I'm struggling to find the significance of $dx$ and $dy$ in terms of the Herschel-Maxwell derivation of Normal Distribution.

My strongest theory so far is that $\rho(x,y)$ is a function which gives the probability of exactly the point $(x,y)$. Then $dx$ and $dy$ are multiplied, giving a small area around the function, and when multiplied by p, results in a probability density function. Therefore $\rho·dx·dy$ is the probability of a small area 'being selected' on the graph.

I invite clarification or confirmation of this theory.


1 Answer 1


First of all: The Normal distribution is (absolutely) continuous. Hence, $\rho(x,y)$ cannot be the the probability of (x,y), since the probability of (x,y) is $0$.

Your basic intuition is pretty good, but not entirely correct. $\rho(x)dxdy$ is a probability measure that is defined by the probability density function $\rho$. A measure is a function that maps certain sets (measurable sets) to positive values. These sets are called events when speaking about probability. To evaluate the measure $\rho(x)dxdy$ for a given event $A \subseteq \mathbb{R}^2$, one just needs to integrate: $$\iint_A \rho(x)dxdy.$$


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