7.2: Central moments of probability density function

5. Calculating the Central Moment of pdf
   1. Expected value function \(E[x]\)
      1. Similar feature to indicate mean value for the variable
         1. What is the expected value of 3? \(E[3] = 3\)
         2. How would you do that for a random variable \(x\)?
            • \(E[x]\) best guess would be the mean value
         3. Allows the argument to be multiplied by the pdf \[ E[f(x)] = \int_{-\infty}^{+\infty}f(x) p(x) dx \]
            • \(f(x)\) is the value person is adding up
            • \(p(x)\) is fraction of cases when \(f(x)\) occurs
      2. Expectation function is a linear operator:
         1. Distribution law applies: \(E[a + b]=E[a]+E[b]\)
         2. Multiply by a constant: \(E[7c] = 7E[c]\) when \(c\) is random variable
      3. Used to find the central moments of pdf \(\mu_{m}\) through \[ \mu_{m} = E[(x-x')^{m}] = \int_{-\infty}^{+\infty}(x-x')^{m} p(x) dx \] where \(f(x) = (x-x')^{m}\) as the argument of the expected value function
   2. Zero\(^{\mathrm{th}}\) moment (\(m=0\)) \[ \mu_{0} = E[(x-x')^{0}] =\int_{-\infty}^{+\infty}(x-x')^{0} p(x) dx \] \[ \mu_{0} = E[1] = \int_{-\infty}^{+\infty}1 p(x) dx = 1 \]
      • Implies that area under pdf must have value of 1
   3. First moment (\(m=1\))
      1. Using expectation function first \[ \mu_{1} = E[(x-x')^{1}] = E[x] - E[x'] \] Which of those is constant? If expectation is like mean of random variable \[ \mu_{1} = E[x] -x' = \mathbf{0} \]
      2. Using in pdf form instead \[ \mu_{1} = \int_{-\infty}^{+\infty}(x-x')^{1} p(x) dx \] \[ \mu_{1} = \int_{-\infty}^{+\infty}x p(x) dx - x'\int_{-\infty}^{+\infty} p(x) dx = 0 \]
      3. In order for pdf version to work, must define true mean as \[ x' \equiv \int_{-\infty}^{+\infty}x p(x) dx \]
      4. If \(m=0\) defined area as 1, then \(x'\) acts as horizontal centroid of the area
   4. Second central moment (\(m=2\))
      1. Using the linearity of expectation function again \[ \mu_{2} = E[(x-x')^{2}] = E[x^{2}-2xx'+x'^{2}] \] \[\mu_{2} = E[x^{2}]-2x'E[x]+x'^{2} \] \[ \mu_{2} = E[x^{2}] - x'^{2} \equiv \sigma^{2} \] where \(\sigma^{2}\) is the variance of the random data
      2. Same expansion could be applied to pdf \[ \mu_{2} = \int_{-\infty}^{+\infty}(x-x')^{2} p(x) dx = \sigma^{2} \]
      3. Note the different terms:
         1. variance \(\sigma^{2}\) is square of units relative to the mean
         2. standard deviation \(\sigma\) is in the same units of random variable
         3. mean square is the non-centered square of the random variable \[ \psi^{2} = \int_{-\infty}^{+\infty}x^{2} p(x) dx = \sigma^{2} + x'^{2} \]
   5. Third central moment (\(m=3\)) \[ \mu_{3} = E[(x-x')^{3}] =\int_{-\infty}^{+\infty}(x-x')^{3} p(x) dx \]
      1. Defines the skewness or asymmetry of the pdf
      2. \(Sk = \frac{\mu_{3}}{\sigma^{3}}\) making it dimensionless and signed value
      3. Right shifted (tail stretches to the right) \(Sk>0\)
      4. Left shifted when \(Sk<0\)
      5. Balanced when \(Sk = 0\)
   6. Fourth central moment (\(m=4\)) \[ \mu_{4} = E[(x-x')^{4}] =\int_{-\infty}^{+\infty}(x-x')^{4} p(x) dx \]
      1. Defines the kurtosis or flatness versus peakedness of the pdf
      2. \(Ku = \frac{\mu_{4}}{\sigma^{4}}\) since scaled is again dimensionless
      3. \(Ku = 3\) for bell curve (normal distribution; next section)
      4. \(Ku > 3\) when sharp top peak (high derivative over the top)
      5. \(0 < Ku < 3\) when more flat on top
   7. Drawn graphically as part of lecture
   8. All moments are independent of one another except for the true mean \(x'\) since that is embedded within \(m=\{2,3,4\}\)

      Exercise \(\PageIndex{1}\)

      Consider a sailboat-like probability density function for acceleration: \[ p(a) = \left\{\begin{array}{cr} 0 & a < 0 \\ & \\
                          ma & 0\leq a < 4 \\ & \\ 0 & 4\leq a < 6 
                          \\ & \\ -0.14a+1.26 & 6 \leq a < 9
                          \\ & \\ 0 & a \geq 9
                      \end{array} \right. \]
      shown in the figure:

       

      1. What value of \(m\) is necessary for the slope to have a valid probability density?
      2. Determine the true mean value \(a'\)
      3. Calculate the probability acceleration falls between the range of \(2.5< a < 6.5\) m/s\(^{2}).
      4. Determine the numerical values of variance, skewness, and kurtosis of the probability density function; justify the magnitude and scale of calculated results.

      Answer

      1. Area under curve has to add to unity: \[\int_{-\infty}^{+\infty} p(a)da =1\] setting up integration or area calculation of triangles \[\int_{0}^{4} ma\ da + \int_{6}^{9} \left(-0.14a + 1.26\right)\ da=1\] completing integration yields the result \(m = 0.04625\)
      2. The distance of variable \(a\) now multiplied by probability \[a'= \int a p(a)da=  \int_{0}^{4} 0.04625a^{2}\ da + \int_{6}^{9} \left(-0.14a^{2} + 1.26a\right)\ da\] which is equivalent to the centroid calculation of right triangles. The integration results in \(a' = 5.397\) m/s\(^{2}\)
      3. Calculate area under limited range of the function \[$P(2.5< a < 6.5) =\int_{2.5}^{4} 0.04625a\ da + \int_{6}^{6.5} \left(-0.14a + 1.26\right)\ da\] produces probability \(P = 42\)%
      4. First, the variance will need to be determined: \[\sigma^{2} = \int_{0}^{4} ma^{3}\ da + \int_{6}^{9}\left( -0.14a^{3} + 1.26a^{2}\right)\ da - a'^{2} \] making \(\sigma^{2} =  5.021\) m\(^{2}\)/s\(^{4}\)
      5. The value of skewness is \[Sk = \left[\int_{0}^{4} (a-a')^{3} ma\ da + \int_{6}^{9} (a-a')^{3}(-0.14a+1.26)\ da\right]/\sigma^{3} = -0.547 \] indicating a left shift due to \(Sk<0\) (longer stretch to the left relative to 5.397)
      6. The value of kurtosis is \[Ku =  \left[\int_{0}^{4} (a-a')^{4} ma\ da + \int_{6}^{9} (a-a')^{4}(-0.14a+1.26)\ da\right]/\sigma^{4}=  1.91\] indicating somewhat "flat" form near the mean region of the pdf.