Hints on proving existence of $(prod_{i=1}^{n}X_{n})^{frac{1}{n}}$












1












$begingroup$


Let $(X_{n})_{n}$ be independent random variables that are $mathcal{U}{[1,2]}$



Prove $(prod_{i=1}^{n}X_{n})^{frac{1}{n}}$ exists for $n to infty$ and that $exists c in mathbb R$ such that



$(prod_{i=1}^{n}X_{n})^{frac{1}{n}}to c$, a.s. for $n to infty$



I honestly do not even know where to begin because I've never worked with the product of random variables.



Any ideas and tips?






probability-theory random-variables independence uniform-distribution







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint. Strong Law of Large Numbers.
    $endgroup$
    – Sangchul Lee
    Jan 7 at 13:49










  • $begingroup$
    But strong law of large numbers pertains to a sum of random variables $sum X_{i}$ and not the Product $prod X_{i}$
    $endgroup$
    – SABOY
    Jan 7 at 13:51
















1












$begingroup$


Let $(X_{n})_{n}$ be independent random variables that are $mathcal{U}{[1,2]}$



Prove $(prod_{i=1}^{n}X_{n})^{frac{1}{n}}$ exists for $n to infty$ and that $exists c in mathbb R$ such that



$(prod_{i=1}^{n}X_{n})^{frac{1}{n}}to c$, a.s. for $n to infty$



I honestly do not even know where to begin because I've never worked with the product of random variables.



Any ideas and tips?






probability-theory random-variables independence uniform-distribution







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint. Strong Law of Large Numbers.
    $endgroup$
    – Sangchul Lee
    Jan 7 at 13:49










  • $begingroup$
    But strong law of large numbers pertains to a sum of random variables $sum X_{i}$ and not the Product $prod X_{i}$
    $endgroup$
    – SABOY
    Jan 7 at 13:51














1












1








1





$begingroup$


Let $(X_{n})_{n}$ be independent random variables that are $mathcal{U}{[1,2]}$



Prove $(prod_{i=1}^{n}X_{n})^{frac{1}{n}}$ exists for $n to infty$ and that $exists c in mathbb R$ such that



$(prod_{i=1}^{n}X_{n})^{frac{1}{n}}to c$, a.s. for $n to infty$



I honestly do not even know where to begin because I've never worked with the product of random variables.



Any ideas and tips?






probability-theory random-variables independence uniform-distribution







share|cite|improve this question









$endgroup$




Let $(X_{n})_{n}$ be independent random variables that are $mathcal{U}{[1,2]}$



Prove $(prod_{i=1}^{n}X_{n})^{frac{1}{n}}$ exists for $n to infty$ and that $exists c in mathbb R$ such that



$(prod_{i=1}^{n}X_{n})^{frac{1}{n}}to c$, a.s. for $n to infty$



I honestly do not even know where to begin because I've never worked with the product of random variables.



Any ideas and tips?






probability-theory random-variables independence uniform-distribution




probability-theory random-variables independence uniform-distribution






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 13:22









SABOYSABOY

531311




531311








  • 1




    $begingroup$
    Hint. Strong Law of Large Numbers.
    $endgroup$
    – Sangchul Lee
    Jan 7 at 13:49










  • $begingroup$
    But strong law of large numbers pertains to a sum of random variables $sum X_{i}$ and not the Product $prod X_{i}$
    $endgroup$
    – SABOY
    Jan 7 at 13:51














  • 1




    $begingroup$
    Hint. Strong Law of Large Numbers.
    $endgroup$
    – Sangchul Lee
    Jan 7 at 13:49










  • $begingroup$
    But strong law of large numbers pertains to a sum of random variables $sum X_{i}$ and not the Product $prod X_{i}$
    $endgroup$
    – SABOY
    Jan 7 at 13:51








1




1




$begingroup$
Hint. Strong Law of Large Numbers.
$endgroup$
– Sangchul Lee
Jan 7 at 13:49




$begingroup$
Hint. Strong Law of Large Numbers.
$endgroup$
– Sangchul Lee
Jan 7 at 13:49












$begingroup$
But strong law of large numbers pertains to a sum of random variables $sum X_{i}$ and not the Product $prod X_{i}$
$endgroup$
– SABOY
Jan 7 at 13:51




$begingroup$
But strong law of large numbers pertains to a sum of random variables $sum X_{i}$ and not the Product $prod X_{i}$
$endgroup$
– SABOY
Jan 7 at 13:51










1 Answer
1






active

oldest

votes


















2












$begingroup$

It is unclear what $mathcal{U}[1,2]$ means, so I'll presume $mathcal{U}[1,2]$ refers to the uniform distribution on $[1,2]$.



