If we have a vector of normally distributed random variables $x^T = (x_1,x_2)$ with mean $\mathbf{\mu}^T = (10, 14)$ and a covariance matrix $$S_1 =\begin{bmatrix}13 & 12\\12 & 13\end{bmatrix}.$$
I would first like to calculate the marginal distribution
My Thoughts
The random variables are normally distributed, so the distributions are simply:
$x_1 \sim N(10,13)$ and $x_2 \sim N(14, 13)$? Is this correct? It seems to simple, don't I need to use the covariance?
Suppose now that we assume that $x_1$ and $x_2$ are returns to financial assets, we need to calculate the distribution of the portfolio that gives equal weight to both assets
My thoughts
I'm not quite sure what to do here? Any thoughts? Do they mean that for every $x_2$ we need $\frac{14}{10}$ assets of $x_2$ to get the same amount of money? Any help is must appreciated.
Best Answer
That is correct. You could stop here and say if we marginalize over $x_2$, we get $x_1 \sim N(10,13)$. Similarly, if we marginalize over $x_1$, we get $x_2 \sim N(14,13)$
The following tells you why you're allowed to say that.
Marginalization
Let $\mu_k,\sigma_k^2$ be the mean and variances of $x_k$ where $k = \lbrace 1,2\rbrace$. Also let $\rho = 12$ be the coefficient of correlation.
When you want to marginalize, you have to get the joint distribution $f(x_1,x_2)$, which is given to us by the bivariate normal distribution, \begin{equation} f_{X_1,X_2}(x_1,x_2) = \frac{1}{2\pi \sigma_1 \sigma_2 \sqrt{1-\rho^2}} \exp\Big( - \frac{1}{2} Q(x_1,x_2) \Big) \end{equation} where \begin{equation} Q(x_1,x_2) = \frac{1}{1-\rho^2} [\frac{(x_1 - \mu_1)^2}{\sigma_1^2}+\frac{(x_2 - \mu_1)^2}{\sigma_2^2} - \frac{2\rho(x_1 - \mu_1)(x_2 - \mu_2)}{\sigma_1\sigma_2}] \end{equation} which could be written as \begin{equation} Q(x_1,x_2) = \frac{1}{1-\rho^2} \Big( (\frac{x_1 - \mu_1}{\sigma_1} - \rho \frac{x_2 - \mu_2}{\sigma_2} )^2 + (1-\rho^2) (\frac{x_2 - \mu_2}{\sigma_2})^2 \Big) \end{equation} and even more \begin{equation} Q(x_1,x_2) = \frac{(x_1 - a(x_2))^2}{(1-\rho^2)\sigma_1^2} + \frac{(x_2 - \mu_2)^2}{\sigma_2^2} \end{equation} where $a(x_2) = \mu_1 + \rho\frac{\sigma_1}{\sigma_2}(x_2 - \mu_2)$
Then integrate with respect to one variable, \begin{equation} f(x_2) = \int_{-\infty}^{\infty} f(x_1,x_2) dx_1 = K \exp(- \frac{(x_2 - \mu_2)^2}{2\sigma_2^2} ) \int_{-\infty}^{\infty} \exp(\frac{(x_1 - a(x_2))^2}{(1-\rho^2)\sigma_1^2}) \ dx_1 \end{equation} But $ \int_{-\infty}^{\infty} \exp(\frac{(x_1 - a(x_2))^2}{(1-\rho^2)\sigma_1^2}) \ dx_1 = \sqrt{2\pi} \sqrt{1- \rho^2}\sigma_1$ so \begin{equation} f(x_2) = \frac{1}{\sqrt{2\pi} \sigma_2} \exp(- \frac{(x_2 - \mu_2)^2}{2\sigma_2^2} ) \end{equation} So \begin{equation} X_2 \sim N(\mu_2, \sigma_2^2) \end{equation} In the same way, you would also get \begin{equation} X_1 \sim N(\mu_1, \sigma_1^2) \end{equation} So yes marginalizing a bivariate normal distribution is also a normal distribution.