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I have tried to understand this myself but what I have found on the internet so far has not helped.

I have a likelihood function that for part of it has the following statement:

d0 is the Dirac delta function at zero

and it enters in the likelihood function as the following (there is more but it is not relevant at the moment) $$ e^{-L}d_0(y) + \ldots $$ I am not sure what it means that $d_0(y)$ is a Dirac delta function, because what the Internet has told me that it will effectively always take the value of 0, meaning that the above term contributes nothing to the likelihood.

Could someone offer an explanation for what a Dirac delta function could be in this case?

Thank you

Tim
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user3742335
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  • hi, it seems weird, can you please give more details about the problem and the likelihood function you have? – chuse Dec 15 '14 at 12:26
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    A Dirac delta function should not be used in a density but in the dominating measure: that is, if you have a variable that takes continuous values _and_ a specific value like zero, the dominating measure is the sum of the Lebesgue measure and of the Dirac measure at zero. The density is then the sum of a function $f(x)$ defined everywhere and of a mass $p_0$ multiplied by the indicator $\mathbb{I}_0(x)$. – Xi'an Dec 15 '14 at 12:54

2 Answers2

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A typical example where mixed (continuous and finite support) distributions occur is in censoring: the simplest model is to have a continuous variable, say $x\sim\text{N}(\mu,\sigma^2)$ observed, unless it is larger than a fixed value, say $\omicron$, in which case the bound $\omicron$ is reported. In such a case, the density of the reported random variable is defined against a measure that includes the Lebesgue measure on $\mathbb{R}$ and a Dirac mass at $\omicron$, which is a measure that gives a mass of $1$ to any set containing $\omicron$ and $0$ otherwise. With respect to this dominating measure the density is \begin{align*} f(x|\mu,\sigma,\omicron) &= \frac{1}{\sqrt{2\pi}\sigma}\exp\{-(x-\mu^2)/2\sigma^2\}\\ & + \int_{\omicron}^{+\infty} \frac{1}{\sqrt{2\pi}\sigma}\exp\{-(y-\mu^2)/2\sigma^2\}\,\text{d}y\,\mathbb{I}(x=\omicron) \end{align*} which involves an indicator, not a Dirac mass.

Xi'an
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The Dirac $\delta$ function is zero except for one point, where it is infinite. It is used with integrals, and is defined to have an integral of 1 if the special point is included. The effect of this is to "pull" specific point values through the integral.

This lets you assign probability weight to precise point values. @Xian's answer has a nice example of why you might do this.

Technically, it isn't a real-valued function and doesn't work with the normal definition of integrals -- which are the Riemann and Lebesgue measures. It's just convenient to write it as a function inside an integral.

In your example, $d_0$ is the Dirac delta function at zero, which means $\int_X f(y)\delta_0(y)$ is $f(0)$ -- i.e. $e^{−L}$ in your example -- if zero is inside the region integrated $X$, and 0 otherwise.

Daniel Winterstein
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