Since the total mass of a Dirac function is unity, can it be a PDF of a given random variable?
I think people are speaking of different things here.
If we speak of the definition of a probability density function in possibly all probability textbooks, then the answer is NO.
If we speak of the language usage in the physics community, then it is clear from many of the answers that the answer is very probably YES (I am not a physicist).
Why NO or YES? NO, because a PDF is by definition a real function, i.e. it cannot take an infinite value. In that language, not every probability distribution has a PDF: only distributions which are absolutely continuous with respect to the Lebesgue measure have one. The Dirac distribution has a jump and so is not absolutely continuous.
YES, because, starting with (of course) Dirac, physicists noticed that they could use the Dirac function as if it were a PDF and always get the right answer. Therefore, it would be an unnatural contortion for them not to call it a PDF as well.
John von Neumann wrote his famous book on the mathematical foundations of quantum mechanics with the explicit project of disposing of Dirac's mathematically unacceptable trick.
Later, Laurent Schwarz proposed a more general theory of distributions within which Dirac's intuitive manipulations make fully rigorous mathematical sense.
And even later developments in non-standard analysis make the whole controversy appear as quite banal, as Costas Drossos has aptly remarked here.
Yes. If the probability density is delta(x) where x is the position in R^3, the particle is located at the origin with probability 1.
Thank you for the answer.
How can we define the repartition function in this case?
I am not completely in agreement that Diracdelta function can be a probability density function (pdf). If X is a random variable with fX as pdf, fX must satisfy 1. fX>0, Int(-inf,inf)fXdx=1 and 3. Int(a,b)fXdx=P(a
@ MIguel: We use the Dirac delta as a probability distribution all the time in physics. Certainly x is equal to zero with probability one for the probability distribution P(x)=delta(x). Note that P(x)=delta(x) is not equal to one for x = 0 --- it is undefined.
It cannot be a valid probability distribution. At best, it describes a certain event!!
Using Nonstandard analysis, we let dx be a positive infinitesimal. Define a function:
f(x):= 1/dx if x \in (-dx, dx) and f(x)=0 otherwise. Then this is a Dirac delta function, which has integral 1 over the *reals and of course is everywhere zero over ordinary R. One might prove that this is p.d.f of the d.f. F(x)=1 if x>= to 1 and zero otherwise.
thus using nonstandard analysis, Schwartz distribution became almost trivial!
Yes, Abdallah, it can be a random variable. However, this random variable does not have anything to do with probability. A random variable need not follow a probability law, while a variable following a probability law is random by definition.
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the Dirac distribution can be considered as a proper probability density ; its associated cumulative density function is the Heaviside function (H(x) = 0 for x0)
see http://en.wikipedia.org/wiki/Dirac_delta_function#Probability_theory
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Thank you so much for all your helpful answers!
So if we can consider this Dirac delta function as a probability density function for a given random variable and its repartition function is the Heaviside function since it is its primitive as Fabrice had indicated, one now may ask the following questions:
1. Can we really consider this random variable as continuous or it is discrete as Sergio has mentioned in the previous answer?
2. What is the mathematical expectation of this random variable and how its standard variation is defined?
Thank you in advance!
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Abdallah, your questions above can be answered from a very simple application of the distribution theory (distribution in the "mathematical analysis" sense, not in the probabilistic sense ...)
you can start here :
http://en.wikipedia.org/wiki/Distribution_%28mathematics%29
alternatively, googling "distribution mathématique" will lead you to quite a lot of introductory courses
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Yes Abdallah, a particular example of this use is in the classical density functional formulation of statistical mechanics. In this case particle positions may be represented by the dirac delta function and summations may be converted to integrals.
If you are interested in seeing this application I would recommend the book "Theory of simple liquids" by Hansen and McDonald.
Dirac delta function is integral killing function , transform a wide range to a single point, while the probability density function is a malty valued.
I think people are speaking of different things here.
If we speak of the definition of a probability density function in possibly all probability textbooks, then the answer is NO.
If we speak of the language usage in the physics community, then it is clear from many of the answers that the answer is very probably YES (I am not a physicist).
Why NO or YES? NO, because a PDF is by definition a real function, i.e. it cannot take an infinite value. In that language, not every probability distribution has a PDF: only distributions which are absolutely continuous with respect to the Lebesgue measure have one. The Dirac distribution has a jump and so is not absolutely continuous.
YES, because, starting with (of course) Dirac, physicists noticed that they could use the Dirac function as if it were a PDF and always get the right answer. Therefore, it would be an unnatural contortion for them not to call it a PDF as well.
John von Neumann wrote his famous book on the mathematical foundations of quantum mechanics with the explicit project of disposing of Dirac's mathematically unacceptable trick.
Later, Laurent Schwarz proposed a more general theory of distributions within which Dirac's intuitive manipulations make fully rigorous mathematical sense.
And even later developments in non-standard analysis make the whole controversy appear as quite banal, as Costas Drossos has aptly remarked here.
Dear colleagues,
Let me add my contribution to your valuable ones by noting that, since the Delta function, or rather distribution in a more rigorous functional analytic jargon, is the distributional limit, as sigma-->0, of the distribution
(1/sigma)*f(x/sigma)
where f is any positive integrable function with L^1 norm equal to 1 (PDF in probability theory), it can, quite naturally, be regarded as the limit of the Gaussian PDF of the normal distribution with standard deviation shrinking to 0.
Simply stated, and if I understood correctly what you meant, the answer is YES but…..
The meaning of a probability density of a variable that is a Dirac delta function is clear: it is a deterministic variable since it can just take one particular value. In fact, the solutions of all ODEs can be formally expressed in terms of a PDF using a Dirac delta. Yet, if it is a deterministic quantity is NOT a stochastic variable!
The answer is most certainly NO. There seems to be some confusion about terminology throughout many parts of this post. The "dirac delta function" usually refers to a probability measure \delta_x for which \delta_x(A)=0 for any Borel set A not containing the point x, and =1 for any Borel set A containing the point x. As Pedro mentioned, you need absolute continuity with respect to Lebesgue measure to have a well-defined PDF. Clearly, the dirac measure doesn't satisfy this condition. However, all random variables can characterized by their cummulative distribution function (CDF). Fabrice stated the correct CDF (not PDF) for the random variable X associated with the dirac measure with x=0 in one dimensional Euclidean space i.e. the heavy-side function
Quite right, Dirac's delta is not of bounded variation and hence can not be absolutely continuous.
Dirac delta function can be probability density function for some continuous random variable because the area under the curve and the x-axis is 1, that is the total probability is 1.
Bhavin,
What is the "curve" that you are referring to? The whole point of my answer is that there is no PDF for a dirac measure and, consequently, no continuous curve enclosing an area of one. However, you can get the dirac measure as a weak* limit point of a sequence of absolutely continuous probability measures e.g. normal random variables with mean 0 and variance 1/n as n goes to infinity.
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