The answer is straightforward once we note that $$ln left([prod_{i=1}^n X_i]^{1/n}right) = frac{1}{n}sum_{i=1}^nln X_i$$ is an averaged sum of i.i.d. random variables $(ln X_n)_{n=1}^infty$. By strong law of large numbers, it follows
$$
frac{1}{n}sum_{i=1}^nln X_i to E[ln X_1]= int_1^2 ln x ;dx = (xln x-x)big|^2_1=2ln 2-1
$$almost surely as $ntoinfty$. Therefore, we have
$$
left[prod_{i=1}^n X_iright]^{1/n}=expleft(frac{1}{n}sum_{i=1}^nln X_iright)to exp(2ln 2-1) = frac{4}{e}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would never have though of that $operatorname{ln}$ trick thank you!
    $endgroup$
    – SABOY
    Jan 7 at 13:54










  • $begingroup$
    Just one question: How do we know that $operatorname{ln}(X)$ ~ $mathcal{U}{[1,2]}$ if $X$ ~ $mathcal{U}{[1,2]}$?
    $endgroup$
    – SABOY
    Jan 7 at 17:38












  • $begingroup$
    Umm ... ? There's a misunderstanding, I guess. Let me explain. Of course $ln Xsim mathcal{U}[1,2]$ is false, and I didn't use it. What I used is just well-known expectation formula that $E[g(X)]=int g(x)f(x)dx$ where $f(x)$ is a p.d.f. of $X$. Yes, over the same interval $[1,2]$ where $X$ is defined! I hope this makes it clear.
    $endgroup$
    – Song
    Jan 7 at 17:41












  • $begingroup$
    So over the same interval $[1,2]$
    $endgroup$
    – SABOY
    Jan 7 at 17:42










  • $begingroup$
    Quick question: you say that $(ln(X_{i}))_{i=1,...,n}$ are i.d.d. random variables if $(X_{i}))_{i=1,...,n}$ are i.i.d. But why would $(exp(lambda X_{i}))_{i=1,...,n}$ not be i.i.d. ? As stated here: math.stackexchange.com/a/3068631/512018
    $endgroup$
    – SABOY
    2 days ago













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

It is unclear what $mathcal{U}[1,2]$ means, so I'll presume $mathcal{U}[1,2]$ refers to the uniform distribution on $[1,2]$.



The answer is straightforward once we note that $$ln left([prod_{i=1}^n X_i]^{1/n}right) = frac{1}{n}sum_{i=1}^nln X_i$$ is an averaged sum of i.i.d. random variables $(ln X_n)_{n=1}^infty$. By strong law of large numbers, it follows
$$
frac{1}{n}sum_{i=1}^nln X_i to E[ln X_1]= int_1^2 ln x ;dx = (xln x-x)big|^2_1=2ln 2-1
$$almost surely as $ntoinfty$. Therefore, we have
$$
left[prod_{i=1}^n X_iright]^{1/n}=expleft(frac{1}{n}sum_{i=1}^nln X_iright)to exp(2ln 2-1) = frac{4}{e}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would never have though of that $operatorname{ln}$ trick thank you!
    $endgroup$
    – SABOY
    Jan 7 at 13:54










  • $begingroup$
    Just one question: How do we know that $operatorname{ln}(X)$ ~ $mathcal{U}{[1,2]}$ if $X$ ~ $mathcal{U}{[1,2]}$?
    $endgroup$
    – SABOY
    Jan 7 at 17:38












  • $begingroup$
    Umm ... ? There's a misunderstanding, I guess. Let me explain. Of course $ln Xsim mathcal{U}[1,2]$ is false, and I didn't use it. What I used is just well-known expectation formula that $E[g(X)]=int g(x)f(x)dx$ where $f(x)$ is a p.d.f. of $X$. Yes, over the same interval $[1,2]$ where $X$ is defined! I hope this makes it clear.
    $endgroup$
    – Song
    Jan 7 at 17:41












  • $begingroup$
    So over the same interval $[1,2]$
    $endgroup$
    – SABOY
    Jan 7 at 17:42










  • $begingroup$
    Quick question: you say that $(ln(X_{i}))_{i=1,...,n}$ are i.d.d. random variables if $(X_{i}))_{i=1,...,n}$ are i.i.d. But why would $(exp(lambda X_{i}))_{i=1,...,n}$ not be i.i.d. ? As stated here: math.stackexchange.com/a/3068631/512018
    $endgroup$
    – SABOY
    2 days ago


















2












$begingroup$

It is unclear what $mathcal{U}[1,2]$ means, so I'll presume $mathcal{U}[1,2]$ refers to the uniform distribution on $[1,2]$.



