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Consider a random variable $V$ with variance $\sigma_V^2$.

Since the covariance between a random variance and a constant is zero, I think, if $\sigma_V^2=0$, the covariance between $V$ and another random variable, say $U$, should be zero.

That is, $\sigma_V^2=0 \implies \sigma_{VU}=0$ for any random variable $U$.

I am strongly sure. But, I want to check this one more again.

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Yes, this is the consequence of the Cauchy-Schwarz inequality: \begin{align} \operatorname{Cov}(U, V)^2 \leq \operatorname{Var}(U)\operatorname{Var}(V). \end{align}

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    $\begingroup$ And another way to get the result is from the definition of covariance as the expected value of the product of the deviation of U from its mean with the deviation of V from its mean. Variance of V = 0 means that the deviation of V from its mean is always zero. Thus Cov is the expectation of a product of two functions, one of which is zero. Thus the expectation is zero. $\endgroup$ May 22, 2023 at 3:30

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