Probability and Statistics Lecture 11

Continued

Probability and Statistics Lecture 10

Snippet on general form for $U.V$

$$\int u.vdx=u\int v.dx-\frac{du}{dv}\int v.dx+\frac{d^{2}u}{d{x}^2}\int v+....$$

Question on Raw Moment

Question 1

$$F(X)=\{\begin{array}{ll}\frac{1}{4}{e}^{-\frac{x}{4}},& x>0\\ 0,& \text{otherwise}\end{array}$$

Calculating ${\mu }_1^{\prime }$

The first raw moment is:

$${\mu }_1^{\prime }=E(X)={\int }_0^{\infty }xf(x)dx$$

Substituting the PDF:

$${\mu }_1^{\prime }={\int }_0^{\infty }x\frac{e^{-\frac{x}{4}}}{4}dx$$

Simplifying:

$${\mu }_1^{\prime }=\frac{1}{4}{\int }_0^{\infty }x{e}^{-\frac{x}{4}}dx$$

Using integration by parts:

$${\mu }_1^{\prime }=\frac{1}{4}{\left[(-4)x{e}^{-\frac{x}{4}}-\int e^{-\frac{x}{4}}\right]}_0^{\infty }$$

Evaluating the integral:

$${\mu }_1^{\prime }=\frac{1}{4}{\left[0+16e^{-\frac{x}{4}}\right]}_0^{\infty }$$

Final value:

$${\mu }_1^{\prime }=4$$

Calculating ${\mu }_2^{\prime }$ (Second Raw Moment)

The second raw moment is:

$${\mu }_2^{\prime }=E(X^2)={\int }_0^{\infty }{x}^{2}f(x)dx$$

Substituting the PDF:

$${\mu }_2^{\prime }={\int }_0^{\infty }{x}^2\frac{e^{-\frac{x}{4}}}{4}dx$$

Simplifying:

$${\mu }_2^{\prime }=\frac{1}{4}{\int }_0^{\infty }{x}^2{e}^{-\frac{x}{4}}dx$$

Using integration by parts twice:

$${\mu }_2^{\prime }=\frac{1}{4}{\left[(-4)x^2{e}^{-\frac{x}{4}}-8x{e}^{-\frac{x}{4}}+8e^{-\frac{x}{4}}\right]}_0^{\infty }$$

Evaluating:

$${\mu }_2^{\prime }=32$$

Calculating ${\mu }_3^{\prime }$ (Third Raw Moment)

The third raw moment is:

$${\mu }_3^{\prime }=E(X^3)={\int }_0^{\infty }{x}^{3}f(x)dx$$

Substituting the PDF:

$${\mu }_3^{\prime }=\frac{1}{4}{\int }_0^{\infty }{x}^3{e}^{-\frac{x}{4}}dx$$

Using integration by parts three times:

$${\mu }_3^{\prime }=\frac{1}{4}{\left[(-4)x^3{e}^{-\frac{x}{4}}-12x^2{e}^{-\frac{x}{4}}+48x{e}^{-\frac{x}{4}}-48e^{-\frac{x}{4}}\right]}_0^{\infty }$$

Evaluating:

$${\mu }_3^{\prime }=192$$

Calculating ${\mu }_4^{\prime }$ (Fourth Raw Moment)

The fourth raw moment is:

$${\mu }_4^{\prime }=E(X^4)={\int }_0^{\infty }{x}^{4}f(x)dx$$

Substituting the PDF:

$${\mu }_4^{\prime }=\frac{1}{4}{\int }_0^{\infty }{x}^4{e}^{-\frac{x}{4}}dx$$

Using integration by parts four times:

$${\mu }_4^{\prime }=\frac{1}{4}{\left[(-4)x^4{e}^{-\frac{x}{4}}-16x^3{e}^{-\frac{x}{4}}+48x^2{e}^{-\frac{x}{4}}-192x{e}^{-\frac{x}{4}}+192e^{-\frac{x}{4}}\right]}_0^{\infty }$$

Evaluating:

$${\mu }_4^{\prime }=768$$

Central Moments

The central moment ${\mu }_r$ is given by:

$${\mu }_r=E[(X-{\mu }_1^{\prime }{)}^r]={\int }_0^{\infty }(x-{\mu }_1^{\prime }{)}^{r}f(x)dx$$

Calculating the Central Moment ${\mu }_2$ (Variance)

$${\mu }_2=E[(X-{\mu }_1^{\prime }{)}^2]={\mu }_2^{\prime }-({\mu }_1^{\prime }{)}^2$$

Using the values for ( \mu’_1 ) and ( \mu’_2 ):

$${\mu }_2=32-4^2=32-16=16$$

Calculating the Central Moment ${\mu }_3$

$${\mu }_3=E[(X-{\mu }_1^{\prime }{)}^3]={\mu }_3^{\prime }-3{\mu }_2^{\prime }{\mu }_1^{\prime }+2({\mu }_1^{\prime }{)}^3$$

Substitute the values:

$${\mu }_3=192-3\cdot 32\cdot 4+2\cdot 4^3=192-384+128=-64$$

Calculating the Central Moment ${\mu }_4$

$${\mu }_4=E[(X-{\mu }_1^{\prime }{)}^4]={\mu }_4^{\prime }-4{\mu }_3^{\prime }{\mu }_1^{\prime }+6{\mu }_2^{\prime }({\mu }_1^{\prime }{)}^2-3({\mu }_1^{\prime }{)}^4$$

Substitute the values:

$${\mu }_4=768-4\cdot (-64)\cdot 4+6\cdot 32\cdot 4^2-3\cdot 4^4$$ $${\mu }_4=768+1024+1536-768=2560$$

Continued

Probability and Statistics Lecture 12