The answer is straightforward once we note that $$ln left([prod_{i=1}^n X_i]^{1/n}right) = frac{1}{n}sum_{i=1}^nln X_i$$ is an averaged sum of i.i.d. random variables $(ln X_n)_{n=1}^infty$. By strong law of large numbers, it follows
$$
frac{1}{n}sum_{i=1}^nln X_i to E[ln X_1]= int_1^2 ln x ;dx = (xln x-x)big|^2_1=2ln 2-1
$$almost surely as $ntoinfty$. Therefore, we have
$$
left[prod_{i=1}^n X_iright]^{1/n}=expleft(frac{1}{n}sum_{i=1}^nln X_iright)to exp(2ln 2-1) = frac{4}{e}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would never have though of that $operatorname{ln}$ trick thank you!
    $endgroup$
    – SABOY
    Jan 7 at 13:54










  • $begingroup$
    Just one question: How do we know that $operatorname{ln}(X)$ ~ $mathcal{U}{[1,2]}$ if $X$ ~ $mathcal{U}{[1,2]}$?
    $endgroup$
    – SABOY
    Jan 7 at 17:38












  • $begingroup$
    Umm ... ? There's a misunderstanding, I guess. Let me explain. Of course $ln Xsim mathcal{U}[1,2]$ is false, and I didn't use it. What I used is just well-known expectation formula that $E[g(X)]=int g(x)f(x)dx$ where $f(x)$ is a p.d.f. of $X$. Yes, over the same interval $[1,2]$ where $X$ is defined! I hope this makes it clear.
    $endgroup$
    – Song
    Jan 7 at 17:41












  • $begingroup$
    So over the same interval $[1,2]$
    $endgroup$
    – SABOY
    Jan 7 at 17:42










  • $begingroup$
    Quick question: you say that $(ln(X_{i}))_{i=1,...,n}$ are i.d.d. random variables if $(X_{i}))_{i=1,...,n}$ are i.i.d. But why would $(exp(lambda X_{i}))_{i=1,...,n}$ not be i.i.d. ? As stated here: math.stackexchange.com/a/3068631/512018
    $endgroup$
    – SABOY
    2 days ago
















2












2








2





$begingroup$

It is unclear what $mathcal{U}[1,2]$ means, so I'll presume $mathcal{U}[1,2]$ refers to the uniform distribution on $[1,2]$.



The answer is straightforward once we note that $$ln left([prod_{i=1}^n X_i]^{1/n}right) = frac{1}{n}sum_{i=1}^nln X_i$$ is an averaged sum of i.i.d. random variables $(ln X_n)_{n=1}^infty$. By strong law of large numbers, it follows
$$
frac{1}{n}sum_{i=1}^nln X_i to E[ln X_1]= int_1^2 ln x ;dx = (xln x-x)big|^2_1=2ln 2-1
$$almost surely as $ntoinfty$. Therefore, we have
$$
left[prod_{i=1}^n X_iright]^{1/n}=expleft(frac{1}{n}sum_{i=1}^nln X_iright)to exp(2ln 2-1) = frac{4}{e}.
$$






share|cite|improve this answer









$endgroup$



It is unclear what $mathcal{U}[1,2]$ means, so I'll presume $mathcal{U}[1,2]$ refers to the uniform distribution on $[1,2]$.



The answer is straightforward once we note that $$ln left([prod_{i=1}^n X_i]^{1/n}right) = frac{1}{n}sum_{i=1}^nln X_i$$ is an averaged sum of i.i.d. random variables $(ln X_n)_{n=1}^infty$. By strong law of large numbers, it follows
$$
frac{1}{n}sum_{i=1}^nln X_i to E[ln X_1]= int_1^2 ln x ;dx = (xln x-x)big|^2_1=2ln 2-1
$$almost surely as $ntoinfty$. Therefore, we have
$$
left[prod_{i=1}^n X_iright]^{1/n}=expleft(frac{1}{n}sum_{i=1}^nln X_iright)to exp(2ln 2-1) = frac{4}{e}.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 13:51









SongSong

7,343422




7,343422












  • $begingroup$
    Would never have though of that $operatorname{ln}$ trick thank you!
    $endgroup$
    – SABOY
    Jan 7 at 13:54










  • $begingroup$
    Just one question: How do we know that $operatorname{ln}(X)$ ~ $mathcal{U}{[1,2]}$ if $X$ ~ $mathcal{U}{[1,2]}$?
    $endgroup$
    – SABOY
    Jan 7 at 17:38












  • $begingroup$
    Umm ... ? There's a misunderstanding, I guess. Let me explain. Of course $ln Xsim mathcal{U}[1,2]$ is false, and I didn't use it. What I used is just well-known expectation formula that $E[g(X)]=int g(x)f(x)dx$ where $f(x)$ is a p.d.f. of $X$. Yes, over the same interval $[1,2]$ where $X$ is defined! I hope this makes it clear.
    $endgroup$
    – Song
    Jan 7 at 17:41












