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In deriving first the bivariate case, I have come across this limit that confuses me, (7). It should be easy enough for a mathematically mature person to understand and explain.

Is the limit a typo? Should one of the lines instead be $\Delta x \to 0$ instead of two $\Delta y \to 0$'s?

In general, why does the limit go to zero and what's the significance of this in learning the multivariate normal distribution?

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There is certainly a typo and it should be $\Delta x\rightarrow 0$. The result being quoted is the integral mean value theorem. What's going on is as follows. In (6), they derive that the probability equals $f(x_0,y_0)\Delta x\Delta y$ for $x_0\in (x,x+\Delta)$ and $y_0\in (y,y+\Delta)$. This is precisely what the integral mean value says, that the integral of a continuous function is equal to the area on which you are integrating multiplied by the value of the function somewhere in that area (exactly where is somewhat immaterial as we'll see below). It's a sort of average value result.

The point is that in $(7)$, you are no longer looking just at $x_0,y_0$ but $f(x,y)$. But, if we interpret the mean value theorem as in $(6)$ then we realize that as $\Delta x$ and $\Delta y$ go to 0, then $x_0\rightarrow x$ and $y_0\rightarrow y$, so by continuity of $f$, we get $f(x_0,y_0)\rightarrow f(x,y)$. Again, here $x_0$ and $y_0$ depend both on $x,y$ and $\Delta x$, $\Delta y$, but we are guruanteed they lie somewhere in $(x,x+\Delta x)$ and $(y,y+\Delta y)$ respectively so as the intervals shrink to zero the $x_0$ and $y_0$ are squeezed arbitrarly close to $x,y$.

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Thanks for the explanation. Although, what's the point of having (1/(delta x * delta y)) in the limit? Wouldn't the difference still go to zero without it? – mathjacks May 25 '12 at 3:24
Careful, the $f$ function carries $\Delta x \Delta y$ but the Pr function does not. Think of it this way, $f$ is a probability density, in that it represents probability per unit area. So if you want to obtain a probability from $f$ you need to multiply it by an area. Hence $f(x,y)\Delta x\Delta y$ has units of probability (a technically uniteless quantity). – Alex R. May 25 '12 at 4:03

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