# Can a Dirac delta function be a probability density function of a random variable?

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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If so, what would you expect the CDF to look like? –  Guess who it is. Jul 28 '11 at 4:48
@J.M. : a unit step function. –  Rajesh D Jul 28 '11 at 4:53
@J.M. : I am not getting what you intend to say. –  Rajesh D Jul 28 '11 at 4:59
@Rajesh: The Dirac delta defines a perfectly good probability measure. However, it is not a function! –  Zhen Lin Jul 28 '11 at 5:04
@Rajesh: As implied in my answer, yes, of course. Modern probability theory allows for distributions which do not have well-defined density functions. –  Zhen Lin Jul 28 '11 at 5:11

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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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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"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 –  Guess who it is. Jul 28 '11 at 14:40

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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How could one convince himself that the probability of getting the number 0 is 1. –  Rajesh D Jul 28 '11 at 9:34
$\lim_{\epsilon\to 0} \int_{a-\epsilon}^{a+\epsilon} \delta(x) dx$ is $1$ if $a=0$, and $0$ if $a\neq 0$. –  Chris Taylor Jul 28 '11 at 11:42
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. –  Chris Taylor Jul 28 '11 at 23:16