Using MGF Technique to prove a theorem Use the MGF technique to prove the theorem:

If X1, X2,...,Xn are
independent random variables where
Xi~Normal(µi, σi2), then Y=Σ aixi follows a normal distribution with parameters µ = Σaiµi and σ2= Σaiσi2

I do know how to use MGF for univariate case, but I am having a hard time understanding how to translate this to multivariate
 A: Statement $µ_y = \sum_i a_iµ_i$ is correct.
However, statement $σ_y^2= \sum_i a_iσ_i^2$ is wrong and should be
$σ_y^2= \sum_i a_i^2σ_i^2$ because $Var(aX) = a^2Var(X).$

You do not show what you have tried to do with MGFs, but I'm guessing
the incorrect statement about the variance of $Y$ may be causing you
trouble. If you are still having difficulty with MGFs, please edit
your question showing some of your work and where you're stuck, so that one of us can
help with that. [To find the MGF of the sum of two independent random variables, take the product of the two MGFs. See my Comment.]

In the example below, independent random variables $X_i$ have
$E(X_1) = 50, Var(X_1) = 5^2 = 25;$ $E(X_2) = 60, Var(X_2) = 10^2 = 100;$
$E(X_3) = 60, Var(X_3) = 3^2 = 9.$
Also $Y = 2X_1 + 3X_2 + 4X_3.$ So $E(Y) = 520, Var(Y) = 1144.$ With a million
iterations of $Y,$ the sample mean $\bar Y \approx E(Y)$ and the sample variance $S_Y^2 \approx Var(Y).$
set.seed(2021)
x1 = rnorm(10^6, 50, 5)   # mean=50, var=25
x2 = rnorm(10^6, 60, 10)  # mean+60, var=100
x3 = rnorm(10^6, 60, 3)   # mean=60, var=9
y = 2*x1 + 3*x2 + 4*x3
mean(y)
[1] 520.0287   # aprx 2(50)+3(60)+4(60) = 520
var(y)
[1]  1144.222  # aprx 4(25)+9(100)+16(9) = 1144

Because a linear combination of independent normal random variables is normal
with means and variances according to the formulas above, a histogram [blue] of the
million values of $Y$ is well approximated by the density function [orange] of
$\mathsf{Norm}(\mu_Y=520, \sigma_Y=\sqrt{1144}).$
hdr = "Simulated values of Y with Normal Density"
hist(y, prob=T, br=50, col="skyblue2", main=hdr)
curve(dnorm(x, mean(y), sd(y)), add=T, col="orange", lwd=2)


