$V$ has a probability mass function. \begin{align} p_V (0) &= 0.5 \\ p_V (1) &= 0.3 \\ p_V (2) &= 0.2 \end{align} Find $E(V −\mu) $
The textbook gives the answer $0$ but no working. Is this a standard definition?
I know $ E(V) = \mu $ but it's confusing as it is contained within the function (within the brackets) of expectation.
So you know that
$$ \mu = E[V] $$ By linearity of expectation you have $$ E[V - \mu] = E[V] - E[\mu] = \mu - \mu = 0 $$
If you want the details of the underlying computation : $$\mu = 0\cdot 0.5 + 1\cdot 0.3 + 2\cdot 0.2 = 0.7$$
Now define $Y = V-\mu = V-0.7$ hence $Y$ takes now values in $\{-0.7,0.3,1.3 \}$. The probabilities of $Y$ haven't changed. Only the value that $Y$ can take : $$ \begin{cases} P_Y(-0.7) = 0.5 \\ P_Y(0.3) = 0.3 \\ P_Y(1.3) = 0.2 \end{cases} $$ Now computing the expectation of $Y$ we have : $$ E[V - \mu] = E[Y] = -0.7 \cdot 0.5 + 0.3 \cdot 0.3 + 1.3 \cdot 0.2 = 0 $$