Consider the plane ${ \mathbb{R} ^2 . }$ Consider the origin ${ (0, 0) . }$

Consider the natural error generating process:

• The random error vector is ${ \mathcal{E} = (\varepsilon _1, \varepsilon _2) . }$
• The components ${ \varepsilon _1, \varepsilon _2 }$ are independent and identically distributed.
• The distribution of ${ \mathcal{E} }$ is rotationally symmetric.

Q) What is the distribution of the random error vector ${ \mathcal{E} }$?

Let ${ f(x, y) }$ be the density of ${ \mathcal{E} . }$

Let ${ f(x) }$ be the density of the components ${ \varepsilon _1 , \varepsilon _2 . }$

Note that by independence of components

\[{ f(x, y) = f(x) f(y) . }\]

Note that by rotational symmetry of ${ f(x, y) }$

\[{ f(x, y) = g(x ^2 + y ^2) . }\]

Hence

\[{ \boxed{f(x, y) = f(x) f(y) = g(x ^2 + y ^2) }. }\]

Note that it suffices to find the expression of ${ g . }$

Note that

\[{ f(x) f(0) = g(x ^2) . }\]

Note that

\[{ f(0) ^2 = g(0) . }\]

Hence

\[{ f(x) = \frac{1}{\sqrt{g(0)}} g (x ^2) . }\]

Hence

\[{ \frac{1}{\sqrt{g(0)}} g (x ^2) \frac{1}{\sqrt{g(0)}} g (y ^2) = g(x ^2 + y ^2) . }\]

Hence

\[{ g(x ^2) g(y ^2) = g(0) g(x ^2 + y ^2) . }\]

Hence

\[{ g(t _1) g(t _2) = g(0) g(t _1 + t _2) }\]

for ${ t _1, t _2 > 0 . }$

Hence

\[{ g(t) = A e ^{kt} . }\]

Hence ${ f(x, y) }$ is of the form

\[{ \boxed{ f(x, y) = A e ^{k(x ^2 + y ^2)} } . }\]

Note that ${ k < 0 . }$

Hence ${ f(x, y) }$ is of the form

\[{ \boxed{f(x, y) = \frac{1}{2 \pi \sigma ^2} e ^{- \frac{x ^2 + y ^2}{2 \sigma ^2}} } . }\]

We call this a Gaussian distribution.