Appendix III: Log-Normal and Log-Laplace Appendix III: Log-Normal and Log-Laplace
The following section provides detailed formulas for growth distributions, focusing on the log-normal and log-Laplace distributions.
Normal Distribution
The formula for a normal distribution is given by:
N ( μ , σ ^ , x ) = 1 2 π σ ^ 2 exp ( − ( x − μ ) 2 2 σ ^ 2 ) N(\mu, \hat \sigma, x) = \frac{1}{\sqrt{ 2 \pi \hat \sigma^2 }} \exp\left( - \frac{ (x - \mu)^2 } {2 \hat \sigma^2} \right) N ( μ , σ ^ , x ) = 2 π σ ^ 2 1 exp ( − 2 σ ^ 2 ( x − μ ) 2 ) Lognormal Distribution
The lognormal distribution can be expressed as:
10 N ( μ , σ ^ , x ) = e N ( m , V , ln 10 ( x ) ) 10^{N(\mu, \hat \sigma, x)} = e^{N(m, \sqrt{V}, \ln10(x))} 1 0 N ( μ , σ ^ , x ) = e N ( m , V , l n 10 ( x )) where:
m = μ ln ( 10 ) m = \mu \ln(10) m = μ ln ( 10 ) V = σ 2 ln 2 ( 10 ) V = \sigma^2 \ln^2(10) V = σ 2 ln 2 ( 10 )
Additionally, D ( μ , σ ^ ; n ) D(\mu, \hat \sigma; n) D ( μ , σ ^ ; n ) n n n D ( μ , σ ^ ) D(\mu, \hat \sigma) D ( μ , σ ^ )
Moments of a Log-Normal
The expected value, or first moment, is:
E [ 10 N ( μ , σ ^ , x ) ] = e m + V / 2 = 10 μ + σ 2 ln ( 10 ) / 2 E[10^{N(\mu, \hat \sigma, x)}] = e^{m + V/2} = 10^{\mu + \sigma^2 \ln(10) / 2} E [ 1 0 N ( μ , σ ^ , x ) ] = e m + V /2 = 1 0 μ + σ 2 l n ( 10 ) /2 In general, the k-th moment is:
E [ 10 k N ( μ , σ ^ , x ) ] = exp ( k 2 2 σ 2 ln 2 ( 10 ) + k μ ln ( 10 ) ) = 10 ( k 2 2 σ 2 ln ( 10 ) + k μ ) E[10^{k N(\mu, \hat \sigma, x)}] = \exp\left( \frac{k^2}{2}{\sigma}^{2} \ln^2(10) + k \mu \ln(10)\right) = 10^{\left( \frac{k^2}{2}{\sigma}^{2} \ln(10) + k \mu \right)} E [ 1 0 k N ( μ , σ ^ , x ) ] = exp ( 2 k 2 σ 2 ln 2 ( 10 ) + k μ ln ( 10 ) ) = 1 0 ( 2 k 2 σ 2 l n ( 10 ) + k μ ) Expectation Calculation
If the microshocks probability density function (pdf) is Gaussian, i.e., p ( t ) = exp ( − ( t − μ ) 2 / ( 2 σ 2 ) ) / 2 π σ p(t)=\exp(- (t - \mu)^2/(2 \sigma^2))/\sqrt{2 \pi} \sigma p ( t ) = exp ( − ( t − μ ) 2 / ( 2 σ 2 )) / 2 π σ 10 k t 10^{k t} 1 0 k t
E [ 10 k t ] = ∫ p ( t ) 10 k t d t = 1 2 π σ ∫ exp ( − ( t − μ ) 2 2 σ 2 + ln ( 10 ) k t ) d t E[10^{k t}] = \int p(t) 10^{k t} dt = \frac{1}{\sqrt{2 \pi} \sigma} \int \exp\left( \frac{- (t - \mu) ^2}{2 {\sigma}^2} + \ln(10) k t \right) dt E [ 1 0 k t ] = ∫ p ( t ) 1 0 k t d t = 2 π σ 1 ∫ exp ( 2 σ 2 − ( t − μ ) 2 + ln ( 10 ) k t ) d t Completing the square in the exponential, we have:
E [ 10 k t ] = exp ( k μ ln ( 10 ) + k 2 2 σ 2 ln 2 ( 10 ) ) E[10^{k t}] = \exp\left( k \mu \ln(10) + \frac{k^2}{2}{\sigma}^{2} \ln^2(10) \right) E [ 1 0 k t ] = exp ( k μ ln ( 10 ) + 2 k 2 σ 2 ln 2 ( 10 ) ) Special Cases
For a normal distribution t t t
E [ 10 t ] = 10 μ + 1 2 σ ^ 2 ln ( 10 ) ≈ 10 μ ( 1 + 1 2 σ ^ 2 ln 2 ( 10 ) ) E[10^t] = 10^{\mu + \frac{1}{2} \hat \sigma^2 \ln(10)} \approx 10^{\mu} \left(1 + \frac{1}{2} \hat \sigma^2 \ln^2(10)\right) E [ 1 0 t ] = 1 0 μ + 2 1 σ ^ 2 l n ( 10 ) ≈ 1 0 μ ( 1 + 2 1 σ ^ 2 ln 2 ( 10 ) ) E [ 10 2 t ] = 10 2 μ + 2 σ ^ 2 ln ( 10 ) E[10^{2t}] = 10^{2 \mu + 2 \hat \sigma^2 \ln(10)} E [ 1 0 2 t ] = 1 0 2 μ + 2 σ ^ 2 l n ( 10 ) Variance Calculation
The variance, defined as v a r [ 10 N ( ⋅ ) ] = E [ 10 2 N ( ⋅ ) ] − E 2 [ 10 N ( ⋅ ) ] var[10^{N(\cdot)}] = E[10^{2 N(\cdot)}] - E^2[10^{N(\cdot)}] v a r [ 1 0 N ( ⋅ ) ] = E [ 1 0 2 N ( ⋅ ) ] − E 2 [ 1 0 N ( ⋅ ) ]
E [ 10 2 t ] = 10 2 μ + σ ^ 2 ln ( 10 ) ( 10 σ ^ 2 ln ( 10 ) − 1 ) ≈ 10 2 μ ( ( σ ^ ln ( 10 ) ) 2 + o ( σ ^ 4 ) ) E[10^{2t}] = 10^{2 \mu + \hat \sigma^2 \ln(10)} (10^{\hat \sigma^2 \ln(10)} - 1) \approx 10^{2 \mu} \left( (\hat \sigma \ln(10))^2 + o(\hat \sigma^4) \right) E [ 1 0 2 t ] = 1 0 2 μ + σ ^ 2 l n ( 10 ) ( 1 0 σ ^ 2 l n ( 10 ) − 1 ) ≈ 1 0 2 μ ( ( σ ^ ln ( 10 ) ) 2 + o ( σ ^ 4 ) ) This section provides a comprehensive understanding of the moments and variance of log-normal distributions, which are crucial for various statistical analyses.