ELEC 321

Continous Random Vectors

Updated 2017-12-04

Continuous Random Vectors

Take \(m=2\) for instance, the corresponding joint distribution function is a multivariable function given by \(F(x_1, x_2)\).

\[\begin{align}F(x_1,x_2)&=\mathbb P(X_1\leq x_1, X_2\leq x_2)\\ &=\int_{-\infty}^{x_2} \int_{-\infty}^{x_1}f(t_1, t_2)\mathrm dt_1\mathrm dt_2 \end{align}\]

By the FTC, we obtain the joint density function:

\[f(x_1,x_2)=\frac{\partial^2}{\partial x_2\partial x_2}F(x_1, x_2)\]

Bivariate Normal (More in module 6)

The main model is given by:


Marginal Density Function

Suppose we have the joint density function, \(f(x_1, x_2)\), for random variables \(X_1, X_2\).

Similar to discrete random variables, the marginal density function of one random variable is obtained by integrating the other random variables out.

\[f_1(x_1)=\int_{-\infty}^\infty f(x_1, x_2)\mathrm dx_2\]

Conditional Density Function

The conditional densities work identical to the discrete version:

\[f(x_2\vert x_1)=\frac{f(x_1, x_2)}{f_1(x_1)}\]

Where \(f(x_1, x_2)\) is the joint density and \(f_1(x_1)\) is the marginal density.

It follows that given conditional and joint densities, we can obtain the joint density:

\[f(x_1, x_2)=f(x_2\vert x_1)f_1(x_1)=f(x_1\vert x_2)f_2(x_2)\]

Conditional Mean

Let \(\mu_{y\vert x}\) denote the mean of \(Y\) given some realization of \(X\). It is defined as

\[\mu_{y\vert x}=\mathbb E(Y\vert X=x)=\int_{-\infty}^\infty y\cdot f(y\vert x)\mathrm dy\]

Conditional Variance

The variance of \(Y\) given some \(X=x\) is defined as

\[\sigma_{y\vert x}^2=\text{Var}(Y\vert X=x)=\int_{-\infty}^{\infty}(y-\mu_{y\vert x})^2f(y\vert x)\mathrm dy\]


If the continuous random variables in the random vector are independent, then the joint density is all the marginal densities multiplied together.

\[f(x_1,x_2,\dots,x_m)=f_1(x_1)f_2(x_2)\cdots f_m(x_m)\]

Thus, any conditional density is zero.

Furthermore, given its covariance matrix, all non-diagonal elements is 0. In particular, \(\sigma_{ij}=0\) for \(i\neq j\).


Suppose \(X\sim \text{Unif}(0,10)\) and that \(Y\vert X=x \sim\text{Exp}(\frac{1}{x})\). What is the mean and variance of \(Y\)? What fractional of the total variance is explained by \(X\)?

Recall exponential random variables: the expected value is \(1/\lambda\) and variance is \(1/\lambda^2\). Since \(Y\vert X=x\sim\text{Exp} (\frac{1}{x})\), then the expected value of \(Y\vert X\), an exponential random variable is

\[\mathbb E(Y\vert X)=X\]

Recall uniform random variables: the expected value is \(\frac{(a+b)}{2}\) and the variance is \(\frac{(b-a)^2}{12}\). Plugging \(a=0, b=10\), we get \(\mathbb E(X)=5\) and \(\text{Var}(X)=\frac{100}{12}\).

Putting everything together, we get

\[\mathbb E(Y)=\mathbb\{\mathbb E(Y\vert X\}=\mathbb E\{X\}=5\\\]

We compute the total variance:

\[\begin{align} \text{Var}(Y)&=\mathbb E\{\text{Var}(Y\vert X)\}+\text{Var}\{\mathbb E(Y\vert X)\}\\ &=\mathbb E\{X^2\}+\text{Var}\{X\}\\ &=\text{Var}(X)+[\mathbb E(X)]^2+\text{Var}(X)\\ &=\frac{200}{12}+25\\ &=41.7 \end{align}\]

To compute the percentage of explained variance, we look at the explained variance, which is \(\text{Var}(\mathbb E(Y\vert X))\), and divide it by the total variance, which gives

\[\frac{\text{Var}\{\mathbb E(Y\vert X)\}}{\text{Var}(Y)}=\frac{100/12}{41.7}=0.20\]

Functions of Continuous Random Vectors

Suppose we have a function \(\mathbf h\) that is bijective, then for some random vectors \(\mathbf x\) and \(\mathbf y\),

\[\mathbf y=\mathbf h(\mathbf x)\\ \mathbf x=\mathbf h^{-1}(\mathbf y)\]

If given the density \(f_{\mathbf X}\), then the density for \(\mathbf y\) is

\[f_{\mathbf Y(\mathbf y)}=f_{\mathbf X}(\mathbf h^{-1}(\mathbf y))\cdot \underbrace{\left\vert \det\left(\frac{\partial x_i}{\partial y_j}\right)\right\vert }_{J(\mathbf y)}\]

Where \(J(\mathbf y)\) is the Jacobian which is used for transformations.

Linear Transformations

Suppose we have random vector \(\mathbf Y=\mathbf A\mathbf X + \mathbf b\), where \(\mathbf Y\) and \(\mathbf X\) is a \(n\)-dimensional random vector, \(\mathbf A\) is an \(n\times n\) matrix, and \(\mathbf b\) is an \(n\)-dimensional vector.

Then the expected value and covariance is transformed as follows:

\[\boldsymbol\mu_{\mathbf Y}=\mathbb E\{\mathbf Y\}=\boxed{\mathbf{A}\boldsymbol \mu_{\mathbf X}+\mathbf b}\\ \text{Cov}(\mathbf Y)=\boxed{\mathbf A \text{Cov}(\mathbf X) \mathbf A^T}\]