  • $begingroup$
    So over the same interval $[1,2]$
    $endgroup$
    – SABOY
    Jan 7 at 17:42










  • $begingroup$
    Quick question: you say that $(ln(X_{i}))_{i=1,...,n}$ are i.d.d. random variables if $(X_{i}))_{i=1,...,n}$ are i.i.d. But why would $(exp(lambda X_{i}))_{i=1,...,n}$ not be i.i.d. ? As stated here: math.stackexchange.com/a/3068631/512018
    $endgroup$
    – SABOY
    2 days ago




















  • $begingroup$
    Would never have though of that $operatorname{ln}$ trick thank you!
    $endgroup$
    – SABOY
    Jan 7 at 13:54










  • $begingroup$
    Just one question: How do we know that $operatorname{ln}(X)$ ~ $mathcal{U}{[1,2]}$ if $X$ ~ $mathcal{U}{[1,2]}$?
    $endgroup$
    – SABOY
    Jan 7 at 17:38












  • $begingroup$
    Umm ... ? There's a misunderstanding, I guess. Let me explain. Of course $ln Xsim mathcal{U}[1,2]$ is false, and I didn't use it. What I used is just well-known expectation formula that $E[g(X)]=int g(x)f(x)dx$ where $f(x)$ is a p.d.f. of $X$. Yes, over the same interval $[1,2]$ where $X$ is defined! I hope this makes it clear.
    $endgroup$
    – Song
    Jan 7 at 17:41












  • $begingroup$
    So over the same interval $[1,2]$
    $endgroup$
    – SABOY
    Jan 7 at 17:42










  • $begingroup$
    Quick question: you say that $(ln(X_{i}))_{i=1,...,n}$ are i.d.d. random variables if $(X_{i}))_{i=1,...,n}$ are i.i.d. But why would $(exp(lambda X_{i}))_{i=1,...,n}$ not be i.i.d. ? As stated here: math.stackexchange.com/a/3068631/512018
    $endgroup$
    – SABOY
    2 days ago


















$begingroup$
Would never have though of that $operatorname{ln}$ trick thank you!
$endgroup$
– SABOY
Jan 7 at 13:54




$begingroup$
Would never have though of that $operatorname{ln}$ trick thank you!
$endgroup$
– SABOY
Jan 7 at 13:54












$begingroup$
Just one question: How do we know that $operatorname{ln}(X)$ ~ $mathcal{U}{[1,2]}$ if $X$ ~ $mathcal{U}{[1,2]}$?
$endgroup$
– SABOY
Jan 7 at 17:38






$begingroup$
Just one question: How do we know that $operatorname{ln}(X)$ ~ $mathcal{U}{[1,2]}$ if $X$ ~ $mathcal{U}{[1,2]}$?
$endgroup$
– SABOY
Jan 7 at 17:38














$begingroup$
Umm ... ? There's a misunderstanding, I guess. Let me explain. Of course $ln Xsim mathcal{U}[1,2]$ is false, and I didn't use it. What I used is just well-known expectation formula that $E[g(X)]=int g(x)f(x)dx$ where $f(x)$ is a p.d.f. of $X$. Yes, over the same interval $[1,2]$ where $X$ is defined! I hope this makes it clear.
$endgroup$
– Song
Jan 7 at 17:41






$begingroup$
Umm ... ? There's a misunderstanding, I guess. Let me explain. Of course $ln Xsim mathcal{U}[1,2]$ is false, and I didn't use it. What I used is just well-known expectation formula that $E[g(X)]=int g(x)f(x)dx$ where $f(x)$ is a p.d.f. of $X$. Yes, over the same interval $[1,2]$ where $X$ is defined! I hope this makes it clear.
$endgroup$
– Song
Jan 7 at 17:41














$begingroup$
So over the same interval $[1,2]$
$endgroup$
– SABOY
Jan 7 at 17:42




$begingroup$
So over the same interval $[1,2]$
$endgroup$
– SABOY
Jan 7 at 17:42












$begingroup$
Quick question: you say that $(ln(X_{i}))_{i=1,...,n}$ are i.d.d. random variables if $(X_{i}))_{i=1,...,n}$ are i.i.d. But why would $(exp(lambda X_{i}))_{i=1,...,n}$ not be i.i.d. ? As stated here: math.stackexchange.com/a/3068631/512018
$endgroup$
– SABOY
2 days ago






$begingroup$
Quick question: you say that $(ln(X_{i}))_{i=1,...,n}$ are i.d.d. random variables if $(X_{i}))_{i=1,...,n}$ are i.i.d. But why would $(exp(lambda X_{i}))_{i=1,...,n}$ not be i.i.d. ? As stated here: math.stackexchange.com/a/3068631/512018
$endgroup$
– SABOY
2 days ago




















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