I am doing some practice problems to prepare for my statistics exam, and I just want to know if my reasoning is correct on one problem, and if not, I want to know how I should reason through this. The question is as follows:

Let X and Y follow a bivariate normal distribution with means (3, 2), variances
(1, 4) and covariance c.

$\begin{bmatrix}X\\Y\end{bmatrix} = N(\begin{bmatrix}3\\2\end{bmatrix}, \begin{bmatrix}1&c\\c&4\end{bmatrix})$

What is the Marginal Distribution of X?

My initial reasoning is that one method I could go about it is the calculation of the joint PDF and integrating it, but that doesn't sound very clean. A cleaner method could possibly be that since the joint is determined by two normal distributions, then $f(X,Y) = f(X)f(Y)$, and $f_x(X)$ (the marginal distribution of X) equals $\int_{-\infty}^{\infty}f(X)f(Y)dY = f(X)\int_{-\infty}^{\infty}f(Y)dY = f(X)$ since $f(Y)$ is a normal distribution, with the final answer being $f(X) = N(\mu_1, \sigma_1^2) = N(3,1)$. This seems to make sense but seems a little bit too simplified. Could anyone give me insight on how to tackle this problem, and if I am wrong anywhere, point it out? Thank you.


1 Answer 1


Bivariate is a special case of jointly normality for 2D, which means these variables are also marginally normal. So, above $X$ and $Y$ are normal RVs. From the mean vector and the covariance matrix, we know that $\mu_X=3$ and $\sigma^2_X=1$, so $X\sim N(3,1)$. Similarly $Y\sim N(2,4)$. So, your final answer is correct.

But, you cannot say $f(x,y)=f(x)f(y)$ when $c\neq 0$, since it implies independence. Check the bivariate distribution formula here.

And, it is always true that $\int_y{f(x,y)dy}=f(x)$, not only when you factorize the joint PDF out, i.e. like $f(x,y)=f(x)f(y)$.


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