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I understand a concept best when I see it being applied in the real world.

moonman239
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  • This is another case where a better answer might from from stats.stackexchange.com than from here. ${}\qquad{}$ – Michael Hardy Jun 11 '15 at 19:41
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    I can try to concoct one example. The GPS data at each time is a 2D random variable, which is assumed by many apps to be normally distributed. – Tunococ Jun 11 '15 at 19:41
  • I don't understand how an example from the real word can help here... How would the answer differ if you had asked for an explanation of a different probability distribution? – Mariano Suárez-Álvarez Jun 11 '15 at 19:43
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    @MarianoSuárez-Alvarez It just helps me digest the information. If information is food, a real-world example is the fiber that helps me digest the information. – moonman239 Jun 11 '15 at 19:57
  • Assuming you have two real world independent «real-world» examples of random variables $A$ and $B$ with normal distribution, then the vector $(A,B)$ is distributed with a bivariate normal distribution. This example is as «real world» as the variables $A$ and $B$ are... and, in fact, essentially every example is of this form. – Mariano Suárez-Álvarez Jun 12 '15 at 09:10
  • The «real world» is very much over-estimated as a source of enlightment, I am afraid. A probability distribution is not a concept, and is not something one can understand in any sensible meaning of the word, much as there is nothing tto understand about the sine function (apart from its definition, and the real world is not going to help with that!) – Mariano Suárez-Álvarez Jun 12 '15 at 09:12

2 Answers2

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I'll give you an example from Data Communication theory.

The input/output relations of a narrowband, single-user MIMO link (multi-input multi-output) can be represented by the complex baseband vector notation $\mathbf{y} = \mathbf{H}\mathbf{x} + \mathbf{n}$, where $\mathbf{x}$ is the $(n_T ×1)$ transmit vector, $\mathbf{y}$ is the $(n_R×1)$ receive vector, $\mathbf{H}$ is the $(n_R × n_T)$ channel matrix, and $\mathbf{n}$ is the $(n_R × 1)$ additive white Gaussian noise (AWGN) vector at a given instant in time.

A general entry of the channel matrix is denoted by ${h_{ij}}$. This represents the complex gain of the channel between the $j$th transmitter and the $i$th receiver. With a MIMO system consisting of $n_T$ transmit antennas and $n_R$ receive antennas, the channel matrix is written as \begin{equation} \mathbf{H} = \left[ \begin{matrix} h_{11} & h_{12} & h_{13} & \dots & h_{1n_T} \\ h_{21} & h_{22} & h_{23} & \dots & h_{2n_T} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ h_{n_R1} & h_{n_R2} & h_{n_R3} & \dots & h_{n_Rn_T} \end{matrix} \right] \end{equation}
where $h_{ij} = \alpha + j\beta = \left|h_{ij}\right| e^{\phi_{ij}}$.

In a rich scattering environment with no line-of-sight (LOS), the channel gains $\left|h_{ij}\right|$ are usually Rayleigh distributed. If $\alpha$ and $\beta$ are independent and normal distributed random variables, then $\left|h_{ij}\right|$ is a Rayleigh distributed random variable.

Now the question is how much information can be reliably exchanged through this system. The answer is \begin{equation} I = \log_2\left[\det\left( \mathbf{H} \mathbf{\Phi} \mathbf{H}^* \left( \mathbf{K}^n\right)^{-1} + \mathbf{I}_{n_R} \right) \right] \end{equation}

where $E \left\{ \mathbf{y}\mathbf{y}^* \right\} = \mathbf{K}$

I'm not going into more details but just mentioning that the multivariate normal distribution is the underlying assumption to develop a very rich subject.

Ali
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Here is an introduction to a course chapter I found online. Hope this helps!:

"Let the random variable Y denote the weight of a randomly selected individual, in pounds. Then, suppose we are interested in determining the probability that a randomly selected individual weighs between 140 and 160 pounds. That is, what is P(140 < Y < 160)?

But, if we think about it, we could imagine that the weight of an individual increases (linearly?) as height increases. If that's the case, in calculating the probability that a randomly selected individual weighs between 140 and 160 pounds, we might find it more informative to first take into account a person's height, say X. That is, we might want to find instead $P(140 \le Y \le 160 \mid X = x)$. To calculate such a conditional probability, we clearly first need to find the conditional distribution of Y given X = x. That's what we'll do in this lesson, that is, after first making a few assumptions.

First, we'll assume that (1) Y follows a normal distribution, (2) $E(Y \mid x)$, the conditional mean of Y given x is linear in x, and (3) $Var(Y \mid x)$, the conditional variance of Y given x is constant. Based on these three stated assumptions, we'll find the conditional distribution of Y given $X = x$.

Then, to the three assumptions we've already made, we'll then add the assumption that the random variable X follows a normal distribution, too. Based on the now four stated assumptions, we'll find the joint probability density function of X and Y.

Objectives

To find the conditional distribution of Y given X = x, assuming that (1) Y follows a normal distribution, (2) $E(Y \mid x)$, the conditional mean of Y given x is linear in x, and (3) $Var(Y \mid x)$, the conditional variance of Y given x is constant.

To learn how to calculate conditional probabilities using the resulting conditional distribution.

To find the joint distribution of X and Y, assuming that (1) X follows a normal distribution, (2) Y follows a normal distribution, (3) $E(Y \mid x)$, the conditional mean of Y given x is linear in x, and (4) $Var(Y \mid x)$, the conditional variance of Y given x is constant."

Note this is for bi-variate normal distributions. You can add more variables that approx. normally distributed.

BLAZE
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Anne
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