Central Limit Theorem - Different Forms











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Given the following function:



$W_n = frac{1}{sqrt{n}}Pi_{k=1}^{infty}log(U_k)$ where $U_k$ is uniformly distributed from $1$ to $e$.



Does ${W_n}_{ngeq 1}$ converge in distribution?



I found that: $log(Pi_{k=1}^{infty}U_k)$ is equivalent to $sum_{k=1}^infty log(U_k)$ which can be called $S_n$.



$W_n$ can then be written as $frac{S_n}{sqrt{n}}$. The answers says that this function has the distribution $mathcal{N}(sqrt{n}E[log(U_k)],V[log(U_k])$ where $V$ is variance. The final result is that it does not converge because its mean is dependent on $n$.



I am having trouble understanding where the multiplication by $sqrt{n}$ comes from.



I know that $frac{S_n-nmu}{sqrt{nsigma^2}}$ has the distribution $mathcal{N}(0,1)$ by the central limit theorem. I also saw in my book that $frac{S_n-nmu}{sqrt{n}}$ has the distribution $mathcal{N}(0,sigma^2)$.



I am just having trouble understanding how modifying that equation has the corresponding effect on the distribution.






probability-theory convergence summation weak-convergence central-limit-theorem







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  • Just checking, the product in outside of logarithm?
    – Siong Thye Goh
    Nov 28 at 16:46










  • No sorry, Ill edit the question.
    – jackana3
    Nov 28 at 16:56















up vote
1
down vote

favorite












Given the following function:



$W_n = frac{1}{sqrt{n}}Pi_{k=1}^{infty}log(U_k)$ where $U_k$ is uniformly distributed from $1$ to $e$.



Does ${W_n}_{ngeq 1}$ converge in distribution?



I found that: $log(Pi_{k=1}^{infty}U_k)$ is equivalent to $sum_{k=1}^infty log(U_k)$ which can be called $S_n$.



$W_n$ can then be written as $frac{S_n}{sqrt{n}}$. The answers says that this function has the distribution $mathcal{N}(sqrt{n}E[log(U_k)],V[log(U_k])$ where $V$ is variance. The final result is that it does not converge because its mean is dependent on $n$.



I am having trouble understanding where the multiplication by $sqrt{n}$ comes from.



I know that $frac{S_n-nmu}{sqrt{nsigma^2}}$ has the distribution $mathcal{N}(0,1)$ by the central limit theorem. I also saw in my book that $frac{S_n-nmu}{sqrt{n}}$ has the distribution $mathcal{N}(0,sigma^2)$.



I am just having trouble understanding how modifying that equation has the corresponding effect on the distribution.






probability-theory convergence summation weak-convergence central-limit-theorem







share|cite|improve this question
























  • Just checking, the product in outside of logarithm?
    – Siong Thye Goh
    Nov 28 at 16:46










  • No sorry, Ill edit the question.
    – jackana3
    Nov 28 at 16:56













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Given the following function:



$W_n = frac{1}{sqrt{n}}Pi_{k=1}^{infty}log(U_k)$ where $U_k$ is uniformly distributed from $1$ to $e$.



Does ${W_n}_{ngeq 1}$ converge in distribution?



I found that: $log(Pi_{k=1}^{infty}U_k)$ is equivalent to $sum_{k=1}^infty log(U_k)$ which can be called $S_n$.



$W_n$ can then be written as $frac{S_n}{sqrt{n}}$. The answers says that this function has the distribution $mathcal{N}(sqrt{n}E[log(U_k)],V[log(U_k])$ where $V$ is variance. The final result is that it does not converge because its mean is dependent on $n$.



I am having trouble understanding where the multiplication by $sqrt{n}$ comes from.



I know that $frac{S_n-nmu}{sqrt{nsigma^2}}$ has the distribution $mathcal{N}(0,1)$ by the central limit theorem. I also saw in my book that $frac{S_n-nmu}{sqrt{n}}$ has the distribution $mathcal{N}(0,sigma^2)$.



I am just having trouble understanding how modifying that equation has the corresponding effect on the distribution.






probability-theory convergence summation weak-convergence central-limit-theorem







share|cite|improve this question















Given the following function:



$W_n = frac{1}{sqrt{n}}Pi_{k=1}^{infty}log(U_k)$ where $U_k$ is uniformly distributed from $1$ to $e$.



Does ${W_n}_{ngeq 1}$ converge in distribution?



I found that: $log(Pi_{k=1}^{infty}U_k)$ is equivalent to $sum_{k=1}^infty log(U_k)$ which can be called $S_n$.



$W_n$ can then be written as $frac{S_n}{sqrt{n}}$. The answers says that this function has the distribution $mathcal{N}(sqrt{n}E[log(U_k)],V[log(U_k])$ where $V$ is variance. The final result is that it does not converge because its mean is dependent on $n$.



I am having trouble understanding where the multiplication by $sqrt{n}$ comes from.



I know that $frac{S_n-nmu}{sqrt{nsigma^2}}$ has the distribution $mathcal{N}(0,1)$ by the central limit theorem. I also saw in my book that $frac{S_n-nmu}{sqrt{n}}$ has the distribution $mathcal{N}(0,sigma^2)$.



