# Why is variance of a non-random variable zero? [closed]

In deriving ${Var}(\hat{\beta_1})$ in simple linear regression model, I came across following derivation. Why are the variances of non-random terms zero below?

\begin{align} \text{Var}(\hat{\beta_1}) & = \text{Var} \left(\frac{\sum_i (x_i - \bar{x})(y_i - \bar{y})}{\sum_i (x_i - \bar{x})^2} \right) \\ &= \text{Var} \left(\frac{\sum_i (x_i - \bar{x})(\beta_0 + \beta_1x_i + u_i )}{\sum_i (x_i - \bar{x})^2} \right)\;\;\; \\ &= \text{Var} \left(\frac{\sum_i (x_i - \bar{x})u_i}{\sum_i (x_i - \bar{x})^2} \right), \;\;\;\text{noting only $u_i$ is a random variable} \\ \end{align}

## closed as unclear what you're asking by Michael Chernick, Tim♦, kjetil b halvorsen, Nick Cox, Peter Flom♦Jun 7 '17 at 11:38

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• Where did you get the second equation? – Michael Chernick Jun 5 '17 at 21:19
• Because they are never different from their mean. – AdamO Jun 5 '17 at 21:32
• You don't actually need a definition or concept of variance of "non-random terms". This calculation merely uses the fact that $\operatorname{Var}(X)=\operatorname{Var}(X+c)$ for any random variable $X$ and constant $c$. That fact is readily derived from any of the usual definitions of variance. It is also intuitively obvious: since the variance is supposed to be a measure of the spread of a random variable, and shifting random variables (by adding a constant to them) doesn't change their spread, the variance shouldn't change. – whuber Jun 5 '17 at 23:17

$\delta(t) = \lim_{\sigma\to0} \frac{e^{-t^2/2 \sigma^2}}{\sqrt{2 \pi} \sigma}$.
Since $\text{Var} = \sigma^2$, variance also goes to zero as standard deviation goes to zero. Thus, in this view non-random variables are the same as random variables, but with zero variance.