Bivariate normal exercise - check please

Question

I am trying to self learn some probability and wanted to ensure I was getting these exercises correct.

If $X \sim N(20,2^2)$ and $Y \sim N(10,1)$ and $X$ and $Y$ are independent,

then find:

$a)$ the distribution vector $(X,Y)$

$b)$ give the distribution of $U = 2X-3Y$ and specify the mean and variance

For $a) $ $\mu = (20,10) $ and $\Sigma = \begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix} $

and $b)$ $E[U] = E[2X-3Y] = 2E[X] - 3E[Y]$

$=2(20) - 3(10) = 10$

Would this be denoted as $\mu_{XY} = 10$ ?

Also:

$VAR(2X-3Y) = VAR(2X) + VAR(-3Y) = 4VAR(X) + 9VAR(Y) = 4(4) - 9(1) = 25$

So

$U \sim N(10,25)$

Are my solutions correct?

Answer 1

Well you are also probably asked to say which are the distributions of the random variables of points a) and b). The distribution is not generally uniquely determined by its moments but is uniquely determined by its cdf (cumulative distribution function) or pdf (probability density function) or characteristic function. (Although in this case the distribution of (X, Y) is bivariate normal and the distribution of U is again normal, due to the properties of the normal distribution)