Chebychev inequailty on random variable $X_1 + ldots + X_n$
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Let $X_i$ be independent random variable such that each $X_i$ is symmetric about 0 and Var($X_i) = 2i -1$, for $i geq 1$. Then $$lim_{n mapsto
infty} P(X_1 + ldots + X_n > n {rm log};n) = ?$$
Options:
a) doesn't exists
b) equals 1/2
c) equals 1
d) equals 0
Right answer b)
My approach:
Since question involved inequality, I thought I could apply Chebychev inequality. But before that I made few observations,
1) $E(X_i) = 0$ as it is symmetric about $0$.
2) $P(mid X mid geq epsilon) = 2 P(X geq epsilon)$ because of symmetry around $0$.
Let $S = X_1 + ldots + X_n$. So I wrote the equation $$2 times P(S geq ksigma )leq 1/ k^2$$
Identifying $n$ log $n = k sigma$ where $sigma = $Var$(S)$, my answer should be $sigma^2 / (2 times n$ log $n)^2$. To calculate $sigma$ observe that $X_i$ are independent and hence $sigma = sum_{i=1}^{n}(2i -1)$ which is $n^2$. Hence the final expression should be $$lim_{n mapsto infty}big(n^2/(2 times n {rm log} n) big)^2 = 0$$ which is different from the answer given. Examining the expression again, there is $1/2$ term present which would mean that maybe I made mistake in caluclating the limits. So can anyone help me to sort this out?
Reference: CSIR NET DEC 2015 Paper Code A Q.no. 53
probability probability-limit-theorems
asked Nov 20 at 19:41
henceproved
795
add a comment |
up vote
3
down vote
favorite
Let $X_i$ be independent random variable such that each $X_i$ is symmetric about 0 and Var($X_i) = 2i -1$, for $i geq 1$. Then $$lim_{n mapsto
infty} P(X_1 + ldots + X_n > n {rm log};n) = ?$$
Options:
a) doesn't exists
b) equals 1/2
c) equals 1
d) equals 0
Right answer b)
My approach:
Since question involved inequality, I thought I could apply Chebychev inequality. But before that I made few observations,
1) $E(X_i) = 0$ as it is symmetric about $0$.
2) $P(mid X mid geq epsilon) = 2 P(X geq epsilon)$ because of symmetry around $0$.
Let $S = X_1 + ldots + X_n$. So I wrote the equation $$2 times P(S geq ksigma )leq 1/ k^2$$
Identifying $n$ log $n = k sigma$ where $sigma = $Var$(S)$, my answer should be $sigma^2 / (2 times n$ log $n)^2$. To calculate $sigma$ observe that $X_i$ are independent and hence $sigma = sum_{i=1}^{n}(2i -1)$ which is $n^2$. Hence the final expression should be $$lim_{n mapsto infty}big(n^2/(2 times n {rm log} n) big)^2 = 0$$ which is different from the answer given. Examining the expression again, there is $1/2$ term present which would mean that maybe I made mistake in caluclating the limits. So can anyone help me to sort this out?
Reference: CSIR NET DEC 2015 Paper Code A Q.no. 53
probability probability-limit-theorems
asked Nov 20 at 19:41
henceproved
795
I believe $sigma = sqrt{mathrm{Var}(S)}$ in the context of Chebyshev's inequality, although I am not sure if that is the mistake that you have made here.
– rubikscube09
Nov 20 at 20:29
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $X_i$ be independent random variable such that each $X_i$ is symmetric about 0 and Var($X_i) = 2i -1$, for $i geq 1$. Then $$lim_{n mapsto
infty} P(X_1 + ldots + X_n > n {rm log};n) = ?$$
Options:
a) doesn't exists
b) equals 1/2
c) equals 1
d) equals 0
Right answer b)
My approach:
Since question involved inequality, I thought I could apply Chebychev inequality. But before that I made few observations,
1) $E(X_i) = 0$ as it is symmetric about $0$.
