Multivariate Distributions

Let X = (X_1 \ldots X_n)^T be an n-dimensional vector of random variables.
For all x = (x_1, \ldots, x_n) \in \mathbb{R}^n, the joint cumulative distribution function of X satisfies
F_{X_i}(x_i) = F_X (\infty, \ldots, \infty, x_i, \infty, \ldots, \infty)

Clearly it is straightforward to generalise the previous definition to join marginal distributions. For example, the join marginal distribution of X_i and X_j satisfies
F_X(x_1, \ldots, x_n) = \int_{-\infty}^{x_1} \ldots \int_{-\infty}^{x_n} f_x (u_1, \ldots, u_n) du_1 \ldots du_n

If X_1 = (X_1 \ldots X_k)^T and X_2 = (X_{k+1} \ldots X_n)^T is a partition of X then the conditional CDF of X_2 given X_1 satisfies
F_{X_2|X_1} (X_2|X_1) = P(X_2 \le x_2 | X_1 = x_1).  If X has a PDF,latex f_X (\bullet), then the conditional PDF oflatex X_2givenlatex X_1satisfieslatex f_{X_2 | X_1} (X_2 | X_1) = \frac{f_X (X)}{f_{X_1}(X_1)} = \frac{f_{X_1 | x_2}(X_1 | X_2) f_{X_2}(X_2)}{f_{X_1}(X_1)}and the conditional CDF is then given bylatex F_{X_2 | X_1}(X_2 |X_1) = \int_{- \infty}^{x_{k+1}} \ldots \int_{- \infty}^{x_n} \frac{f_X (x_1, \ldots , x_k, u_{k+1}, \ldots , u_n)}{f_{X_1}(X_1)} du_{k+1} \dots du_nwherelatex f_{X_1}(\bullet)is the joint marginal PDF oflatex X_1which is given bylatex f_{X_1} (x_1, \ldots , x_k) = \int_{- \infty}^{\infty} \ldots \int_{- \infty}^{\infty} f_X (x_1, \ldots , x_K u_{k+1}, \ldots , u_n) du_{k+1} \ldots du_nWe next look at independence, which is something H2 Mathematics can easily relate to.   Here, we say the collection X is independent if joint CDF can be factored into the product of the marginal CDFs so thatlatex F_X (x_1 \ldots , x_n) = F_{X_1} (x_1) \ldots F_{X_n}(x_n)If X has a PDF,latex f_X(\bullet)then independence implies that the PDF also factories into the product of marginal PDFs so thatLatex f_X(x) = f_{X_1} (x_1) \ldots f_{X_n}(x_n).  Using the above results, we have that iflatex X_1andlatex X_2are independent thenlatex f_{x_2|x_1}(x_2 | x_1) = \frac{f_X (X)}{f_{X_1}(X_1)} = \frac{f_{X_1}(X_1) f_{X_2}(X_2)}{f_{X_1}(X_1)} = f_{X_2}(X_2)The above results tell us that having information aboutlatex X_1tells us nothing aboutlatex X_2.  Let's continue to look further on the implications of independence.   Let X and Y be independent random variables. Then for any events A and B,latex P(X \in A, Y \in B) = P(X \in A) P(Y \in B)We can check this with,latex P(X \in A, Y \in B)latex = \mathbb{E}[1_{X \in A} 1_{Y \in B}]latex = P(X \in A) P(Y \in B)In general, iflatex X_1, \ldots, X_nare independent random variables thenlatex \mathbb{E}[f_1 (X_1) f_2(X_2) \ldots f_n(X_n)] = \mathbb{E}[f_1(X_1)] \mathbb{E}[f_2(X_2)] \ldots \mathbb{E}[f_n(X_n)]Moreover, random variables can also be conditionally independent. For example, we say that X and Y are conditionally independent given Z iflatex \mathbb{E}[f(X)g(Y)|Z] = \mathbb{E}[f(X)|Z]\mathbb{E}[g(Y)|Z]. The above will be used in the Gaussian copula model for pricing of collateralised debt obligation (CDO).  [caption id="attachment_2829" align="alignnone" width="300"]<a href="" rel="attachment wp-att-2829"><img src="" alt="Source:" width="300" height="168" class="size-medium wp-image-2829" /></a> Source:[/caption]  We letlatex D_ibe the event that thelatex i^{th}bond in a portfolio defaults. Not reasonable to assume that thelatex D_i's are independent though they are conditionally independent given Z solatex P(D_1, \ldots, D_n |Z ) = P(D_1|Z) \ldots P(D_n|Z)is often easy to compute.  Lastly, we consider the mean and covariance. The mean vector of X is given byLatex \mathbb{E}[X]:=(\mathbb{E}[X_1] \ldots \mathbb{E}[X_n])^Tand the covariance matrix of X satisfieslatex \sum := \mathrm{Cov}(X) := \mathbb{E}[(X- \mathbb{E}[X])(X – \mathbb{E}[X])^T]so thelatex (i,j)^{th}element oflatex \sumis simply the covariance ofLatex X_iandlatex X_j. The covariance matrix is symmetric and its diagonal elements satisfylatex \sum_{i, i} \ge 0and is also positive semi definite so thatlatex x^T \sum x \ge 0for alllatex x \in \mathbb{R}^nThen the correlation matrixlatex \rho (X)haslatex (i, j)^{th}elementlatex \rho_{ij} := \mathrm{Corr}(X_i, X_j). This is also symmetric, postiie semi-definite and has 1's along the diagonal.  For any matrixlatex A \in \mathbb{R}^{k \times n}and vectorlatex a \in \mathbb{R}^k,Latex \mathbb{E}[AX +a] = A \mathbb{E} [X] + a(distributes linearly)latex \mathrm{Cov}(AX +a) = A \mathrm{Cov}(X) A^Tlatex \Rightarrow \mathrm{Var} (aX + bY) = a^2 \mathrm{Var}(X) + b^2 \mathrm{Var}(Y) + 2ab \mathrm{Cov}(X, Y)Recall that if X and Y are independent, thenlatex \mathrm{Cov}(X, Y)=0$, and the converse is not true in general.

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