Find the marginal distribution given the mean and the covariance matrix

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.

• If portfolio with equal weights means what I think it means, you are asked to calculate the distribution of $(x_1 + x_2)/2$. A linear combination of jointly normal components is normal with $\mu = (1/2, 1/2) \cdot \boldsymbol \mu^t = 12, \,\sigma^2 = (1/2, 1/2) S_1 (1/2, 1/2)^t = 25/2$. – Maxim Oct 1 '18 at 22:55

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?

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.

• Great explanation, thanks. Do you have some thoughts on the second part of the question? What does it mean 'equal weight'? – Whizkid95 Sep 28 '18 at 6:32
• Any thoughts please? – Whizkid95 Oct 1 '18 at 9:47