Assume that $Z_{1}$ and $Z_{2}$ are independent standard normally distributed random variables. Show that $(X,Y)$ has bivariate normal distribution when $X = Z_{1}$ and $Y = Z_{1} + Z_{2}$.
MY SOLUTION
To begin with, notice that $Z_{1} = X$ and $Z_{2} = Y - X$. According to the change of variables theorem, we have \begin{align*} f_{X,Y}(x,y) & = f_{Z_{1},Z_{2}}(x,y-x)|\det J(x,y)| = f_{Z_{1},Z_{2}}(x,y-x) = f_{Z_{1}}(x)f_{Z_{2}}(y-x)\\\\ & = \frac{1}{2\pi}\exp\left(-\frac{x^{2}}{2}\right)\exp\left(-\frac{(x-y)^{2}}{2}\right) = \frac{1}{2\pi}\exp\left(-x^{2} + xy - \frac{y^{2}}{2}\right) \end{align*}
But I do not know how check whether this is the correct result or not. Can somebody help me out? Thanks in advance!
You are almost there. Now you must recall that a multivariate variable ${\bf X}$ has a normal distribution if and only if its density can be written in the form
$$ \alpha \exp{\left(- \frac{1}{2}(X- \mu)' C (X- \mu)\right)}$$ where $\mu = E(X)$ and $C= \Sigma^{-1}$ is a positive definite matrix.
You must show that in our case $\mu = (0,0)'$ , and then find $C$.