Expected value of random variables that are generated by substraction from the mean If suppose we have $ X_1,X_2,\ldots,X_N$ that are independent Normal random variables with mean $\mu$ and variance $\sigma^2$, $X\sim N(\mu,\sigma^2)$.
And if we have $Y=\mu-X,$ then is the mean of $Y$ equal to $0$, $Y\sim N(0,\sigma^2)$?
If so, how do we derive that?
[EDIT]
I should have realized earlier but now I do. Using the same approach, I can find the mean and variance of $Y$.
 A: The result of any linear operation on a RV of normal distribution is a normal RV. That is, if $Y = aX + b$, where $a$ and $b$ are constants, and $X \sim N(μ,σ^2)$, then $Y \sim N(b+aμ,(aσ)^2)$. In your case, $a = -1$ and $b = μ$.
If you really want to prove it, you may note that $P(Y≤y) = P(X≥μ-y)$, and write out the integral expression of $P(X≥μ-y)$, then write out the expression for its derivative respect to $y$ to obtain the pdf of $Y$.
A: 
Edit: It seems I'm wrong about the variance of $Y$, but I can't see why.
I'll leave this here out of interest to others.

It's clear enough that the expectation (mean) of $Y$ is $0$,
since $E(Y) = E(\mu) - E(X) = E(\mu) - E(\mu) = 0$, where $E(.)$ is the expectation.
Since $\text{var}(a - b) = \text{var}(a) + \text{var}(b)$
if $a$ and $b$ aren't correlated
(where $\text{var}(.)$ is the variance),
the variance of $Y$ is
$$
\begin{align}
\text{var}(Y) 
&= \text{var}(\mu) + \text{var}(X)\\
&= \frac{\sigma^2}{n^2} + \sigma^2
\end{align}
$$
In other words, the variance of $Y$ is the variance of $X$,
plus the variance in the estimate of the $\mu$.
Putting this together, $Y \sim N(0, \frac{\sigma^2}{n^2} + \sigma^2)$.

Edit: This is what I thought should be the case,
but simulations in R seem to show that I'm wrong, and var$(Y) = \sigma^2$.

f = function(i){
  X = rnorm(10, 0, sqrt(25))
  mu = mean(X)
  Y = X - mu
  data.frame(mu = mu, ey = mean(Y), vy=var(Y))
}
df = purrr::map_df(1:5000, f)
# mu and E(y) are uncorrelated
plot(df$mu, df$ey)
cor(df$mu, df$ey) # ≈ 0

mean(df$vy) #≈ 25


