Bivariate Normal Distribution and sub vector

Question

Let $$(X,Y)\sim N(\mu^\rightarrow ,\Sigma)$$ we cannot assume anything about the dependancy between X and Y. Can we assume the following? $$X\sim N(\mu_x,\sigma ^2_x)~,~Y\sim N(\mu_y, \sigma^2_y)$$note:(X,Y) is not necessarily standard bivariate normal distributed.

Answer 1

Yes! They are both marginally normally distributed, just, as you say, you do not know the relation between the two.

The density funciton of $N(\mu, \Sigma)$ can be found at this Wikipedia link. The specific case where you have a vector of length $2$, ie your $(X,Y)$, rather than a general $(X_1, ..., X_n)$ is given its own subsection. If you take this density function, say $f(x,y)$, and integrate over all $Y = y$, then you get the density function of the marginal distribution for $X$: $f_X(x) = \int_{\mathbb R} f(x,y) dy$.

If you do this integral yourself, you'll find that $f_X(x)$ is precisely the density function of a $N(\mu_x, \sigma_x^2)$ distribution! The same argument holds, analogously, for $f_Y(y) = \int_{\mathbb R} f(x,y) dx$.

So, the diagonal terms in $\Sigma$ describe the variance of the marginals of $X$ and $Y$ and the relationship between $X$ and $Y$ is described by the off-diagonal terms.

You can also prove this by writing $(X,Y)$ as a matrix times a vector of independent standard normals, plus a mean shift. See this Wikipedia link for details of this representation.