laplace experiment probability

• Report an issue

Laplace probability

Experiments in which all results are equally likely are called Laplace experiments .

The probability of an event is calculated in a Laplace experiment with the formula:

$|E| ...$ Number of results where $E$ occurs $|\Omega| ...$ Total number of results

Examples of Laplace experiments are the throwing of a coin, a dice or the turning of a wheel of fortune with fields of equal size.

A dice is thrown. Your are interested in the probability of an even number.

Sample space : $\Omega=\{1,2,3,4,5,6\}$ Event : $E=\{2, 4, 6\}$ Probability: $P(E) = \frac{|E|}{|\Omega|}$ $=\frac{3}{6}$

Non-Laplace experiment

In non-Laplace experiments , the probabilities for each possible outcome can not be determined by e.g. symmetry considerations or the like. However, after many experiments have been carried out, estimates of the probabilities can be determined.

Examples of non-Laplace experiments are the throwing of thumbtacks, a LEGO stone or a crown cork. It is not exactly possible to say which event occurs with which probability.

• Help with math
• Visual illusions
• Problem solving
• Index/Glossary
• Elementary geometry
• Tell a friend
• Talk about it

Laplace's Rule of Succession

Assume you run $n$ independent equiprobable failure/success trials which ended up with $s$ successful outcomes. In the absence of additional information, it is natural (or at least customary) to view $\displaystyle p=\frac{s}{n}$ as the probability of success. Laplace's Rule of Succession suggests that, in some circumstances, an estimate $\displaystyle p=\frac{s+1}{n+2}$ is more useful. Of course, for large $n$, the two estimates are hardly distinguishable. However, for a small number of trials, the latter is (often) more meaningful. For example, assume a few trials have all ended in failure; $\displaystyle p=\frac{s}{n}$ in this case will be $0$, implying that a trial has no chance of success. However, it is rather obvious that to reach such a clear-cut conclusion on the basis of a small number of trials would be imprudent. Laplace's formula shows one way around that difficulty. Similarly, of a small number of trials all end up with successful outcomes, Laplace's formula $\displaystyle \frac{n+1}{n+2}$ leaves door open to a possibility of failure, while the usual $\displaystyle \frac{n}{n}=1$ does not.

I shall start with following [ Feller ] in deriving Laplace's formula for $s=n$.

Let there be $N+1$ urns each containing $N$ balls such that the urn #$k$ contains $k$ red and $N-k$ blue balls ($k=0,1,\ldots , N$.) At the first stage of the experiment we choose a random urn (with the probability of $\frac{1}{N+1}$) and then proceed to pick up balls from the chosen urn. After the ball's color has been recorded, the ball is returned back to the urn. Assume the red ball showed up $s$ times (event $A$) and - on the basis of that observation - predict the probability of the red ball showing up on the next trial (event $B$). We are thus looking into the conditional probability $P(B|A)$. Note that

For the urn with $k$ red balls, the probability $P_{k}(s, n)$ of having $s$ red balls in $n$ trials is

where $\displaystyle {n \choose s}$ is the number of combinations of $s$ out of $n$ symbols. In case, where $s = n$, $\displaystyle P_{k}(n, n) = (\frac{k}{N})^{n}$ and $P(AB) = P(B)$, making $P(B|A)=P(B)/P(A)$ so that what is needed is to evaluate the two probabilities $P(A)$ and $P(B)$ separately.

Define $Q(k)$ as the probability of having a string of $k$ successes in $k$ trials. Then $P(A) = Q(n)$ and $P(B) = Q(n+1)$. But

which can be viewed as a Riemann sum approximation of the integral $\displaystyle \int_{0}^{1}t^{n}dt = \frac{1}{n+1}$. It follows that, for $s=n$, indeed $\displaystyle P(B|A)=\frac{Q(n+1)}{Q(n)}=\frac{n+1}{n+2}$.

In case, where $s\lt n$, no longer $P(AB)=P(B)$, and we have to look for other ways to evaluate $P(B|A)$.

... to be continued ...

The way to proceed now is to view the sums as the Riemann sums of a couple of integrals, so that, for example,

Plain (or repeated) integration by parts yields the formula

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

The Laplace (or double exponential ) distribution has the form of two exponential distributions joined back-to-back around a location parameter μ. It arises naturally as the difference between two independent and identically distributed exponential random variables.

Probability Density Function

X ∼ Laplace(μ, β)

E(X) = , Var(X) =

Note that the pdf of the Laplace distribution is symmetric about the location parameter μ.

Since the pdf of the Laplace distribution is symmetric about the location parameter μ, the cdf has the property that F(μ - x) = 1 - F(μ + x).

The illustration above shows two values X 1 and X 2 chosen independently and at random from an exponential(β) distribution. The random variable X 1 − X 2 + μ has a Laplace(μ, β) distribution, where μ denotes a location parameter.

The simulation above shows two values X 1 and X 2 chosen independently and at random from an exponential(β) distribution. The light blue line shows the value of X 1 − X 2 + μ, where μ denotes a location parameter. The distribution of this value has a Laplace(μ, β) distribution. The histogram accumulates the results of each simulation.

Y = |X − μ| ∼ Exponential(β)

E(Y) = , Var(Y) =

This can be thought of as "folding" the distribution about the line x = μ and shifting back to the origin. Since the Laplace pdf is symmetric about x = μ, this has the effect of doubling the value of the probability density in the folded distribution.

Advertisement

Voice of the Engineer

Laplace’s analytical theory of probabilities

French mathematician and astronomer, Pierre-Simon Laplace brought forth the first major treatise on probability that combined calculus and probability theory in 1812.

A single roll of the dice can be considered a random event, but after many rolls, certain statistical patterns emerge. These patterns, when studied, can be used to eventually make predictions!

Stephen Hawking calls Laplace’s treatise a “masterpiece” and states that, “Laplace held that because the world is determined, there can be no probabilities in things. Probability results in our lack of knowledge.”

So nothing will be uncertain for a sufficiently advanced being according to Laplace. This idea and conceptual model remained true in the face of twentieth century chaos theory and quantum mechanics!

So how do probabilistic processes give results that can be predicted? Imagine several urns arranged in a circle, says Laplace. One of the urns contains all black balls and another urn only white balls. All of the other urns are comprised of some mixture of black and white balls.

Let’s withdraw a ball from any urn and place it in an adjacent urn and continue this process around the circle. Eureka! Eventually the ratio of black to white balls will be approximately the same in all of the urns!

Thus Laplace demonstrates how random “natural forces” can create results with order and predictability.  He said, “It is remarkable that this science, which originated in the consideration of games of chance, should become the most important object of human knowledge….The most important questions in life are, for the most part, really only problems of probability.”

So pack your bags for Vegas and go make your fortune! (Laplace would be rolling over in his grave at this)

One of my best professors, Athanasios Papoulis, during my Master’s term at Polytechnic University, would be proud of me for publishing this treatise in my blog! Professor Papoulis wrote a classic textbook, “Probability, Random Variables, and Stochastic Processes” from which he taught me back in the mid-1980s. We lost a great and caring man and professor on April 25, 2002!

div-gpt-ad-inread

0 comments on “ Laplace’s analytical theory of probabilities ”

Leave a reply cancel reply.

You must Sign in or Register to post a comment.

[ninja_form id=2]