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How do you show a variance matrix is nonnegative definite? - Answers

I assume you mean covariance matrix and I assume that you are familiar with the definition:C = E[(X-u)(X-u)T]where X is a random vector and u = E(X) is the meanThe definition of non-negative definite is:xTCx ≥ 0 for any vector x Є RSo is xTE[(X-u)(X-u)T]x ≥ 0?Then, from one of the covariance properties:E[(xT(X-u))((X-u)Tx)] = E[xT([(X-u)(X-u)T]x)] = E[((xTI)x)] = E[xTx]Finally, since we've already defined x to have only real values, xTx is therefore non-negative definite by definition.



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How do you show a variance matrix is nonnegative definite? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_show_a_variance_matrix_is_nonnegative_definite

I assume you mean covariance matrix and I assume that you are familiar with the definition:C = E[(X-u)(X-u)T]where X is a random vector and u = E(X) is the meanThe definition of non-negative definite is:xTCx ≥ 0 for any vector x Є RSo is xTE[(X-u)(X-u)T]x ≥ 0?Then, from one of the covariance properties:E[(xT(X-u))((X-u)Tx)] = E[xT([(X-u)(X-u)T]x)] = E[((xTI)x)] = E[xTx]Finally, since we've already defined x to have only real values, xTx is therefore non-negative definite by definition.



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https://math.answers.com/math-and-arithmetic/How_do_you_show_a_variance_matrix_is_nonnegative_definite

How do you show a variance matrix is nonnegative definite? - Answers

I assume you mean covariance matrix and I assume that you are familiar with the definition:C = E[(X-u)(X-u)T]where X is a random vector and u = E(X) is the meanThe definition of non-negative definite is:xTCx ≥ 0 for any vector x Є RSo is xTE[(X-u)(X-u)T]x ≥ 0?Then, from one of the covariance properties:E[(xT(X-u))((X-u)Tx)] = E[xT([(X-u)(X-u)T]x)] = E[((xTI)x)] = E[xTx]Finally, since we've already defined x to have only real values, xTx is therefore non-negative definite by definition.

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      I assume you mean covariance matrix and I assume that you are familiar with the definition:C = E[(X-u)(X-u)T]where X is a random vector and u = E(X) is the meanThe definition of non-negative definite is:xTCx ≥ 0 for any vector x Є RSo is xTE[(X-u)(X-u)T]x ≥ 0?Then, from one of the covariance properties:E[(xT(X-u))((X-u)Tx)] = E[xT([(X-u)(X-u)T]x)] = E[((xTI)x)] = E[xTx]Finally, since we've already defined x to have only real values, xTx is therefore non-negative definite by definition.

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