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 of $X_2$ given $X_1$ satisfies
$f_{X_2 | X_1} (X_2 | X_1) = \frac{f_X (X)}{f_{X_1}(X_1)} = \frac{f_{X_1 | x_2}(X_1 | X_2) f_{X_2}(X_2)}{f_{X_1}(X_1)}$

and the conditional CDF is then given by
$F_{X_2 | X_1}(X_2 |X_1) = \int_{- \infty}^{x_{k+1}} \ldots \int_{- \infty}^{x_n} \frac{f_X (x_1, \ldots , x_k, u_{k+1}, \ldots , u_n)}{f_{X_1}(X_1)} du_{k+1} \dots du_n$
where $f_{X_1}(\bullet)$ is the joint marginal PDF of $X_1$ which is given by
$f_{X_1} (x_1, \ldots , x_k) = \int_{- \infty}^{\infty} \ldots \int_{- \infty}^{\infty} f_X (x_1, \ldots , x_K u_{k+1}, \ldots , u_n) du_{k+1} \ldots du_n$

We 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 that
$F_X (x_1 \ldots , x_n) = F_{X_1} (x_1) \ldots F_{X_n}(x_n)$

If X has a PDF, $f_X(\bullet)$ then independence implies that the PDF also factories into the product of marginal PDFs so that
$f_X(x) = f_{X_1} (x_1) \ldots f_{X_n}(x_n)$.

Using the above results, we have that if $X_1$ and $X_2$ are independent then
$f_{x_2|x_1}(x_2 | x_1) = \frac{f_X (X)}{f_{X_1}(X_1)} = \frac{f_{X_1}(X_1) f_{X_2}(X_2)}{f_{X_1}(X_1)} = f_{X_2}(X_2)$
The above results tell us that having information about $X_1$ tells us nothing about $X_2$.

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, $P(X \in A, Y \in B) = P(X \in A) P(Y \in B)$

We can check this with,
$P(X \in A, Y \in B)$
$= \mathbb{E}[1_{X \in A} 1_{Y \in B}]$
$= P(X \in A) P(Y \in B)$

In general, if $X_1, \ldots, X_n$ are independent random variables then
$\mathbb{E}[f_1 (X_1) f_2(X_2) \ldots f_n(X_n)] = \mathbb{E}[f_1(X_1)] \mathbb{E}[f_2(X_2)] \ldots \mathbb{E}[f_n(X_n)]$

Moreover, random variables can also be conditionally independent. For example, we say that X and Y are conditionally independent given Z if
$\mathbb{E}[f(X)g(Y)|Z] = \mathbb{E}[f(X)|Z]\mathbb{E}[g(Y)|Z]$.
The above will be used in the Gaussian copula model for pricing of collateralised debt obligation (CDO).

We let $D_i$ be the event that the $i^{th}$ bond in a portfolio defaults. Not reasonable to assume that the $D_i$‘s are independent though they are conditionally independent given Z so $P(D_1, \ldots, D_n |Z ) = P(D_1|Z) \ldots P(D_n|Z)$ is often easy to compute.

Lastly, we consider the mean and covariance.
The mean vector of X is given by $\mathbb{E}[X]:=(\mathbb{E}[X_1] \ldots \mathbb{E}[X_n])^T$
and the covariance matrix of X satisfies
$\sum := \mathrm{Cov}(X) := \mathbb{E}[(X- \mathbb{E}[X])(X - \mathbb{E}[X])^T]$ so the $(i,j)^{th}$ element of $\sum$ is simply the covariance of $X_i$ and $X_j$.
The covariance matrix is symmetric and its diagonal elements satisfy $\sum_{i, i} \ge 0$ and is also positive semi definite so that $x^T \sum x \ge 0$ for all $x \in \mathbb{R}^n$
Then the correlation matrix $\rho (X)$ has $(i, j)^{th}$ element $\rho_{ij} := \mathrm{Corr}(X_i, X_j)$. This is also symmetric, postiie semi-definite and has 1’s along the diagonal.

For any matrix $A \in \mathbb{R}^{k \times n}$ and vector $a \in \mathbb{R}^k$,
$\mathbb{E}[AX +a] = A \mathbb{E} [X] + a$ (distributes linearly)
$\mathrm{Cov}(AX +a) = A \mathrm{Cov}(X) A^T$
$\Rightarrow \mathrm{Var} (aX + bY) = a^2 \mathrm{Var}(X) + b^2 \mathrm{Var}(Y) + 2ab \mathrm{Cov}(X, Y)$
Recall that if X and Y are independent, then $\mathrm{Cov}(X, Y)=0$, and the converse is not true in general.

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