Probability and Statistics Lecture 10

Continued

Probability and Statistics Lecture 9

$$E(X)={\int }_0^{\infty }x\cdot x{e}^{-x}dx={\int }_0^{\infty }{x}^2{e}^{-x}dx$$

Syllabus

• Unit 1 Complete
• Unit 2 ( R.V, P.M.F, P.D.F, C.D.F, Mean, Variance )

  Raw and Central Moments

  INFO

  In some snippets I have used $M$ to represent capital $\mu$ Since there is no support for capital mu try to interpret it as such.

  Moments

  Moments are quantities that describe the shape of a distribution. These moments can be used to calculate $mean$, $variance$, $skewness$, $kurtosis$, and more details regarding them.

  Raw Moment

  Moments are always described relative to a point. When a moment is relative to its origin $(O)$ It is given as

  $r^{th}rawmoment{\mu }_r^{\prime }=E((X-0{)}^r)=E(X^r)$ The zero here denotes the value around which the moment is calculated. Here zero is the relative point. For Raw Moments the relative point to which moments are calculated is zero ${\mu }_o=E(1)=1$ The property of expectation says that the expectation of constant is a constant.

  ${\mu }_1^{{}^{\prime }}=E(X)$ ${\mu }_2^{{}^{\prime }}=E(X^2)$ ${\mu }_2^{{}^{\prime }}=E(X^3)$ and so on… capital $\mu$ is used to represent the raw moments

  ---

  Central Moment

  When a moment is relative to its mean its moments are said to be central moment

  The $r^{th}$ Central Moment is given by ${\mu }_r=E[(X-\mu {)}^r]$

  $0^{th}$ Central Moment

  ${\mu }_0=1$

  $1^{st}$ Central Moment

  ${\mu }_1=E[X-\mu ]=[E(X)-M]=0$

  $2^{nd}$ Central Moment

  1. Solving for 2nd Central Moment The given equation is: $E[(X-\mu {)}^2]=E[X^2-2X\mu +{\mu }^2]=E(X^2)-2E(X)\mu +{\mu }^2$

  2. Identify $\mu$ as the Mean: Since $\mu$ represents the mean of $X$, we have: $E(X)=\mu$

  3. Substitute $E(X)$ with $\mu$: Substitute $E(X)$ with $\mu$ in the equation: $E[(X-\mu {)}^2]=E(X^2)-2{\mu }^2+{\mu }^2$

  4. Simplify the Expression: Simplify the right-hand side: $E[(X-\mu {)}^2]=E(X^2)-{\mu }^2$

  5. Relate to Variance: The Variance ${\sigma }^2$ of a random variable $X$ is defined as: ${\sigma }^2=E[(X-\mu {)}^2]$

     Therefore, we have: ${\sigma }^2=E(X^2)-{\mu }^2$ So, the rewritten equation with $\mu$ and the derivation showing it as variance is: ${\sigma }^2=E[(X-\mu {)}^2]=E(X^2)-{\mu }^2$ This shows that the expected value of the squared deviation of $X$ from its mean $\mu$ is indeed the variance of $X$.

  $=E(X^2)-M^2$ $=E(X^2)$

  Relation B/w Raw Moments and Central Moment

  $M_2=M_2^{\prime }-(M_1^{\prime }{)}^2$ $M_2=M_2^{\prime }-(M_1^{\prime }{)}^2$ $M_3=E[(X-M{)}^3]$

  $M_3=M_3^{\prime }-3M_2^{\prime }{M}_1^{\prime }+2(M_1^{\prime }{)}^3$

  $M_4=M_4^{\prime }-4M_3{}^{\prime }{M}_1^{\prime }+6M_2^{\prime }(M_1^{\prime }{)}^2+3(M_1^{\prime }{)}^4$

  References

  Information

  • date: 2025.03.13
  • time: 08:18

  Link to original

References

Information

• date: 2025.01.29
• time: 14:04

Continued

Probability and Statistics Lecture 11