Central Limit Theorem and convergence of transformed
I have the following exercise to solve:
Let $X_n, n geq1$ be a sequence of i.i.d random variables where each $X_n$ is a discrete random variable with distribution $P(X_n=1)=1-p$ and $P(X_n=2)=p$, with $0<p<1$.
Determine the asymptotic behaviour for $n rightarrow infty$ of
$Z_n=n frac{(prod_{i=1}^{n}Xi)^{(1/n)}}{X_1^2+X_2^2+....X_n^2}$
I know I should apply the log transofrmation in the numerator but I get very confused after that.
Thank you in advance
central-limit-theorem
asked Nov 30 at 19:42
Matteo
23
add a comment |
I have the following exercise to solve:
Let $X_n, n geq1$ be a sequence of i.i.d random variables where each $X_n$ is a discrete random variable with distribution $P(X_n=1)=1-p$ and $P(X_n=2)=p$, with $0<p<1$.
Determine the asymptotic behaviour for $n rightarrow infty$ of
$Z_n=n frac{(prod_{i=1}^{n}Xi)^{(1/n)}}{X_1^2+X_2^2+....X_n^2}$
I know I should apply the log transofrmation in the numerator but I get very confused after that.
Thank you in advance
central-limit-theorem
asked Nov 30 at 19:42
Matteo
23
add a comment |
I have the following exercise to solve:
Let $X_n, n geq1$ be a sequence of i.i.d random variables where each $X_n$ is a discrete random variable with distribution $P(X_n=1)=1-p$ and $P(X_n=2)=p$, with $0<p<1$.
Determine the asymptotic behaviour for $n rightarrow infty$ of
$Z_n=n frac{(prod_{i=1}^{n}Xi)^{(1/n)}}{X_1^2+X_2^2+....X_n^2}$
I know I should apply the log transofrmation in the numerator but I get very confused after that.
Thank you in advance
central-limit-theorem
asked Nov 30 at 19:42
Matteo
23
I have the following exercise to solve:
Let $X_n, n geq1$ be a sequence of i.i.d random variables where each $X_n$ is a discrete random variable with distribution $P(X_n=1)=1-p$ and $P(X_n=2)=p$, with $0<p<1$.
Determine the asymptotic behaviour for $n rightarrow infty$ of
$Z_n=n frac{(prod_{i=1}^{n}Xi)^{(1/n)}}{X_1^2+X_2^2+....X_n^2}$
I know I should apply the log transofrmation in the numerator but I get very confused after that.
Thank you in advance
central-limit-theorem
central-limit-theorem
asked Nov 30 at 19:42
Matteo
23
asked Nov 30 at 19:42
Matteo
23
asked Nov 30 at 19:42
Matteo
23
asked Nov 30 at 19:42
Matteo
23
asked Nov 30 at 19:42
Matteo
23
23
add a comment |
add a comment |
1 Answer
1
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votes
Some steps:
- Using the law of large numbers, we know that $n^{-1}sum_{i=1}^nX_i^2to mathbb Eleft[X_1^2right]$ almost surely.
- Since $lnleft(left(prod_{i=1}^nX_iright)^{1/n}right)=frac 1nsum_{i=1}^nln X_i$, the law of large numbers applied to the i.i.d. sequence $left(ln X_iright)_{igeqslant 1}$ gives the asymptotic behavior of $left(prod_{i=1}^nX_iright)^{1/n}$.
answered Nov 30 at 23:14
Davide Giraudo
125k16150259
Yes that is what I was trying too. However, I should apply the central limit theorem to find the limiting distribution, and this is my problem since I am not sure about how to procede
– Matteo
Dec 1 at 6:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Some steps:
- Using the law of large numbers, we know that $n^{-1}sum_{i=1}^nX_i^2to mathbb Eleft[X_1^2right]$ almost surely.
- Since $lnleft(left(prod_{i=1}^nX_iright)^{1/n}right)=frac 1nsum_{i=1}^nln X_i$, the law of large numbers applied to the i.i.d. sequence $left(ln X_iright)_{igeqslant 1}$ gives the asymptotic behavior of $left(prod_{i=1}^nX_iright)^{1/n}$.
answered Nov 30 at 23:14
Davide Giraudo
125k16150259
Yes that is what I was trying too. However, I should apply the central limit theorem to find the limiting distribution, and this is my problem since I am not sure about how to procede
– Matteo
Dec 1 at 6:32
add a comment |
Some steps:
- Using the law of large numbers, we know that $n^{-1}sum_{i=1}^nX_i^2to mathbb Eleft[X_1^2right]$ almost surely.
- Since $lnleft(left(prod_{i=1}^nX_iright)^{1/n}right)=frac 1nsum_{i=1}^nln X_i$, the law of large numbers applied to the i.i.d. sequence $left(ln X_iright)_{igeqslant 1}$ gives the asymptotic behavior of $left(prod_{i=1}^nX_iright)^{1/n}$.
answered Nov 30 at 23:14
Davide Giraudo
125k16150259
Yes that is what I was trying too. However, I should apply the central limit theorem to find the limiting distribution, and this is my problem since I am not sure about how to procede
– Matteo
Dec 1 at 6:32
add a comment |
Some steps:
- Using the law of large numbers, we know that $n^{-1}sum_{i=1}^nX_i^2to mathbb Eleft[X_1^2right]$ almost surely.
- Since $lnleft(left(prod_{i=1}^nX_iright)^{1/n}right)=frac 1nsum_{i=1}^nln X_i$, the law of large numbers applied to the i.i.d. sequence $left(ln X_iright)_{igeqslant 1}$ gives the asymptotic behavior of $left(prod_{i=1}^nX_iright)^{1/n}$.
answered Nov 30 at 23:14
Davide Giraudo
125k16150259
Some steps:
- Using the law of large numbers, we know that $n^{-1}sum_{i=1}^nX_i^2to mathbb Eleft[X_1^2right]$ almost surely.
- Since $lnleft(left(prod_{i=1}^nX_iright)^{1/n}right)=frac 1nsum_{i=1}^nln X_i$, the law of large numbers applied to the i.i.d. sequence $left(ln X_iright)_{igeqslant 1}$ gives the asymptotic behavior of $left(prod_{i=1}^nX_iright)^{1/n}$.
answered Nov 30 at 23:14
Davide Giraudo
125k16150259
answered Nov 30 at 23:14
Davide Giraudo
125k16150259
answered Nov 30 at 23:14
Davide Giraudo
125k16150259
answered Nov 30 at 23:14
Davide Giraudo
125k16150259
125k16150259
Yes that is what I was trying too. However, I should apply the central limit theorem to find the limiting distribution, and this is my problem since I am not sure about how to procede
– Matteo
Dec 1 at 6:32
add a comment |
Yes that is what I was trying too. However, I should apply the central limit theorem to find the limiting distribution, and this is my problem since I am not sure about how to procede
– Matteo
Dec 1 at 6:32
Yes that is what I was trying too. However, I should apply the central limit theorem to find the limiting distribution, and this is my problem since I am not sure about how to procede
– Matteo
Dec 1 at 6:32
Yes that is what I was trying too. However, I should apply the central limit theorem to find the limiting distribution, and this is my problem since I am not sure about how to procede
– Matteo
Dec 1 at 6:32
add a comment |
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