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If $f(\vec{x})$ is a vector of normal distribution function and assuming that $\sigma$ is same in all dimension, can we say that

$ f(\vec{x}) = \prod_{1\leq i \leq k} \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}\exp\big(-{(x_{i}-\mu_{x,i})^2}/{2\sigma^2}\big) \mathrm{d}x_{i}\ $

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    @shaik: please, next time edit the question with the new information, so as to minimize duplication.2011-08-23

2 Answers 2

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Maybe you mean "$\sigma$ is the same in all directions" -- a spherically symmetric distribution. That would force the above density, and independence of the coordinates $x_i$.

If you mean the joint normal distribution of several coordinate "dimensions" $X_i$, where each dimension considered by itself is normal with the same $\sigma$, then the formula holds if and only if all these dimensions are statistically independent of each other (no correlation between different coordinates).

Edit: ... that is all true if you remove the integral signs from the formula. For the formula as it was posted, $f(x)$ is independent of $x$ because of the integration.

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    Yes. $ $ $ $ $ $2013-02-16
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The formula as written is absurd: the integral makes the rhs equal to one... Without the integral sign, this is a special case of the multivariate normal density.