Can I assume the joint density of two normals to be bivariate normal? I am looking for the joint distribution of two normal, could I assume it is bivariate?  If no, what should I do?  
$X_{1}$~$N(0,1)$
$X_{2}$~$N(1,2)$
cov($X_{1}, X_{2})=1$
How do I find the joint density? And $P(X_{1}<X_{2})?$
 A: The joint distribution of two normals is not necessarily normal; see, e.g:
http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Two_normally_distributed_random_variables_need_not_be_jointly_bivariate_normal
And here:
https://stats.stackexchange.com/questions/33354/when-are-two-normally-distributed-random-variables-jointly-bivariate-normal
for a more full discussion.
For $P(X_1<X_2)$, you can rewrite as $P(X_1-X_2 <0)$. For this, you can find the distribution of $-X_2$ , by composing $X_2$ with a function, and then find the
distribution of $X_1-X_2$ by using the convolution of $X_1$ and $-X_2$.
You can see here:( in the "Functions of Random Variables" section) http://en.wikipedia.org/wiki/Random_variables
How to get the distribution of $-X_2$ from that of $X_2$.
The idea is this: if $g(x)$ is an invertible ( or locally-invertible, but f(x)=-x is globally-invertible) , then we use that:
$P(g(X)\leq y)=P(X \leq g^{-1}(y))$. In our case, $g(x)=-x=g^{-1}(x)$ , so that 
$P(-X_2>y)=P(X_2 \leq -y)$
Looking at this geometrically, notice this just changes the region of integration.
EXAMPLE:
Say you're in $\mathbb R $, with a cdf $P_{X_2}$ for $X_2$. Then $P(X_2>x)$ is given by
the integral from $x_2$ to $\infty$ , but $P( -X_2>x)=P(X_2\leq -x)$ is the integral 
from $-x$ to -$\infty$ . The regions of integration are complementary, in that their
union is the entire space.
After knowing the distribution of $-X_2$, you can find the distribution of $X_1-X_2$ by doing the convolution of $X_1$ with $-X_2$.
