Let's say we have a set of $N$ observations $D = \{\bf X, t\}$ where ${\bf X} = [{\bf x}_1, ..., {\bf x}_N]^T$ are the locations and ${\bf t} = [t_1, ..., t_N]^T$ are the targets.

When applying a Gaussian process for the problem of regression, we start with an assumption that

\begin{equation} t_i = f_i + \epsilon_i \end{equation}

where $t_i$ is the observed targets, $f_i$ is a random variable that is the output at location ${\bf x}_i$, and $\epsilon \sim \mathcal{N}(0, \sigma^2)$ is assumed noise. We are then interested in the joint marginal distribution of ${\bf t}$, which we find to be

\begin{equation} {\bf t} \sim \mathcal{N}({\bf t}|{\bf 0}, {\bf C}) \end{equation}

where the covariance matrix $\bf C$ is defined by

\begin{equation} {\bf C}_{n,m} = k({\bf x}_n, {\bf x}_m) + \sigma^2\delta_{n,m} \end{equation}

The book I'm reading this from (Pattern recognition and Machine Learning, Bishop 2006, pp 306) does not explain what $\delta_{n,m}$ is, but from this thread I've come to understand that this is the Dirac delta function

\begin{equation} \delta_{n,m} = \delta({\bf x}_n - {\bf x}_m) \end{equation}

Any definition I've found assumes the input is univariate (one-dimensional), but the book strongly suggests that the input is multivariate.

My question is this: what is the definition of the Dirac delta function for multivariate input?

  • $\begingroup$ I believe this $\delta_{n,m}$ must be the Kronecker delta, not the Dirac delta. $\endgroup$
    – whuber
    May 14, 2019 at 18:59

1 Answer 1


Dirac's delta notation is $\delta(x)$. When you see indices used $\delta_{ij}$, it must be Kroneker delta. In Bishop's book you can see how these two notation are used. For instance, check Exercise 3.4 for the latter, and Eq.4.146 for the former.

  • $\begingroup$ Just to be clear, the person (Alexey Zaytsev) who referred to this function as the 'Dirac delta function' in the question I linked to was wrong? $\endgroup$
    – Mossmyr
    May 14, 2019 at 20:18
  • $\begingroup$ @Mossmyr, I wanted to say "No", but the way he defines Dirac delta $\delta(x)$ is wrong. I'm going to comment on his answer now. $\endgroup$
    – Aksakal
    May 14, 2019 at 20:26
  • $\begingroup$ @Akasal But Alexey is talking about the same thing as Bishop (pp 306-307): the covariance of the marginal distribution of the target in a Gaussian process. They use different variable names (Alexey uses $y_i = f_i + \epsilon_i$ while Bishop uses $t_i = y_i + \epsilon_i$), but it's the same thing. Edit: I see that you changed your comment now, disregard my comment. $\endgroup$
    – Mossmyr
    May 14, 2019 at 20:30
  • $\begingroup$ @Mossmyr, yes, I think he meant to use Kronecker delta. Dirac delta is not an ordinary function, it's a generalized function and is defined through integrals $\endgroup$
    – Aksakal
    May 14, 2019 at 20:32

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