Expectation for a $chi^2_n$ distributed random variable.
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Consider $X_1,ldots, X_nsimmathcal{N}(mu,sigma^2)$ iid, where $mu$, $sigma^2$ are unknown and $c>0$. Define
$$S_n^2:=sum_{k=1}^nBig(X_k-frac{1}{n}sum_{i=1}^nX_iBig)^2$$
Now I want to calculate
$$E[cS_n^2-sigma^2]$$
Actually I was reading that $S_n^2$ might be $chi_n^2$-distributed, but I do not know why. So does someone has a hint on this problem?
Thanks in advance!
probability-theory statistics stochastic-calculus stochastic-integrals descriptive-statistics
asked Nov 30 '18 at 17:28
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
Consider $X_1,ldots, X_nsimmathcal{N}(mu,sigma^2)$ iid, where $mu$, $sigma^2$ are unknown and $c>0$. Define
$$S_n^2:=sum_{k=1}^nBig(X_k-frac{1}{n}sum_{i=1}^nX_iBig)^2$$
Now I want to calculate
$$E[cS_n^2-sigma^2]$$
Actually I was reading that $S_n^2$ might be $chi_n^2$-distributed, but I do not know why. So does someone has a hint on this problem?
Thanks in advance!
probability-theory statistics stochastic-calculus stochastic-integrals descriptive-statistics
asked Nov 30 '18 at 17:28
user408858user408858
482213
$endgroup$
$begingroup$
Any source explaining the Bonferroni correction will prove $ES_n^2=(n-1)sigma^2$, which is enough for your purposes.
$endgroup$
– J.G.
Nov 30 '18 at 23:38
$begingroup$
I was able to show it by simple calculation. Thanks for your comment!
$endgroup$
– user408858
Dec 1 '18 at 12:08
add a comment |
$begingroup$
Consider $X_1,ldots, X_nsimmathcal{N}(mu,sigma^2)$ iid, where $mu$, $sigma^2$ are unknown and $c>0$. Define
$$S_n^2:=sum_{k=1}^nBig(X_k-frac{1}{n}sum_{i=1}^nX_iBig)^2$$
Now I want to calculate
$$E[cS_n^2-sigma^2]$$
Actually I was reading that $S_n^2$ might be $chi_n^2$-distributed, but I do not know why. So does someone has a hint on this problem?
Thanks in advance!
probability-theory statistics stochastic-calculus stochastic-integrals descriptive-statistics
asked Nov 30 '18 at 17:28
user408858user408858
482213
$endgroup$
Consider $X_1,ldots, X_nsimmathcal{N}(mu,sigma^2)$ iid, where $mu$, $sigma^2$ are unknown and $c>0$. Define
$$S_n^2:=sum_{k=1}^nBig(X_k-frac{1}{n}sum_{i=1}^nX_iBig)^2$$
Now I want to calculate
$$E[cS_n^2-sigma^2]$$
Actually I was reading that $S_n^2$ might be $chi_n^2$-distributed, but I do not know why. So does someone has a hint on this problem?
Thanks in advance!
probability-theory statistics stochastic-calculus stochastic-integrals descriptive-statistics
probability-theory statistics stochastic-calculus stochastic-integrals descriptive-statistics
asked Nov 30 '18 at 17:28
user408858user408858
482213
asked Nov 30 '18 at 17:28
user408858user408858
482213
edited Nov 30 '18 at 23:29
user408858
asked Nov 30 '18 at 17:28
user408858user408858
482213
asked Nov 30 '18 at 17:28
user408858user408858
482213
asked Nov 30 '18 at 17:28
user408858user408858
482213
482213
$begingroup$
Any source explaining the Bonferroni correction will prove $ES_n^2=(n-1)sigma^2$, which is enough for your purposes.
$endgroup$
– J.G.
Nov 30 '18 at 23:38
$begingroup$
I was able to show it by simple calculation. Thanks for your comment!
$endgroup$
– user408858
Dec 1 '18 at 12:08
add a comment |
$begingroup$
Any source explaining the Bonferroni correction will prove $ES_n^2=(n-1)sigma^2$, which is enough for your purposes.
$endgroup$
– J.G.
Nov 30 '18 at 23:38
$begingroup$
I was able to show it by simple calculation. Thanks for your comment!
$endgroup$
– user408858
Dec 1 '18 at 12:08
$begingroup$
Any source explaining the Bonferroni correction will prove $ES_n^2=(n-1)sigma^2$, which is enough for your purposes.
$endgroup$
– J.G.
Nov 30 '18 at 23:38
$begingroup$
Any source explaining the Bonferroni correction will prove $ES_n^2=(n-1)sigma^2$, which is enough for your purposes.
$endgroup$
– J.G.
Nov 30 '18 at 23:38
$begingroup$
I was able to show it by simple calculation. Thanks for your comment!
$endgroup$
– user408858
Dec 1 '18 at 12:08
$begingroup$
I was able to show it by simple calculation. Thanks for your comment!
$endgroup$
– user408858
Dec 1 '18 at 12:08
add a comment |
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$begingroup$
Any source explaining the Bonferroni correction will prove $ES_n^2=(n-1)sigma^2$, which is enough for your purposes.
$endgroup$
– J.G.
Nov 30 '18 at 23:38
$begingroup$
I was able to show it by simple calculation. Thanks for your comment!
$endgroup$
– user408858
Dec 1 '18 at 12:08