5
$\begingroup$

I have a question about concepts:

What is the practical meaning of these common concepts of the probability theory and mathematical statistics. These concepts often go in tandem with each other. Everywhere are just dull formulas and cold formal definitions, that does not help to perceive what is the physical meaning/significance and how it can be applied on the real-world representative example, rather than set of similar problems given in textbooks and websites.

Usually, such cold formulations are given: "expectation of the squared deviation of a random variable from its mean", "amount of variation or dispersion of a set of data values", accompanied by the formulas.

Can anyone explain what is the practical meaning of these concepts on a real-world example?

|cite|improve this question
$\endgroup$
  • 3
    $\begingroup$ What do you mean by "physical meaning" and "practical meaning" and "real meaning"? What kind of explanation are you seeking? What examples are you reading and in what way are they not giving you what you seek? As it stands the question seems to invite a plethora of different answers (which would be grounds to close the question as either unclear or too broad), without necessarily getting any that do something different from what may be found in a good textbook. $\endgroup$ – Glen_b♦ Sep 29 '16 at 1:13
5
$\begingroup$

I'll go with the cliche example - coin flipping. Note that I'm abandoning rigor and some important assumptions in this example, but that's just fine.

Let's say I have a regular coin - that is, once I flip it, it has a 50% chance of landing heads and a 50% chance of landing tails.

So if I flip it 10 times, I'd expect 5 tails and 5 heads. But I could very well get 6 heads and 4 tails. Or 7 heads and 3 tails.

But wait a second - why would I expect 5 tails and 5 heads? Maybe it's obvious - because each flip has a 50% chance of landing heads - so $50\% \times 10 = 5$. In other words the expected value of my coin flipping exercise is 5 heads (and therefore 5 tails) .

Let's make the example more interesting now. Let's flip the coin 100 times. But check it - nothing changes in terms of my expected value. I still expect half of the tosses to be heads - i.e. 50 heads.

But in reality I might not get 50 heads. Let's say I got 45 heads. Is that far from my expected value of 50? Should I be surprised by that result? Would you be? Probably not. If I told you that I got only 20 heads, then you might think something's up. Why do you think that is?

That's sort of the intuitive notion of variance. How likely is it for our results to deviate from the expected value? Some things (like coin flips) have a pretty good chance of deviating from their expected value. Other things don't.

We can put a number on this. In some instances, for mathematical convenience and interpretability, we can take the square root of this number. That's the standard deviation.

The definitions you refer to above are more technically accurate and have direct mathematical formulations, hence terms like probability weighted and random variable.

But if you understand the coin flipping example, then you'll understand the spirit of the terms.

|cite|improve this answer
$\endgroup$
  • 3
    $\begingroup$ Chichi, isn't coin flipping the perfect archetype of the "meaningless set of similar examples given in textbooks and websites"? In light of your explicit criticism of them, how could you possibly accept such an explanation? $\endgroup$ – whuber♦ Sep 30 '16 at 16:57
  • $\begingroup$ Perhaps because it explains in simple to understand and intuitive ways the concepts he/she is trying to understand? $\endgroup$ – ilanman Sep 30 '16 at 18:22
  • 1
    $\begingroup$ I don't disagree--but there is a sharp disconnect between what the OP was requesting and this kind of explanation. I'm not criticizing you. I just think that the acceptance of this answer casts much doubt on what the original question was actually asking. $\endgroup$ – whuber♦ Sep 30 '16 at 18:58
  • $\begingroup$ I agree -- I read the question as saying "whatever you do, don't give me this kind of answer", which makes choosing an answer like that seem very strange. A pity the OP didn't clarify as requested. $\endgroup$ – Glen_b♦ Oct 3 '16 at 7:16
3
$\begingroup$

A random variable is a quantity whose value appears to be random when measured. As @ilan man describes, the observation of Heads or Tails in a coin flip experiment is one simple example. You could also define a random variable to be the average value of the last ten tosses of the coin (mapping H to 1 and T to 0, for example). The time that you get to work each day is also a random variable. Your weight every morning is a random variable. Etc.

We characterize random variables by their distribution, which, broadly speaking, gives a description of how likely particular values are to be observed. The Expected Value of the random variable is a measure of the center of this distribution and the Variance is a measure of its spread. (The Standard Deviation is the square root of the variance, which is a nice measure since it has the same units as the variable - roughly speaking it is measures the "width" of the distribution.)

Note that both of these measures are "non-robust", meaning that they are sensitive to outlying values. There are also "robust" statistics for the "center" and "width". The median, which is the value for which there are as many points larger and smaller, is a robust measure of the centroid. Since it only depends on the count of values larger or smaller, the details very far from the median do not have an affect on the median. There are a number of robust width measures but the one most similar to the median is the "median absolute delta", which is $$ median(|X - median(X)|) $$ where X is the random variable in question.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

May be for someone it will be helpful to see how expected value $\largeμ$, variance $\largeσ^2$ and standard deviation $\largeσ$ are related in normal distribution for random variables.

In formula:

$ f(x) = \Large\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} $

On the graph:

For the normal distribution, the values less than one standard deviation away from the mean account for 68.27% of the set; while two standard deviations from the mean account for 95.45%; and three standard deviations account for 99.73%. enter image description here enter image description here Source: Normal distribution

And real-world example with random variable. The heights of NBA players all follow a normal distribution:

enter image description here

|cite|improve this answer
$\endgroup$
0
$\begingroup$

I suspect that probability density function is called after the density in physics. Hence, the mean corresponds to a center of mass of a body.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.