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For two stochastic variables x and y, the rule for independence is given as $$ p(x,y) = p(x) \, p(y) $$ On the left hand side of this equation is a 2D quantity, and on the right hand side are two 1D quantities. In order for this equality to work, the product must therefore have the form of an outer product. Say that $p(x)$ is a Gaussian probability density representing the voltage drop measured across a (defective) diode, so that $p(x)$ has units of inverse volts. Next say that $p(y)$ is an exponential probability density representing the amount of light (optical power) hitting the diode, and thus has units of inverse watts.

If we were to plot $p(x,y)$ using a computer, then we could sample the distribution at various values of $y$, and would find the Gaussian distribution $p(x)$ scaled in amplitude, reaching a maximum at $y=0$, and tailing off to zero amplitude as $y \to \infty$, but keeping the same overall position and width for all $y$.

So far all of this makes sense, though I have yet to find a textbook that explicitly treats the dimensionality issue. But what if we invert the equation, as many texts do. Then we have $$ p(x) = p(x,y) / p(y) $$ and the question is: how do we perform this division correctly? We are dividing a 2D quantity by a 1D quantity, and the most logical interpretation would seem to be to do exactly the inverse of an outer product: at each $x$-value, we divide the distribution by $p(y)$, which leaves us with a 2D distribution like $p(x) \cdot \mathbb{1}(y)$. Assuming that $p(y)$ is nowhere zero, we are left with a uniform value in $y$, but we still have a 2D quantity. To get rid of the extra dimension, and to make the equality work, we appear to need an integral over $y$. But $\int \mathbb{1} (y) dy$ is an improper integral, so that we would appear to need to add a $\delta$-function inside the integral.

But this seems absurdly over-reasoned given that this is such a basic point and that the dimensionality issues are treated in no textbook on probability theory that I have been able to find. But how else can one compute the above division, considering that $x$ and $y$ are not in the same space as one another?

nzh
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    These are probability *densities* which therefore have units of 1/volt ($p(x)$), 1/power ($p(y)$), and 1/(volt*power) ($p(x,y)$). All such quantities are *numbers*: "1D" and "2D" make no sense applied to them. The product is the usual product of real numbers, not an outer product. Because this question hinges on confusing probability densities with other things, it appears that the thread at http://stats.stackexchange.com/questions/4220 answers it. – whuber Sep 12 '13 at 18:41
  • @whuber: It is a _probability_ that is purely a number, while it is a _density_ that may have units. And linking this question to one that conflates probabilities with probability densities exacerbates the misunderstanding. So let me answer my own question: the rules of manipulating probabilities formally hold only for probabilities and not densities, so that the latter apply in the limit where each density is integrated across a small range, while taking the limit of that range to approach zero. If the limit operation is done _last_, then the results hold. – nzh Mar 06 '14 at 12:50

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