Gaussian Distribution

Gaussian Distribution

Named after Gaussian, with some assumptions on error, describe the uncertainty of samples.

Distribution of Errors

Assumptions

1. sum up to zero (average of samples is the true value)
2. $p(e\_1)\le p(e\_2)$ if $e\_1\ge e\_2$
3. errors are independent

Solving the probability density of the error

Suppose:

• the probability (density) function of error is $f$
• the true value is $L$
• the observed values are X=\left{ x\_{i} \right}
• the errors are \left{ z = x-L | x \in X \right}

Then the Maximum Likelihood Estimation of the true value $L$ give the first constraint

\begin{align} \ln P =& \sum\_i^n \ln f(x\_i - L) & \text{log-lilkelihood} & \\ \frac{\partial \ln P}{\partial L} =& \sum\_{i}^n \frac{f'(x\_i - L)}{f(x\_i -L)} (-1) \\ \=&\sum\_{i=1}^n g(x\_i - L) = 0& \text{MLE L s.t. maximize P} & \\ \end{align}

This process says: to maximize the likelihood of $L$, function $g(z)=-\frac{f^{\prime }(z)}{f(z)}$ should sums up to $0$.

Now with the assumption that all errors sums to $0$, we have two constraints:

$$\begin{array}{r}\sum \_{i=1}^{n}g(z\_i)=0\\ \sum \_{i=1}^{n}z\_i=0\end{array}$$

$g(z)=Cz$ can be one solution, thus:

$$g(z)=Cz⟹\frac{f^{\prime }(z)}{f(z)}=-Cz$$

Solve $f$ from the differential equation:

$$\begin{array}{rlrl}\int \frac{f^{\prime }(z)}{f(z)}dz& =\int (-Cz)dz& \text{continuous function has antidirevatives}& \\ \int \frac{1}{f(z)}df(z)& =(-\frac{C}{2}{z}^2)+C^{\prime }& \text{solve right side}& \\ \ln (f(z))& =(-\frac{C}{2}{z}^2)+C^{\prime \prime }& \text{solve left side with chain rule}& \\ f(z)& =A\left(e^{-\frac{C}{2}{z}^2}\right)& \text{power of e}& \end{array}$$

So the distribution of error is parameterized by $A$ and $C$, and the parameterized probability density is:

$$f(z)=A\left(e^{-C{z}^2/2}\right)$$

why not $g(x)=x^3$?

IDK, TODO

Solving A

As probability density is the derivative of accumulation function, integration of $f(z)$ on $\text{[}-\infty ,+\infty ]$ should be $1$:

\begin{align} 1 = & \int\_{-\infty}^{+\infty} f(z) dx \\ \=& A \int\_{-\infty}^{+\infty} e^{-\frac{C}{2}z^2} dx \\ \overset{x = \sqrt{\frac{2}{C}}t}{=}& A\sqrt{\frac{2}{C}} \int\_{-\infty}^{+\infty} e^{-t^2} dt \\ \end{align}

with Gaussian Integral:

$$1=A\sqrt{\frac{2\pi }{C}}⟹A=\sqrt{\frac{C}{2\pi }}$$

Interpreting $C$

Variation of Error

\begin{align} \sigma^{2} & = \int\_{-\infty}^{+\infty} x^{2}f(x),dx \\ & = -\frac{1}{C} \int\_{-\infty}^{+\infty} x \left((-Cx) A \exp\left( -\frac{Cx^{2}}{2} \right) \right) , dx \\ & = -\frac{1}{C} \int\_{-\infty}^{\infty} x \left\[ A \exp\left( -\frac{Cx^{2}}{2} \right) \right]' , dx \\ & = -\frac{1}{C} \left{ \left\[ f(x) \right]_{-\infty}^{+\infty} - \int_{-\infty}^{\infty} f(x) , dx \right} \\ & = -\frac{1}{C} \left\[ 0 - 1 \right] = \frac{1}{C} \end{align}

with ${\sigma }^2=\frac{1}{C}$, then:

$$f(x)=\frac{1}{\sigma \sqrt{2\pi }}\exp \left(-\frac{x^2}{2{\sigma }^2}\right)$$

Expectation of Error

For any function symmetric w.r.t. $(a,0)$, its integral on $\text{[}-\infty ,+\infty ]$ is $0$:

$$\begin{array}{rl}\int \_{-\infty }^{\infty }f(x),dx& =\int \_{-\infty }^{a}f(x),dx+\int \_a^{\infty }f(x),dx\\ & =-\int \_0^{\infty }f(a-t),dt+\int \_0^{\infty }f(a+s),ds\\ & =0\end{array}$$

as $xf(x)$ is symmetric w.r.t. $(0,0)$:

$$\mu =\int \_{-\infty }^{\infty }xf(x),dx=0$$

Distribution of the Samples

PDF of Samples

substituting the observed variable $Z=X-L$ back, then:

$$P(X\text{<}x)=P(Z+L\text{<}x)=P(Z\text{<}x-L)=\int \_{-\infty }^{x-L}f(z),dz=\int \_{-\infty }^{x}f(t-L),dt$$

thus the density function of distribution $Z$ is

$$f\_X(x)={\left(\int \_{-\infty }^{x}f(t-L),dt\right)}^{\prime }=f(x-L)=A(e^{-C(x-L{)}^2/2})$$

Expectation of Samples

expectation of $X$ would be:

$$\mu \_X=\mu +L=L$$

Variance of Samples

according to Transformation, variation of $X$ would be the same, which is ${\sigma }^2$.

Conclusion

In conclusion, with the assumption of error, the distribution of sample would be described as:

$$f(x)=\frac{1}{\sqrt{2\pi }\sigma }\exp \left(-\frac{(x-\mu {)}^2}{2{\sigma }^2}\right)$$

where: $\mu$ is the expectation and ${\sigma }^2$ is the variance.