Compute mean and covariance matrix of $bar{X}$ from a simple random sample
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Given ${X_alpha , alpha =1,...N}$ a simple random sample obtained from any p-dimensional distribution with mean $mu$ and covariance matrix $Sigma$, compute the mean and the covariance matrix of $bar{X}$.
Using the linearity of the mean and knowing that for the variance we have $Var(cX)=c^2 Var(X)$ where $c$ is a constant and $X$ a random variable/vector, I obtained that the mean of $bar{X}$ is again $mu$ and the covariance matrix is $frac{1}{N} Sigma$. Is it correct?
multivariable-calculus random-variables covariance means sampling
asked Dec 9 '18 at 20:36
Maggie94Maggie94
746
$endgroup$
add a comment |
$begingroup$
Given ${X_alpha , alpha =1,...N}$ a simple random sample obtained from any p-dimensional distribution with mean $mu$ and covariance matrix $Sigma$, compute the mean and the covariance matrix of $bar{X}$.
Using the linearity of the mean and knowing that for the variance we have $Var(cX)=c^2 Var(X)$ where $c$ is a constant and $X$ a random variable/vector, I obtained that the mean of $bar{X}$ is again $mu$ and the covariance matrix is $frac{1}{N} Sigma$. Is it correct?
multivariable-calculus random-variables covariance means sampling
asked Dec 9 '18 at 20:36
Maggie94Maggie94
746
$endgroup$
$begingroup$
Yes, it is correct.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:09
add a comment |
$begingroup$
Given ${X_alpha , alpha =1,...N}$ a simple random sample obtained from any p-dimensional distribution with mean $mu$ and covariance matrix $Sigma$, compute the mean and the covariance matrix of $bar{X}$.
Using the linearity of the mean and knowing that for the variance we have $Var(cX)=c^2 Var(X)$ where $c$ is a constant and $X$ a random variable/vector, I obtained that the mean of $bar{X}$ is again $mu$ and the covariance matrix is $frac{1}{N} Sigma$. Is it correct?
multivariable-calculus random-variables covariance means sampling
asked Dec 9 '18 at 20:36
Maggie94Maggie94
746
$endgroup$
Given ${X_alpha , alpha =1,...N}$ a simple random sample obtained from any p-dimensional distribution with mean $mu$ and covariance matrix $Sigma$, compute the mean and the covariance matrix of $bar{X}$.
Using the linearity of the mean and knowing that for the variance we have $Var(cX)=c^2 Var(X)$ where $c$ is a constant and $X$ a random variable/vector, I obtained that the mean of $bar{X}$ is again $mu$ and the covariance matrix is $frac{1}{N} Sigma$. Is it correct?
multivariable-calculus random-variables covariance means sampling
multivariable-calculus random-variables covariance means sampling
asked Dec 9 '18 at 20:36
Maggie94Maggie94
746
asked Dec 9 '18 at 20:36
Maggie94Maggie94
746
asked Dec 9 '18 at 20:36
Maggie94Maggie94
746
asked Dec 9 '18 at 20:36
Maggie94Maggie94
746
asked Dec 9 '18 at 20:36
Maggie94Maggie94
746
746
$begingroup$
Yes, it is correct.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:09
add a comment |
$begingroup$
Yes, it is correct.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:09
$begingroup$
Yes, it is correct.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:09
$begingroup$
Yes, it is correct.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:09
add a comment |
1 Answer
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Your conclusion is correct but your reasoning for the variance is incomplete. You must also invoke the fact that ${X_alpha}_{alpha=1}^N$ comprises independent observations from the same distribution; thus we have in particular $$operatorname{Var}left[sum_{alpha=1}^N X_alpha right] overset{text{ind}}{=} sum_{alpha=1}^N operatorname{Var}[X_alpha] overset{text{i.d.}}{=} N boldsymbol Sigma,$$ where the first equality follows from the independence property as just stated, and the second equality follows from the fact that the observations are identically distributed with variance/covariance $boldsymbol Sigma$. Then, as $bar X = frac{1}{N}sum_{alpha=1}^N X_alpha$, its variance is $(Nboldsymbol Sigma)/N^2 = boldsymbol Sigma/N$ using the property you stated.
answered Dec 9 '18 at 21:32
heropupheropup
63.3k762101
$endgroup$
$begingroup$
Ok that’s true!... the fact that the sumation comes out from the mean is due to linearity while only for the variance is for independence, right?
