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Can the Dirac delta function (or distribution) be a probability density function of a random variable. To my knowledge, it seem to satisfy the conditions. To my interpretation getting a positive real number as the outcome is 1 and that for a negative real number is zero. I wonder what could be the expected value. My question is, whether it is a valid probability density function of a random variable.

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    $\begingroup$ If so, what would you expect the CDF to look like? $\endgroup$ Jul 28, 2011 at 4:48
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    $\begingroup$ @J.M. : a unit step function. $\endgroup$
    – Rajesh D
    Jul 28, 2011 at 4:53
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    $\begingroup$ @Rajesh: The Dirac delta defines a perfectly good probability measure. However, it is not a function! $\endgroup$
    – Zhen Lin
    Jul 28, 2011 at 5:04
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    $\begingroup$ @Rajesh: As implied in my answer, yes, of course. Modern probability theory allows for distributions which do not have well-defined density functions. $\endgroup$
    – Zhen Lin
    Jul 28, 2011 at 5:11
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    $\begingroup$ @RajeshD: On the contrary, what you describe means that $X$ is perfectly non-random. $\endgroup$
    – Rasmus
    Oct 14, 2011 at 7:30

5 Answers 5

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As explained in Gortaur's answer a delta function cannot be the probability density function of a real random variable.

Nevertheless sums of delta functions can be viewed as the "missing link" between discrete and continuous random variables / probability distributions, in the following way:

If $X$ is a discrete random variable taking values $x_k\in{\mathbb R}$ $\ (k\in I$, $\ I$ a countable index set) with probabilities $p_k$ then one can replace the probability space $I$ with the probability space ${\mathbb R}$, provided with the probability measure $$\mu\ :=\ \sum_{k\in I} p_k \ \delta_{x_k}\ ,$$ where $\delta_x$ denotes a unit point mass at the point $x$. In this way $X$ now has become a real random variable. If $f:\ {\mathbb R}\to {\mathbb R}$ is a reasonable function then the expectation $E\bigl(f(X)\bigr)$ may be written as an integral: $$E\bigl(f(X)\bigr)\ =\ \int_{-\infty}^\infty f(x)\ d\mu(x)\ .$$

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    $\begingroup$ Gortaur's answer does not exist. Could you update the answer with the relevant content in the original answer? $\endgroup$ Feb 18, 2016 at 10:47
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Wikipedia article on PDF implies that $\delta(x)$ can be used as a generalized PDF. The corresponding CDF would be the Heaviside (unit step) function as already mentioned. Expected value is 0; I would not really call that variable "random".

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    $\begingroup$ "Expected value is 0; I would not really call that variable "random"." - that's what I was sort of getting at in the comments; I guess I was too oblique... :D $\endgroup$ Jul 28, 2011 at 14:40
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As far as I know, pdf of a random variable $X$ w.r.t Lebesgue measure $\mu$ is defined $\mu$-a.e. as a solution of $$ \mathsf P\{X\in A\} = \int\limits_Af(x)\mu(dx) $$ for all $A\in\mathcal B(\mathbb R)$ where the last integral is Lebesgue integral. For sure you can talk about an integration using $\delta$ function, but usually I mention that people distinguish distributions with densities only when talking about absolute continuous distributions. The distribution of $X\equiv0$ is not absolute continuous since $\mathsf P\{X\in \{0\} \} = 1$ while $\mu(\{0\}) = 0$.

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Can the Dirac delta function (or distribution) be a probability density function of a random variable. To my knowledge, it seem to satisfy the conditions.

That depends on your definition. If you insist that you use the Lesbegue measure as a reference measure, then the delta function is not a Radon-Nikodym density with respect to this reference measure. But if you choose a different reference measure like the counting measure, which assigns to every set the number of its elements, then the delta function is a density (it is the characteristic function of the set {0}).

To my interpretation getting a positive real number as the outcome is 1 and that for a negative real number is zero.

No, the probability to get the number zero is 1, the probability to get anything else is zero.

I wonder what could be the expected value.

Since this random variable is 0 with probability 1, the expected value is 0.

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  • $\begingroup$ How could one convince himself that the probability of getting the number 0 is 1. $\endgroup$
    – Rajesh D
    Jul 28, 2011 at 9:34
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    $\begingroup$ $\lim_{\epsilon\to 0} \int_{a-\epsilon}^{a+\epsilon} \delta(x) dx$ is $1$ if $a=0$, and $0$ if $a\neq 0$. $\endgroup$ Jul 28, 2011 at 11:42
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    $\begingroup$ That proves that the probability of getting the number 0 is 1, and the probability of getting any other number is 0. Which is what you asked. $\endgroup$ Jul 28, 2011 at 23:16
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I was thinking about linear transformation of a zero mean gaussian random variable X. Defining random variable Y=aX, one can easily shows that the probability density function of Y is: $$ p_Y(y)=\frac{1}{|a|}.p_X(y/a)$$ It can be showed that as "a" tends to zero, the $$ p_Y(y)\to\delta (y)$$ Therefore, in my opinion, Dirac delta function can be used to describe the PDF of such a random variable.

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  • $\begingroup$ Excellent answer. Thanks for the answer $\endgroup$
    – Rajesh D
    Mar 27, 2019 at 0:48
  • $\begingroup$ a tends to zero or infty? You may want a to go to zero in your answer. $\endgroup$
    – Rajesh D
    Mar 27, 2019 at 0:51
  • $\begingroup$ Yes, you are right. tend to zero. $\endgroup$ Mar 27, 2019 at 19:33

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