variable $X $ is called Infinite divisible if for each $n in N $ there exist iid random variables $X_1,dots X_n $ such that $X=frac1n sum X_i$ or the same without $frac1n $. The latter is the one I am familiar with.
for example if $Xsim poisson (lambda) $ then we can pick $ X_i sim poisson (lambda /n) $ but with the other definition it is not possible.
probability-theory
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1154
add a comment |
Why are both definitions equivalent?
Random variable $X $ is called Infinite divisible if for each $n in N $ there exist iid random variables $X_1,dots X_n $ such that $X=frac1n sum X_i$ or the same without $frac1n $. The latter is the one I am familiar with.
for example if $Xsim poisson (lambda) $ then we can pick $ X_i sim poisson (lambda /n) $ but with the other definition it is not possible.
probability-theory
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1154
add a comment |
Why are both definitions equivalent?
Random variable $X $ is called Infinite divisible if for each $n in N $ there exist iid random variables $X_1,dots X_n $ such that $X=frac1n sum X_i$ or the same without $frac1n $. The latter is the one I am familiar with.
for example if $Xsim poisson (lambda) $ then we can pick $ X_i sim poisson (lambda /n) $ but with the other definition it is not possible.
probability-theory
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1154
Why are both definitions equivalent?
Random variable $X $ is called Infinite divisible if for each $n in N $ there exist iid random variables $X_1,dots X_n $ such that $X=frac1n sum X_i$ or the same without $frac1n $. The latter is the one I am familiar with.
for example if $Xsim poisson (lambda) $ then we can pick $ X_i sim poisson (lambda /n) $ but with the other definition it is not possible.
probability-theory
probability-theory
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1154
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1154
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1154
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1154
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1154
1154
add a comment |
add a comment |
1 Answer
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If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.1k1190191
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
– Joey Doey
Nov 28 '18 at 17:45
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
– Joey Doey
Nov 28 '18 at 17:47
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
– Jorge Fernández
Nov 28 '18 at 17:47
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
– Jorge Fernández
Nov 28 '18 at 17:48
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.1k1190191
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
– Joey Doey
Nov 28 '18 at 17:45
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
– Joey Doey
Nov 28 '18 at 17:47
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
– Jorge Fernández
Nov 28 '18 at 17:47
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
– Jorge Fernández
Nov 28 '18 at 17:48
add a comment |
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.1k1190191
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
– Joey Doey
Nov 28 '18 at 17:45
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
– Joey Doey
Nov 28 '18 at 17:47
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
– Jorge Fernández
Nov 28 '18 at 17:47
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
– Jorge Fernández
Nov 28 '18 at 17:48
add a comment |
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.1k1190191
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.1k1190191
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.1k1190191
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.1k1190191
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.1k1190191
75.1k1190191
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
– Joey Doey
Nov 28 '18 at 17:45
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
– Joey Doey
Nov 28 '18 at 17:47
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
– Jorge Fernández
Nov 28 '18 at 17:47
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
– Jorge Fernández
Nov 28 '18 at 17:48
add a comment |
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
– Joey Doey
Nov 28 '18 at 17:45
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
– Joey Doey
Nov 28 '18 at 17:47
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
– Jorge Fernández
Nov 28 '18 at 17:47
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
– Jorge Fernández
Nov 28 '18 at 17:48
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
– Joey Doey
Nov 28 '18 at 17:45
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
– Joey Doey
Nov 28 '18 at 17:45
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
– Joey Doey
Nov 28 '18 at 17:47
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
– Joey Doey
Nov 28 '18 at 17:47
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
– Jorge Fernández
Nov 28 '18 at 17:47
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
– Jorge Fernández
Nov 28 '18 at 17:47
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
– Jorge Fernández
Nov 28 '18 at 17:48
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
– Jorge Fernández
Nov 28 '18 at 17:48
add a comment |
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