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I have a question on this prior post on the derivation of the normal distribution.

The author had set up the target/dart analogy often used to introduce a translation-invariant bivariate, uncorrelated distribution $f(x,y) = g(x^2 + y^2),$ which ends up helping in the integration through the use of the Pythagoras theorem and change to polar coordinates.

At that point he mentions:

This means that $g$ must be an exponential function

$g(t) = A e^{-Bt}$

based on the independence of $X$ and $Y.$

This seemingly obvious step is not clear to me, and would like to ask why this is so clear - intuitively, yes, assuming a plausible competent dart thrower, the majority of darts will group around the center of the target, and there'll be an exponential drop-off away from the center (perhaps more or less rapid depending on the accuracy of the thrower).

But I don't see it mathematically.

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Antoni Parellada
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    It is a property of the exponential function that it is a solution to the equation $$g(x+y) = g(x)g(y).$$ Many calculus textbooks cover this result. The result in question is not based on the independence of $X$ and $Y$ but rather on the point that $g(x^2+y^2)\proportionalto g(x^2)g(y^2)$ – Dilip Sarwate Mar 09 '18 at 02:00
  • @DilipSarwate Got it! $g(x^2) = e^{x^2}$ because $e^{x^2+y^2}=e^{x^2}e^{y^2}.$ Very close to the Gaussian function: $ae^{-(x-b)^2/2c^2}.$ Thank you! – Antoni Parellada Mar 09 '18 at 02:52
  • @DilipSarwate In MathJax, use \propto for $\propto$ – Graham Kemp Mar 09 '18 at 03:00

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