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Amir
Amir
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$\text{Var}\left(\sum X_i\right) = \sum\limits_i \text{Var}\left( X_i\right) +\sum\limits_i \sum\limits_{j\not=i} \text{Cov}\left( X_i,X_j\right)$ and$$\text{Var}\left(\sum X_i\right) = \sum\limits_i \text{Var}\left( X_i\right) +\sum\limits_i \sum\limits_{j\not=i} \text{Cov}\left( X_i,X_j\right)$$ is maximised when the covariances take their maximum possible positive values, which happens when all the correlations are $+1$.

So the highest variance case for $\sum X_i$ and thus $\bar X$ will be when there is perfect positive correlation between the $X_i$, in which case $\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$$$\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$$ giving $\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2$.$$\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2 .$$

$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$ Then, as can be shown using the Cauchy–Schwarz inequality,:

$$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$$

with equality only when all the $\text{SD}(X_i)$ are equal.

So your $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum \text{Var}(X_i)$ is correct,

with equality only when $X_i-E[X_i]=X_j-E[X_j]$ for all $i,j$ so when you have identical variances and perfect positive correlation (though possibly different expectations).

$\text{Var}\left(\sum X_i\right) = \sum\limits_i \text{Var}\left( X_i\right) +\sum\limits_i \sum\limits_{j\not=i} \text{Cov}\left( X_i,X_j\right)$ and is maximised when the covariances take their maximum possible positive values which happens when all the correlations are $+1$.

So the highest variance case for $\sum X_i$ and thus $\bar X$ will be when there is perfect positive correlation between the $X_i$, in which case $\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$ giving $\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2$.

$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$, as can be shown using the Cauchy–Schwarz inequality, with equality only when all the $\text{SD}(X_i)$ are equal.

So your $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum \text{Var}(X_i)$ is correct,

with equality only when $X_i-E[X_i]=X_j-E[X_j]$ for all $i,j$ so when you have identical variances and perfect positive correlation (though possibly different expectations).

$$\text{Var}\left(\sum X_i\right) = \sum\limits_i \text{Var}\left( X_i\right) +\sum\limits_i \sum\limits_{j\not=i} \text{Cov}\left( X_i,X_j\right)$$ is maximised when the covariances take their maximum possible positive values, which happens when all the correlations are $+1$.

So the highest variance case for $\sum X_i$ and thus $\bar X$ will be when there is perfect positive correlation between the $X_i$, in which case $$\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$$ giving $$\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2 .$$

Then, using the Cauchy–Schwarz inequality:

$$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$$

with equality only when all the $\text{SD}(X_i)$ are equal.

So your $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum \text{Var}(X_i)$ is correct,

with equality only when $X_i-E[X_i]=X_j-E[X_j]$ for all $i,j$ so when you have identical variances and perfect positive correlation (though possibly different expectations).

covariances
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Henry
Henry
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The$\text{Var}\left(\sum X_i\right) = \sum\limits_i \text{Var}\left( X_i\right) +\sum\limits_i \sum\limits_{j\not=i} \text{Cov}\left( X_i,X_j\right)$ and is maximised when the covariances take their maximum possible positive values which happens when all the correlations are $+1$.

So the highest variance case for $\sum X_i$ and thus $\bar X$ will be when there is perfect positive correlation between the $X_i$, in which case $\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$ giving $\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2$.

$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$, as can be shown using the Cauchy–Schwarz inequality, with equality only when all the $\text{SD}(X_i)$ are equal.

So your $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum \text{Var}(X_i)$ is correct,

with equality only when $X_i-E[X_i]=X_j-E[X_j]$ for all $i,j$ so when you have identical variances and perfect positive correlation tough(though possibly different expectations).

The highest variance case for $\sum X_i$ and $\bar X$ will be when there is perfect positive correlation between the $X_i$, in which case $\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$ giving $\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2$.

$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$, as can be shown using the Cauchy–Schwarz inequality, with equality only when all the $\text{SD}(X_i)$ are equal.

So your $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum \text{Var}(X_i)$ is correct,

with equality only when $X_i-E[X_i]=X_j-E[X_j]$ for all $i,j$ so when you have identical variances and perfect positive correlation tough possibly different expectations.

$\text{Var}\left(\sum X_i\right) = \sum\limits_i \text{Var}\left( X_i\right) +\sum\limits_i \sum\limits_{j\not=i} \text{Cov}\left( X_i,X_j\right)$ and is maximised when the covariances take their maximum possible positive values which happens when all the correlations are $+1$.

So the highest variance case for $\sum X_i$ and thus $\bar X$ will be when there is perfect positive correlation between the $X_i$, in which case $\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$ giving $\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2$.

$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$, as can be shown using the Cauchy–Schwarz inequality, with equality only when all the $\text{SD}(X_i)$ are equal.

So your $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum \text{Var}(X_i)$ is correct,

with equality only when $X_i-E[X_i]=X_j-E[X_j]$ for all $i,j$ so when you have identical variances and perfect positive correlation (though possibly different expectations).

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Henry
Henry
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The highest variance case for $\sum X_i$ and $\bar X$ will be when there is perfect positive correlation between the $X_i$, in which case $\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$ giving $\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2$.

$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$, as can be shown using the Cauchy–Schwarz inequality, with equality only when all the $\text{SD}(X_i)$ are equal.

So your $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum \text{Var}(X_i)$ is correct,

with equality only when $X_i-E[X_i]=X_j-E[X_j]$ for all $i,j$ so when you have identical variances and perfect positive correlation tough possibly different expectations.