2) $P(mid X mid geq epsilon) = 2 P(X geq epsilon)$ because of symmetry around $0$.
Let $S = X_1 + ldots + X_n$. So I wrote the equation $$2 times P(S geq ksigma )leq 1/ k^2$$
Identifying $n$ log $n = k sigma$ where $sigma = $Var$(S)$, my answer should be $sigma^2 / (2 times n$ log $n)^2$. To calculate $sigma$ observe that $X_i$ are independent and hence $sigma = sum_{i=1}^{n}(2i -1)$ which is $n^2$. Hence the final expression should be $$lim_{n mapsto infty}big(n^2/(2 times n {rm log} n) big)^2 = 0$$ which is different from the answer given. Examining the expression again, there is $1/2$ term present which would mean that maybe I made mistake in caluclating the limits. So can anyone help me to sort this out?
Reference: CSIR NET DEC 2015 Paper Code A Q.no. 53
probability probability-limit-theorems
asked Nov 20 at 19:41
henceproved
795
Let $X_i$ be independent random variable such that each $X_i$ is symmetric about 0 and Var($X_i) = 2i -1$, for $i geq 1$. Then $$lim_{n mapsto
infty} P(X_1 + ldots + X_n > n {rm log};n) = ?$$
Options:
a) doesn't exists
b) equals 1/2
c) equals 1
d) equals 0
Right answer b)
My approach:
Since question involved inequality, I thought I could apply Chebychev inequality. But before that I made few observations,
1) $E(X_i) = 0$ as it is symmetric about $0$.
2) $P(mid X mid geq epsilon) = 2 P(X geq epsilon)$ because of symmetry around $0$.
Let $S = X_1 + ldots + X_n$. So I wrote the equation $$2 times P(S geq ksigma )leq 1/ k^2$$
Identifying $n$ log $n = k sigma$ where $sigma = $Var$(S)$, my answer should be $sigma^2 / (2 times n$ log $n)^2$. To calculate $sigma$ observe that $X_i$ are independent and hence $sigma = sum_{i=1}^{n}(2i -1)$ which is $n^2$. Hence the final expression should be $$lim_{n mapsto infty}big(n^2/(2 times n {rm log} n) big)^2 = 0$$ which is different from the answer given. Examining the expression again, there is $1/2$ term present which would mean that maybe I made mistake in caluclating the limits. So can anyone help me to sort this out?
Reference: CSIR NET DEC 2015 Paper Code A Q.no. 53
probability probability-limit-theorems
probability probability-limit-theorems
asked Nov 20 at 19:41
henceproved
795
asked Nov 20 at 19:41
henceproved
795
edited Nov 20 at 19:47
asked Nov 20 at 19:41
henceproved
795
asked Nov 20 at 19:41
henceproved
795
asked Nov 20 at 19:41
henceproved
795
795
I believe $sigma = sqrt{mathrm{Var}(S)}$ in the context of Chebyshev's inequality, although I am not sure if that is the mistake that you have made here.
– rubikscube09
Nov 20 at 20:29
add a comment |
I believe $sigma = sqrt{mathrm{Var}(S)}$ in the context of Chebyshev's inequality, although I am not sure if that is the mistake that you have made here.
– rubikscube09
Nov 20 at 20:29
I believe $sigma = sqrt{mathrm{Var}(S)}$ in the context of Chebyshev's inequality, although I am not sure if that is the mistake that you have made here.
– rubikscube09
Nov 20 at 20:29
I believe $sigma = sqrt{mathrm{Var}(S)}$ in the context of Chebyshev's inequality, although I am not sure if that is the mistake that you have made here.
– rubikscube09
Nov 20 at 20:29
add a comment |
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I believe $sigma = sqrt{mathrm{Var}(S)}$ in the context of Chebyshev's inequality, although I am not sure if that is the mistake that you have made here.
– rubikscube09
Nov 20 at 20:29