Finding the pdf from mean vector and covariance matrix

Question

Let X denote a Gaussian column vector with mean vector mx=[2,3]T,and co-variance matrix  
Cx=[1 0,0 1 ].the random vector Y=AX, where A=[ -1  -2,2  3].

a) Determine E[Y]

Y=AX
E(Y)=A*E(X)= [ -1  -2,2  3]*[2,3]= [-8 ]
                                     13

b) Find and simplify the co-variance matrix of Y

cov[Y,Y]=E[(y-my)(Y-my)T]=E[(A*X-my)(A*X-my)T]
        =A*E[(X-mx)(X-mx)T]AT
        =[-1 -2,2 3][1 0,0 1][-1 2,-2 3]
        =[5   -8,-8  13]

c) Completely specify the probability density function of vector Y

how to answer the part c. T is the transpose. I have calculated the expected value of Y and co-variance matrix of Y.

Answer 1

You know that a linear transformation of a gaussian vector, is gaussian. The pdf of a $n-$gaussian vector of mean $μ$ and covariance matrix $Σ$ is:

$$ f(\Bbb{x})= \frac{1}{\sqrt{ (2π)^n \det(Σ)}} \exp\left( {-\dfrac12 ({x-\mu})^T\ \Sigma^{-1}\ {(x-μ)} } \right)$$

So you need to invert the matrix $Σ$ and you can substitute.