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I am trying to solve the following equation,

\begin{equation} = \int_{-\infty}^{\infty} \frac{1}{\sqrt{ (2\pi)^{k_{Y}} | \Sigma |}} \cdot \mathrm{exp} \{ -\frac{1}{2} (Y - Xm)^{T} \Sigma^{-1} (Y - Xm) \} \times \delta(m - \beta) \mathrm{d} m \end{equation}

where $\delta$ is an indicator variable; and $m$ is multidimensional variable. Assume, m is in size $1 \times n$;

I am not sure, if I should deal with $y = \delta(x)$ as a Dirac delta function $y=1$ if and only if $m = \beta$ and $y=0$ for the rest; or as a step function $y = 1$ for $m-\beta > 0$ is one and elsewhere 0;

From what I read about indicator variable in wikipedia, it is a step function, meaning for ; however, 1) It is not intuitive for me to understand the role of indicator variable 2) Reaching the the integral would be more difficult than assuming a Dirac delta function (I might be wrong).

I tried to crack the integral by assuming the indicator variable as Dircal Delta function; I am not sure I have been successful !

\begin{equation} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \frac{1}{\sqrt{ (2\pi)^{k_{Y}} | \Sigma |}} \cdot \mathrm{exp} \{ -\frac{1}{2} (Y - Xm_i)^{T} \Sigma^{-1} (Y - Xm_i) \} \times \delta(m_i - \beta) \mathrm{d} m_i |_{i=\{1\cdots n\}} \end{equation}

\begin{equation} \frac{1}{\sqrt{ (2\pi)^{k_{Y}} | \Sigma |}} \cdot \mathrm{exp} \{ -\frac{1}{2} (Y - X\beta)^{T} \Sigma^{-1} (Y - X\beta) \}) \end{equation}

I am not sure if it is the correct ! I appreciate if you help me to understand how to work with indicator variables in such cases;

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    $\begingroup$ Indeed: merely set $m_i = \beta$ and remove the integral signs, as you have. Just put in your equal sign so your result is typographically correct. $\endgroup$ Commented Jan 25, 2015 at 17:42
  • $\begingroup$ Thank you - but I saw stats.stackexchange.com/questions/129169/… - that the indicator variable bounded the integral (treated as a step function). $\endgroup$
    – Areza
    Commented Jan 25, 2015 at 17:44
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    $\begingroup$ The answer you referred to was for a slightly different problem: censoring on a finite support. You have a function on infinite support (which happens to be a multivariate Gaussian) times a Dirac delta. $\endgroup$ Commented Jan 25, 2015 at 18:00
  • $\begingroup$ What @DavidG.Stork points out in his first comment is the very definition of $\delta$. $\endgroup$
    – whuber
    Commented Jan 25, 2015 at 20:53

1 Answer 1

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In general, the indicator variable $\theta$ of a set $A$, that is, $\theta (A): A \subset \Omega$ is $=1$ on $A$ and $0$ on the complement of $A$.

A distribution satisfies certain properties, such as positivity $\delta(x) > 0, \forall x$, and normalization $\int_\Omega \delta(x) dx = 1$ and the Dirac delta in particular $\delta(x-x_0)$ is "localized" at $x_0$, while retaining the unit integral.

The Heaviside step function is the indicator function of the positive real axis $\mathbb{R}_+$, which is infinitely extended, so how could its integral be normalizable? Clearly, the Dirac delta and the Heaviside cannot be used interchangeably (however, the Dirac delta can be seen as the generalized (distribution-valued) derivative of the Heaviside function with step at $x_0$).

We use this fact when "measuring" a function $f$ over a set $\Omega$, with a distribution (i.e. measure) $\delta$: $\int_\Omega{f(x) \delta(x-x_0)}$. In the vanishing width limit (a number of distribution converge to the Dirac) the previous integral becomes $= f(x_0)$, the "sampling" property of the Dirac measure.

In your example above it amounts to evaluating the multivariate Gaussian at $\beta$, as you correctly did. What this means w.r.t. your problem I have no idea though :D

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