1.1 The maximum likelihood estimation method  (Page 4/2)

Uniform distribution

Let $f\left(x\right)=\frac{1}{\alpha }$ for $0\le x\le \alpha$ and 0 elsewhere, where $\alpha >0.$ A graph of the pdf for this distribution is shown in Figure 1.

Probability distribution function of a uniform distribution.

It is easy to see from the graph that $f\left(x\right)=\frac{1}{\alpha }>0$ and $\mathrm{Pr}\left(a\le x\le b\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)dx}={\displaystyle \underset{0}{\overset{\alpha }{\int }}\frac{1}{\alpha }dx}=1.$ Moreover, as shown in Figure 1, the area under the pdf curve between a and b is equal to the probability that x lies between a and b ; that is, $$\mathrm{Pr}\left(a\le x\le b\right)={\displaystyle \underset{a}{\overset{b}{\int }}\left(\frac{1}{\alpha }\right)dx}={\frac{x}{\alpha }|}_a^b=\frac{b-a}{\alpha }.$$

The calculation of the mean and variance of this distribution is relatively simple. The population mean is given by $${\mu }_x=E\left(x\right)={\displaystyle \underset{0}{\overset{\alpha }{\int }}xf\left(x\right)dx}$$ or $${\mu }_x={\displaystyle \underset{0}{\overset{\alpha }{\int }}x\left(\frac{1}{\alpha }\right)dx}={\frac{x^2}{2\alpha }|}_0^{\alpha }=\frac{\alpha }{2}.$$

The population variance Quite often, as in the exercises at the end of this module, it is easier to calculate the variance of a distribution using the alternative formula for the variance: ${\sigma }_x^2=V\left(x\right)=E{\left(x-\mu \right)}^2=E\left(x^2\right)-{\mu }^2,$ where $E\left(x^2\right)={\displaystyle \int x^{2}f\left(x\right)dx}.$ is given by $V\left(x\right)=E\left[{\left(x-{\mu }_x\right)}^2.\right]$ Thus, $$V\left(x\right)={\displaystyle \underset{0}{\overset{\alpha }{\int }}{\left(x-\frac{\alpha }{2}\right)}^2\left(\frac{1}{\alpha }\right)dx}={\displaystyle \underset{0}{\overset{\alpha }{\int }}\left(x^2-\alpha x+\frac{{\alpha }^2}{4}\right)\left(\frac{1}{\alpha }\right)dx}$$ or $$V\left(x\right)={\frac{x^3}{3\alpha }-\frac{x^2}{2}+\frac{\alpha }{4}x|}_0^{\alpha }=\frac{{\alpha }^2}{3}-\frac{{\alpha }^2}{2}+\frac{{\alpha }^2}{4}=\frac{{\alpha }^2}{12}.$$

The normal distribution.

A random variable with a mean of $\mu$ and a variance of ${\sigma }^2$ that has a normal distribution —that is, $x~N\left(\mu ,{\sigma }^2\right)—$ has the pdf $$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}.$$ A typical graph of this pdf is given in Figure 2. The area under the curve between values of x of a and b is equal to the probability that x falls between a and b .

Probability distribution function of a normal distribution.