I am just having trouble understanding how modifying that equation has the corresponding effect on the distribution.






probability-theory convergence summation weak-convergence central-limit-theorem




probability-theory convergence summation weak-convergence central-limit-theorem






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share|cite|improve this question













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edited Dec 1 at 11:01









Davide Giraudo

124k16150259




124k16150259










asked Nov 28 at 16:20









jackana3

153




153












  • Just checking, the product in outside of logarithm?
    – Siong Thye Goh
    Nov 28 at 16:46










  • No sorry, Ill edit the question.
    – jackana3
    Nov 28 at 16:56


















  • Just checking, the product in outside of logarithm?
    – Siong Thye Goh
    Nov 28 at 16:46










  • No sorry, Ill edit the question.
    – jackana3
    Nov 28 at 16:56
















Just checking, the product in outside of logarithm?
– Siong Thye Goh
Nov 28 at 16:46




Just checking, the product in outside of logarithm?
– Siong Thye Goh
Nov 28 at 16:46












No sorry, Ill edit the question.
– jackana3
Nov 28 at 16:56




No sorry, Ill edit the question.
– jackana3
Nov 28 at 16:56










1 Answer
1






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0
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accepted










I believe by $S_n$, you intend to mean $S_n = sum_{k=1}^n log(U_k).$



begin{align}
Eleft[ frac{S_n}{sqrt{n}}right] &= frac1{sqrt{n}}Eleft[sum_{k=1}^n log(U_k) right] \
&= frac1{sqrt{n}}left(n E(log(U_1 right)))\
&= sqrt{n} E(log(U_1))
end{align}



begin{align}
Vleft[ frac{S_n}{sqrt{n}}right] &= frac1{n}Vleft[sum_{k=1}^n log(U_k) right] \
&= frac1{n}left(n V(log(U_1 right)))\
&= V(log(U_1))
end{align}






share|cite|improve this answer























  • Ok I see it now. It was really very simple, thank you Siong
    – jackana3
    Nov 28 at 17:08











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










I believe by $S_n$, you intend to mean $S_n = sum_{k=1}^n log(U_k).$



begin{align}
Eleft[ frac{S_n}{sqrt{n}}right] &= frac1{sqrt{n}}Eleft[sum_{k=1}^n log(U_k) right] \
&= frac1{sqrt{n}}left(n E(log(U_1 right)))\
&= sqrt{n} E(log(U_1))
end{align}



begin{align}
Vleft[ frac{S_n}{sqrt{n}}right] &= frac1{n}Vleft[sum_{k=1}^n log(U_k) right] \
&= frac1{n}left(n V(log(U_1 right)))\
&= V(log(U_1))
end{align}






share|cite|improve this answer























  • Ok I see it now. It was really very simple, thank you Siong
    – jackana3
    Nov 28 at 17:08















up vote
0
down vote



accepted










I believe by $S_n$, you intend to mean $S_n = sum_{k=1}^n log(U_k).$



begin{align}
Eleft[ frac{S_n}{sqrt{n}}right] &= frac1{sqrt{n}}Eleft[sum_{k=1}^n log(U_k) right] \
&= frac1{sqrt{n}}left(n E(log(U_1 right)))\
&= sqrt{n} E(log(U_1))
end{align}



begin{align}
Vleft[ frac{S_n}{sqrt{n}}right] &= frac1{n}Vleft[sum_{k=1}^n log(U_k) right] \
&= frac1{n}left(n V(log(U_1 right)))\
&= V(log(U_1))
end{align}






share|cite|improve this answer























  • Ok I see it now. It was really very simple, thank you Siong
    – jackana3
    Nov 28 at 17:08













up vote
0
down vote



accepted







up vote
0
down vote



accepted






I believe by $S_n$, you intend to mean $S_n = sum_{k=1}^n log(U_k).$



begin{align}
Eleft[ frac{S_n}{sqrt{n}}right] &= frac1{sqrt{n}}Eleft[sum_{k=1}^n log(U_k) right] \
&= frac1{sqrt{n}}left(n E(log(U_1 right)))\
&= sqrt{n} E(log(U_1))
end{align}



begin{align}
Vleft[ frac{S_n}{sqrt{n}}right] &= frac1{n}Vleft[sum_{k=1}^n log(U_k) right] \
&= frac1{n}left(n V(log(U_1 right)))\
&= V(log(U_1))
end{align}






share|cite|improve this answer














I believe by $S_n$, you intend to mean $S_n = sum_{k=1}^n log(U_k).$



begin{align}
Eleft[ frac{S_n}{sqrt{n}}right] &= frac1{sqrt{n}}Eleft[sum_{k=1}^n log(U_k) right] \
&= frac1{sqrt{n}}left(n E(log(U_1 right)))\
&= sqrt{n} E(log(U_1))
end{align}



begin{align}
Vleft[ frac{S_n}{sqrt{n}}right] &= frac1{n}Vleft[sum_{k=1}^n log(U_k) right] \
&= frac1{n}left(n V(log(U_1 right)))\
&= V(log(U_1))
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 at 3:08

























answered Nov 28 at 17:03









Siong Thye Goh

98k1463116




98k1463116












  • Ok I see it now. It was really very simple, thank you Siong
    – jackana3
    Nov 28 at 17:08


















  • Ok I see it now. It was really very simple, thank you Siong
    – jackana3
    Nov 28 at 17:08
















Ok I see it now. It was really very simple, thank you Siong
– jackana3
Nov 28 at 17:08




Ok I see it now. It was really very simple, thank you Siong
– jackana3
Nov 28 at 17:08


















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