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!

By the change of variables formula, the pdf of $Y$ is given by $$f_Y(y) = f_X\left(T^{-1}(y)\right)\left\vert \frac{\mathrm d}{\mathrm dy}T^{-1}(y)\right\vert,$$
where $T(x) = a + bx$. Thus, $T^{-1}(y) = \frac{y - a}{b}$ and $\frac{\mathrm d}{\mathrm dy}T^{-1}(y) = \frac 1b$. So we have that $$f_Y(y) = f_X\left(\frac {y-a}b\right)\frac{1}{\vert b\vert}.$$ Let $\mu$ and $\nu$ denote the means of $X$ and $Y$, respectively. Note that $\nu = a + b\mu$. Now does it hold that $f_Y(\nu + y) = f_Y(\nu - y)$ for all $y\in\mathbb R$?
It holds that $$f_Y(\nu + y) = \frac{1}{\vert b\vert}f_X\left(\frac{\nu + y - a}{b}\right) = \frac{1}{\vert b\vert}f_X\left(\frac{a+b\mu - y -a}{b}\right) = \frac{1}{\vert b\vert}f_X\left(\mu - \frac yb\right)$$ and $$f_Y(\nu - y) = \frac{1}{\vert b\vert}f_X\left(\frac{\nu - y - a}{b}\right) = \frac{1}{\vert b\vert}f_X\left(\frac{a+b\mu - y -a}{b}\right) = \frac{1}{\vert b\vert}f_X\left(\mu + \frac yb\right).$$ For every $y\in\mathbb R$, let $x = \frac yb$. Then, since by assumption, $f_X(\mu + x) = f_X(\mu-x)$ for all $x\in\mathbb R$, we have that the distribution of $Y$ is symmetric.
Let $\sigma^2$ denote the variance of $X$. Then $b^2\sigma^2$ is the variance of $Y$. The covariance between $X$ and $Y$ is given by $b\sigma^2$, which can be verified easily by applying the usual laws for variance and covariance. Thus, the correlation between $X$ and $Y$ is given by $$\frac{b\sigma^2}{\sqrt{\sigma^2 b^2\sigma^2}} = \pm 1.$$
Note that since $X$ and $Y$ are perfect linear combinations, it's already clear that both variables are perfectly correlated.
