I'm finding the probability density function of the random variable U defined in the following manner:
$$U=\frac{1}{2}(Y_1+3Y_2)$$
CORRECTION: The line above is supposed to be $U=\frac{1}{2}(Y_1-3Y_2)$
Where
$$Y_1 \sim N(\mu,\sigma^2)\\ Y_2 \sim N(\mu,\sigma^2)$$
*$Y_1$ and $Y_2$ are independent and identically-distributed random variables
Am I justified in saying that $U \sim N(-\mu,\frac{5}{2}\sigma^2)$ based on this reasoning?
I defined
$$\mathbf{b}:=[\frac{1}{2},-\frac{3}{2}]$$ $$\mathbf{Y}:=[Y_1,Y_2]$$ $$\mathbf{U}:=\mathbf{bY}=\frac{1}{2}Y_1-\frac{3}{2}Y_2$$
So it seems that
$$E[\mathbf{U}]=E[\mathbf{bY}]=\mathbf{b}E[\mathbf{Y}]=[\frac{1}{2},-\frac{3}{2}]\cdot[\mu,\mu]^\top=\frac{1}{2}\mu-\frac{3}{2}\mu=-\mu$$
and
$$V[\mathbf{U}]=V[\mathbf{bY}]=\mathbf{b}^{2}V[\mathbf{Y}]=[(\frac{1}{2})^{2},(-\frac{3}{2})^{2}]\cdot\left[\begin{array}{cc} \sigma^{2} & 0\\ 0 & \sigma^{2} \end{array}\right]=\frac{1}{4}\sigma^{2}+\frac{9}{4}\sigma^{2}=\frac{5}{2}\sigma^{2}$$
We are independent random variables: $X_1, X_2, \ldots , X_n$ with probability distribution $N(\mu_i,\sigma_i^2)$, i = 1 ~ n.
Then the linear combination $\sum_1^na_iX_i$ has a normal probability distribution
$N(\sum_1^n a_i\mu_i,\sum_1^n a_i^2\mu_i^2)$
So I would say that the random variable $U=\frac{1}{2}(Y_1-3Y_2)$ with $N(\mu, \sigma^2)$
will have a normal probability distribution $N(-\mu,\frac{5}{2}\sigma^2)$.