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cleaned up odd notation in E( sum(X) )^2 vs. E( sum(X)^2 )
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Macro
Macro
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The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] - E \left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2$$$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) - \left[ E\left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] = E \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right] = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$$$ E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) = E \left( \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ E\left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \left( \sum_{i=1}^{n} E(X_i) \right) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$$$ \left[ E\left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \sum_{i=1}^{n} E(X_i) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & 0.4 &-0.6 \\ 0.4 & 1 & 0.2 \\ -0.6 & 0.2 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] - E \left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] = E \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right] = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ E\left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \left( \sum_{i=1}^{n} E(X_i) \right) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & 0.4 &-0.6 \\ 0.4 & 1 & 0.2 \\ -0.6 & 0.2 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) - \left[ E\left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) = E \left( \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ \left[ E\left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \sum_{i=1}^{n} E(X_i) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & 0.4 &-0.6 \\ 0.4 & 1 & 0.2 \\ -0.6 & 0.2 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

changed example matrix for clarity as explained in my comment
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edit approved Jun 27, 2012 at 18:06
Abe
edit approved Jun 27, 2012 at 18:06
Abe
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The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] - E \left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] = E \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right] = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ E\left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \left( \sum_{i=1}^{n} E(X_i) \right) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & .3 &-.6 \\ .3 & 1 & .3 \\ -.6 & .3 & 1 \\ \end{array} \right) $$$$ \left( \begin{array}{ccc} 1 & 0.4 &-0.6 \\ 0.4 & 1 & 0.2 \\ -0.6 & 0.2 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] - E \left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] = E \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right] = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ E\left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \left( \sum_{i=1}^{n} E(X_i) \right) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & .3 &-.6 \\ .3 & 1 & .3 \\ -.6 & .3 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] - E \left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] = E \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right] = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ E\left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \left( \sum_{i=1}^{n} E(X_i) \right) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & 0.4 &-0.6 \\ 0.4 & 1 & 0.2 \\ -0.6 & 0.2 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

corrected minor typos
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Dilip Sarwate
Dilip Sarwate
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The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] - E \left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] = E \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right] = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ E\left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \left( \sum_{i=1}^{n} E(X_i) \right) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma + \rho) \neq 2\sigma$${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & .3 &-.6 \\ .3 & 1 & .3 \\ -.6 & .3 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] - E \left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] = E \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right] = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ E\left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \left( \sum_{i=1}^{n} E(X_i) \right) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma + \rho) \neq 2\sigma$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & .3 &-.6 \\ .3 & 1 & .3 \\ -.6 & .3 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] - E \left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left[ \left( \sum_{i=1}^{n} X_i \right)^2 \right] = E \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right] = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ E\left[ \left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \left( \sum_{i=1}^{n} E(X_i) \right) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

  • If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

  • If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

  • but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & .3 &-.6 \\ .3 & 1 & .3 \\ -.6 & .3 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

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