$endgroup$
– Maggie94
Dec 10 '18 at 7:07
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your conclusion is correct but your reasoning for the variance is incomplete. You must also invoke the fact that ${X_alpha}_{alpha=1}^N$ comprises independent observations from the same distribution; thus we have in particular $$operatorname{Var}left[sum_{alpha=1}^N X_alpha right] overset{text{ind}}{=} sum_{alpha=1}^N operatorname{Var}[X_alpha] overset{text{i.d.}}{=} N boldsymbol Sigma,$$ where the first equality follows from the independence property as just stated, and the second equality follows from the fact that the observations are identically distributed with variance/covariance $boldsymbol Sigma$. Then, as $bar X = frac{1}{N}sum_{alpha=1}^N X_alpha$, its variance is $(Nboldsymbol Sigma)/N^2 = boldsymbol Sigma/N$ using the property you stated.
answered Dec 9 '18 at 21:32
heropupheropup
63.3k762101
$endgroup$
$begingroup$
Ok that’s true!... the fact that the sumation comes out from the mean is due to linearity while only for the variance is for independence, right?
$endgroup$
– Maggie94
Dec 10 '18 at 7:07
add a comment |
$begingroup$
Your conclusion is correct but your reasoning for the variance is incomplete. You must also invoke the fact that ${X_alpha}_{alpha=1}^N$ comprises independent observations from the same distribution; thus we have in particular $$operatorname{Var}left[sum_{alpha=1}^N X_alpha right] overset{text{ind}}{=} sum_{alpha=1}^N operatorname{Var}[X_alpha] overset{text{i.d.}}{=} N boldsymbol Sigma,$$ where the first equality follows from the independence property as just stated, and the second equality follows from the fact that the observations are identically distributed with variance/covariance $boldsymbol Sigma$. Then, as $bar X = frac{1}{N}sum_{alpha=1}^N X_alpha$, its variance is $(Nboldsymbol Sigma)/N^2 = boldsymbol Sigma/N$ using the property you stated.
answered Dec 9 '18 at 21:32
heropupheropup
63.3k762101
$endgroup$
$begingroup$
Ok that’s true!... the fact that the sumation comes out from the mean is due to linearity while only for the variance is for independence, right?
$endgroup$
– Maggie94
Dec 10 '18 at 7:07
add a comment |
$begingroup$
Your conclusion is correct but your reasoning for the variance is incomplete. You must also invoke the fact that ${X_alpha}_{alpha=1}^N$ comprises independent observations from the same distribution; thus we have in particular $$operatorname{Var}left[sum_{alpha=1}^N X_alpha right] overset{text{ind}}{=} sum_{alpha=1}^N operatorname{Var}[X_alpha] overset{text{i.d.}}{=} N boldsymbol Sigma,$$ where the first equality follows from the independence property as just stated, and the second equality follows from the fact that the observations are identically distributed with variance/covariance $boldsymbol Sigma$. Then, as $bar X = frac{1}{N}sum_{alpha=1}^N X_alpha$, its variance is $(Nboldsymbol Sigma)/N^2 = boldsymbol Sigma/N$ using the property you stated.
answered Dec 9 '18 at 21:32
heropupheropup
63.3k762101
$endgroup$
Your conclusion is correct but your reasoning for the variance is incomplete. You must also invoke the fact that ${X_alpha}_{alpha=1}^N$ comprises independent observations from the same distribution; thus we have in particular $$operatorname{Var}left[sum_{alpha=1}^N X_alpha right] overset{text{ind}}{=} sum_{alpha=1}^N operatorname{Var}[X_alpha] overset{text{i.d.}}{=} N boldsymbol Sigma,$$ where the first equality follows from the independence property as just stated, and the second equality follows from the fact that the observations are identically distributed with variance/covariance $boldsymbol Sigma$. Then, as $bar X = frac{1}{N}sum_{alpha=1}^N X_alpha$, its variance is $(Nboldsymbol Sigma)/N^2 = boldsymbol Sigma/N$ using the property you stated.
answered Dec 9 '18 at 21:32
heropupheropup
63.3k762101
answered Dec 9 '18 at 21:32
heropupheropup
63.3k762101
answered Dec 9 '18 at 21:32
heropupheropup
63.3k762101
answered Dec 9 '18 at 21:32
heropupheropup
63.3k762101
63.3k762101
$begingroup$
Ok that’s true!... the fact that the sumation comes out from the mean is due to linearity while only for the variance is for independence, right?
$endgroup$
– Maggie94
Dec 10 '18 at 7:07
add a comment |
$begingroup$
Ok that’s true!... the fact that the sumation comes out from the mean is due to linearity while only for the variance is for independence, right?
$endgroup$
– Maggie94
Dec 10 '18 at 7:07
$begingroup$
Ok that’s true!... the fact that the sumation comes out from the mean is due to linearity while only for the variance is for independence, right?
$endgroup$
– Maggie94
Dec 10 '18 at 7:07
$begingroup$
Ok that’s true!... the fact that the sumation comes out from the mean is due to linearity while only for the variance is for independence, right?
$endgroup$
– Maggie94
Dec 10 '18 at 7:07
add a comment |
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Yes, it is correct.
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– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:09