How is a Gaussian random process different from a Gaussian random variable?

I am from a physics background with lousy mathematical training. I am trying to delve into the subject of stochastic processes as applied to physics. I have a pretty naive question.

A random variable $$x$$ ($$-\infty) for which the normalized probability density function is $$p(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\Big(-\frac{(x-\mu)^2}{2\sigma^2}\Big)$$ with $$\mu$$ being any real number and $$\sigma$$ being a positive number, is called a Gaussian random variable (GRV).

Question $$1$$ How is a GRV different from and related to a continuous Gaussian random process $$x(t)$$ where $$t$$ is a continuous parameter (e.g., time)? Should I think of Gaussian random process $$x(t)$$ as a non-denumerably infinite number of GRVs corresponding to each value of $$t$$?

Question $$2$$ A multivariate Gaussian distribution is of a set of $$n$$ GRVs ($${\bf x}=(x_1,x_2,...x_n)$$) is given by the normalized probability density function $$p({\bf x})=\frac{1}{\sqrt{(2\pi)^n\det({\boldsymbol \sigma})}}\exp\Big[-\frac{1}{2}({\bf x}-{\boldsymbol \mu})^T{\boldsymbol\sigma}^{-1}({\bf x}-{\boldsymbol\mu})\Big]$$ where $${\boldsymbol\mu}$$ is a column vector of the mean values i.e. $$\mu_i=\langle x_i\rangle$$ and $${\boldsymbol\sigma}$$ is a symmetric nonsingular matrix with nonnegative real elements, called the covariance matrix. If the answer to $$(1)$$ is 'yes', how is a multivariate Gaussian different from Gaussian random process if we can identity $$x_1$$ as $$x(t_1)$$, $$x_2$$ as $$x(t_2)$$... etc?

• Sorry, the downvoter was also me because these definitions are basic enough that it's hard to believe the misunderstanding cannot be cleared up by e.g. reading wikipedia or something
– T_M
Aug 6, 2020 at 14:41
• It's fine. No problem. I think I'm confusing multivariate Gaussian distribution with a Gaussian random process.
– SRS
Aug 6, 2020 at 14:43

Should I think of Gaussian random process 𝑥(𝑡) as a non-denumerably infinite number of GRVs corresponding to each value of 𝑡?

Yes, but there is a little more to it than just that.

You need the property that any finite set of two or more random variables from a Gaussian process enjoys a multivariate Gaussian distribution. As T_M points out, it cannot said that a Gaussian random process is the same as a multivariate Gaussian distribution; they are different. However, the requirement about multivariate Gaussianity is very important. Note that a finite collection of (two or more) Gaussian random variables need not necessarily have a multivariate Gaussian distribution. For example, if $$\phi(x)$$ is the standard Gaussian density function, then $$f_{X,Y}(x,y) = \begin{cases} 2\phi(x)\phi(y), & \text{if}~ x \geq 0, y \geq 0, ~\text{or if}~ x < 0, y < 0,\\ 0,& \text{otherwise}.\end{cases}$$ is a joint density function of two standard Gaussian random variables $$X$$ and $$Y$$ but is not a bivariate Gaussian density function. Bivariate Gaussian densities are nonzero everywhere in the plane, but the density above has value $$0$$ in the second and fourth quadrants.

In short, you need to add to your thinking the idea that "a Gaussian process is a non-denumerably infinite number of GRVs" the additional requirement that every finite subset enjoys a multivariate Gaussian distribution.

Should I think of Gaussian random process 𝑥(𝑡) as a non-denumerably infinite number of GRVs corresponding to each value of 𝑡?

Yes

In other words, is a Gaussian random process same as a multivariate Gaussian distribution?

A multivariate Gaussian is not the same as a non-denumerably infinite number of GRVs, so this is not a rephrasing of the previous question. The answer here is 'no'.

• I have tried to add few more details of my current understanding. Am I still talking gibberish?
– SRS
Aug 6, 2020 at 15:04