I am curios about the concept of Variance. I try to get the better understanding of the variance by checking extreme cases.
$Var(X) = E[(X^2)] - (E[X])^2$
question 1. What does it mean when Variance equals 0.
If variance of random variable X equals 0, it means that
$E[(X^2)] = (E[X])^2$
in case of discrete random variable, we can rewrite it as
$\sum_{x \in X}^{} p(x)x^2 = (\sum_{x \in X}^{} p(x)x)^2$
in case when sample set is just one element $x$ with probability 1, we get
$p(x)x^2 = p(x)^2x^2$, which is OK when $p(x)=1$.
In case when sample is two elements with equal probabilities (uniform distribution), variance doesn't equal 0.
So here I didn't answer my question when variance equals 0 (the only obvious case is when sample space consist of one element).
Do you know more cases when variance is 0?
question 2. What does it mean when variance equals 1.
Few times I saw the mentioning of unit variance (I assume is when variance equals 1)
the question is when $E[(X^2)] - (E[X])^2=1$.
I cannot come up with cases when variance is 1.
Can you explain what is so special about unit variance, and when it occurs?
Addendum:
Thanks to the comments, lets try reconsider the case when variance equals 0.
$E[(X-\mu)^2]=0$
$\sum_{x \in X}^{} p(x)(x-\mu)^2= 0$
it happens, when either $p(x)=0$, or $x=\mu$.
Therefore, in order to random variable has the variance 0, it should get the only value (which is of course will be equal to $\mu$), any other value occurs with probability 0.
Is this the right reasoning?
$\endgroup$ 61 Answer
$\begingroup$variance is $E(X-EX)^2$. So when variance is $0$ you can say that the random variable $X$ is identically equal to its expectation $EX$. That is $X$ is a degenerate random variable that takes the value $EX$ with probability $1$. There is no other extreme cases. Variance $=1$ is not an extreme case at all. Variance of a random variable is as large as you can think of. The larger the variance the greater will be the observation on that random variable dispersed (or spread) around the mean value.
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