Is an affine transformation of a symmetric random variable also symmetric?

Question

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!

Answer 1

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.

Answer 2

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.