$\begingroup$

I am a probability theory beginner. The expression for the variance of a random variable $x$ (of a random process is

$$\sigma^2 = E(x^2) - (\mu_{x})^2$$

If $E(x^2) = (\mu_{x})^2$, then $\sigma^2 = 0$. Can this happen ? Can a random variable have a density function whose variance (the second central moment alone) is $0$ (other than the dirac delta function).

$\endgroup$ 2

1 Answer

$\begingroup$

The variance $$ E(X^2)-E(X)^2=E(X-E(X))^2 $$ is equal to $0$ if and only if $X$ is equal to $E(X)$ in all of its support. This can only happen if $X$ is equal to some constant with probability $1$ (known as a degenerate distribution).

$\endgroup$ 2

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy