I have a random variable $Y \sim \mathsf{N}(2,5)$ and we define $Z = 3Y-4$. I want to find the distribution of $Z$.
Intuitively I can see that it is Normal as well due to the nature of the transformation. To show this, my first thought is to scale the variance by 3 and shift the mean by -4, giving $Z \sim \mathsf{N}(-2, 15)$.
However I am uncomfortable with this as it seems too rudimentary. Another thought of mine is to calculate the following.
$$ f_Z(x) = 3f_Y(x) – 4$$
where $f_Z$ and $f_Y$ are the pdfs. I tried subbing in $f_Y$ and manipulating the expression into the usual $\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(x-\mu)^2}{2\sigma^2}}$ but the $-4$ term is unwieldy.
Are either of these approaches correct, or are there better alternative methods?
Best Answer
$$E(3Y-4)=3E(Y)-4=2$$ $$Var(3Y-4)=9Var(Y)=45$$
So $Z$ has the distribution $ N(2,45)$