How to interpret this variance

Question

If I have a probability measure defined by $P( \Omega ) = \int_{\Omega} (1-a) \delta(x) + a \delta(x-a^2) dx,$ then I noticed that the variance is given by $a^5(1-a)$. This is somewhat strange, cause the variance can be negative if $a \in \mathbb{R}\setminus [0,1]$. Is there any way to understand this result? Is there any meaning in this probability distribution for $a \notin [0,1]$?

Answer 1

The way to understand this must include noticing that not only would the "variance" be negative if $a\notin[0,1]$, but one of the "probabilities" would be negative if $a<0$ and the other if $a>1$. This is actually a probability distribution only if $a\in[0,1]$.