Is an affine transformation of a symmetric random variable also symmetric? If the random variable X has a probability distribution function that is symmetric around the mean, does the random variable Y=a+bX has a symmetric probability distribution?
Also, what is the covariance and the correlation between X and Y?
Thanks!

The answer from @Imaosome is already good, but I would like complement the first part and provide an answer that does not rely on the change of variables formula.
We say that $X$ is symmetric in around $\mu$ if $\mathbb{P}(X < \mu - x ) = 1 - \mathbb{P}(X > \mu + x ), \, \forall x \in \, \mathbb{R}$. Notice that $X$ might not have a density function, e.g. can be discrete. If the density exists, it is equivalent to $f_X(\mu - x) = f_X(\mu + x), \, \forall x \in \, \mathbb{R}$.
Assume $X$ is symmetric around $\mu$ and let $Y = a + bX$. We will prove that $Y$ is symmetric around $\gamma = a+b\mu$.
\begin{align}
\mathbb{P}(Y < \gamma - z) &= \mathbb{P}(Y < a + b\mu - z)\\
&= \mathbb{P}(a+bX < a + b\mu - z) \\
&= \mathbb{P}(X < \mu - z/b) \\
&= 1 - \mathbb{P}(X > \mu + z/b) \quad\textbf{(X symmetry)}\\
&= 1 - \mathbb{P}(Y > a + b\mu + z) \\
&= 1 - \mathbb{P}(Y > \gamma + z),
\end{align}
proving the result. Notice that $\mu$ does not necessarily have to be $E[X]$. In fact, $E[X]$ might not even exist, and the result still holds.
