# $k$-dimensional normal distribution function

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}\$$

-
I can't see the difference between this and your previous question... – J. M. Aug 22 '11 at 8:48
@J.M.: in the previous question the OP didn't give a formula that is nonsensical in view of the question. Best I can tell $f(x)$ defined in the question is independent of $x$. – Willie Wong Aug 22 '11 at 12:56
Yes, I did not mention the formula in my last question. that is why I repeat the question to make it more clear – shaikh Aug 23 '11 at 0:53
@shaik: please, next time edit the question with the new information, so as to minimize duplication. – Mariano Suárez-Alvarez Aug 23 '11 at 8:01

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.

-
Yes, I am assuming that all the dimensions are independent of each other. – shaikh Aug 23 '11 at 0:57
...then the formula holds... Seriously? – Did Aug 30 '11 at 14:41
Yes.    – Did Feb 16 '13 at 20:41

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.

-
+1.    – Did Aug 30 '11 at 14:41