How does one formulate continuous probabilities/pdfs?

  • #1
Discrete examples are easy enough. Toss a coin, 1/2, toss a die, 1/6.

Continuous examples, Probability of a nucleus decaying during observation, 1-exp(-λt), Probability of a neutron moves x without interaction, exp(-Σx), where Σ can be assumed to be the inverse of the mean free path i.e. the distance a neutron travels without interaction on average.

My point is that I don't really have an idea as to how these continuous probabilities are derived. Any assistance?
 

Answers and Replies

  • #2
StoneTemplePython
Science Advisor
Gold Member
1,180
578
Discrete examples are easy enough. Toss a coin, 1/2, toss a die, 1/6.

Continuous examples, Probability of a nucleus decaying during observation, 1-exp(-λt),
There's basically two ways to interpret continuous time probability distributions. One is that they are 'merely' a limiting form of discrete cases. The other is that they exist as probability models on their own right. Both interpretations give you some insight.

Former interpretation: your probability of a nucleus decaying during observation -- that is the CDF of an exponential distribution. If you wanted to count the number of these occurrences, you'd count these 'arrivals' via a Poisson Process. Consider tossing a coin that is heads with probability ##p \in (0,1)## and tossing it##n## times. The mean is given by ##\lambda := np##. If you take the limit in such a way that as ##n \to \infty##, ##\lambda## is constant (or bounded in some desired range), then you recover the Poisson distribution. High level the result can be interpreted as tossing an arbitrarily large number of coins at an arbitrarily small probability ##p##, while preserving the essence which is encapsulated in the mean. You may want to look into Le Cam's Theorem which gives a relatively simple setup for something like the above that not only shows the limiting value is Poisson, but gives a useful finite ##n## bound on the (total variation) distance between actual distribution and an idealized one like Poisson.

Latter interpretation: People may have uncovered these distributions as a limiting form of something discrete but they stand on their own two legs. Someone say in physics may have decided classically, that a continuous time model is most appropriate. There may have been a lot of theoretical or physical insights first, or it may have been, basically, an experimental fit. Earthquakes (the big ones, not the aftershocks) are modeled as a Poisson process by the way -- if you want a memoryless counting process in continuous time you really have no other choice.
 
  • Like
Likes random_soldier
  • #3
Stephen Tashi
Science Advisor
7,583
1,472
Discrete examples are easy enough. Toss a coin, 1/2, toss a die, 1/6.
Those are famous examples, but they are not "derived" from any physical theory. They result from assuming each possibility has the same probability of occuring.

Continuous examples, Probability of a nucleus decaying during observation, 1-exp(-λt), Probability of a neutron moves x without interaction, exp(-Σx), where Σ can be assumed to be the inverse of the mean free path i.e. the distance a neutron travels without interaction on average.

My point is that I don't really have an idea as to how these continuous probabilities are derived. Any assistance?
It isn't clear what you mean by "derived". Are you asking whether they can be deduced from some simple mathematical assumption analogous to "all possibilities have the same probability of occurring"? In the case of radioactive decay, the assumption is that the probability of each individual atom of a given type decaying in time t follows the same continuous probability distribution. That is the assumption analogous to "all possibilities have the same probability". To deduce exactly what that continuous probability distribution is, requires finding the continuous probability distribution for individual atoms whose predictions fit experimental data for a large number of atoms.
 
  • Like
Likes random_soldier
  • #4
RPinPA
Science Advisor
Homework Helper
571
319
Probability of a nucleus decaying during observation, 1-exp(-λt),
This one does have a mathematical derivation. It's an example of an exponential distribution. One of the key properties of the exponential distribution is that it is "memoryless", meaning that distribution of the remaining lifetime of the nucleus is the same no matter how long that particular nucleus has already been alive. It is the only continuous distribution with that property, and I believe that it can be derived from assuming that property.

So in some sense it's a naturally-occurring probability distribution that is associated with lifetimes of all kinds of things.
 
  • Like
Likes random_soldier

Related Threads on How does one formulate continuous probabilities/pdfs?

Replies
2
Views
2K
Replies
0
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
4
Views
2K
Replies
12
Views
4K
Top