Scaling normal distribution by Y = 2X I know that when you scale a normal random variable by $a$, you get $a\mu$ and $a^2 \sigma^2$. In this example I have $y=2x$.
$f{_x}(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^-{\frac{(x-\mu)^2}{2\sigma^2}}$
and my solution is
$f{_y}(y)=\frac{1}{\sqrt{8\pi\sigma^2}}e^-{\frac{(\frac{y}{2}-2\mu)^2}{8\sigma^2}}$
is it correct?
Thanks
 A: You actually did the job twice. Let's start from the beginning.
$$f{_x}(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$
Now we substitute $x=\frac{y}2$. We also have to divide the solution by 2, because the integral of the distribution over whole real axis has to stay 1.
$$f{_y}(y)=\frac12\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{\left(\frac{y}2-\mu\right)^2}{2\sigma^2}\right)=
\frac{1}{2\sqrt{2\pi\sigma^2}}\exp\left(-\frac{\frac14\left(y-2\mu\right)^2}{2\sigma^2}\right)=
\frac{1}{\sqrt{8\pi\sigma^2}}\exp\left(-\frac{\left(y-2\mu\right)^2}{8\sigma^2}\right)
$$
This is almost the same as you wrote but the $y$ isn't divided by 2.
A: No.
First of all, it does not make sense to denote the random variable and the variable for the density function both as $x$.
If you have a random variable $X\sim N(\mu,\sigma^2)$, the density function for $X$ is
$$
f_X(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}
$$
If you let $Y=2X$, then $Y\sim N(\mu',\sigma'^2)=N(2\mu,4\sigma^2)$ and thus the density function for $Y$ is
$$
f_Y(y)=\frac{1}{\sqrt{2\pi\sigma'^2}}e^{-\frac{(y-\mu')}{2\sigma'^2}}
=\frac{1}{\sqrt{2\pi\cdot 4\sigma^2}}e^{-\frac{(y-2\mu)}{2\cdot 4\sigma^2}}
$$
