Moments of a Log-Laplace
The Laplace distribution is defined as:
L ( μ , b , x ) = 1 2 b exp ( − ∣ x − μ ∣ b ) = 1 2 σ ^ exp ( − ∣ x − μ ∣ σ ^ / 2 ) L(\mu, b, x) = \frac{1}{2b} \exp\left(-\frac{|x - \mu|}{b}\right) = \frac{1}{\sqrt{2} \hat{\sigma}} \exp\left(-\frac{|x - \mu|}{\hat{\sigma}/\sqrt{2}}\right) L ( μ , b , x ) = 2 b 1 exp ( − b ∣ x − μ ∣ ) = 2 σ ^ 1 exp ( − σ ^ / 2 ∣ x − μ ∣ ) The parameters σ ^ = 2 b \hat{\sigma} = \sqrt{2} b σ ^ = 2 b b b b σ ^ \hat{\sigma} σ ^
The log-Laplace in base 10 is expressed as 10 L ( μ , σ ^ , x ) 10^{L(\mu, \hat{\sigma}, x)} 1 0 L ( μ , σ ^ , x )
To derive the first moment in the simplified case of a symmetric shocks distribution and mean zero, we compute the expected value E [ 10 t ] E[10^t] E [ 1 0 t ] t t t
Derivation of the First Moment
To compute E [ 10 t ] E[10^t] E [ 1 0 t ]
E [ 10 t ] = ∫ − ∞ 0 p ( t ) 10 t d t + ∫ 0 ∞ p ( t ) 10 t d t E[10^t] = \int\limits_{-\infty}^{0} p(t) 10^t \, dt + \int\limits_{0}^{\infty} p(t) 10^t \, dt E [ 1 0 t ] = − ∞ ∫ 0 p ( t ) 1 0 t d t + 0 ∫ ∞ p ( t ) 1 0 t d t Knowing the expression of the indefinite integral of 10 t p ( t ) 10^t p(t) 1 0 t p ( t )
∫ 10 t exp ( ± t / b ) / ( 2 b ) d t = 10 t exp ( ± t / b ) / ( 2 ( b ln ( 10 ) ± 1 ) ) \int 10^t \exp(\pm t/b)/(2b) \, dt = 10^t \exp(\pm t/b)/(2 (b \ln(10) \pm 1)) ∫ 1 0 t exp ( ± t / b ) / ( 2 b ) d t = 1 0 t exp ( ± t / b ) / ( 2 ( b ln ( 10 ) ± 1 )) Thus, the expected value is:
E [ 10 t ] = 1 2 ( b ln ( 10 ) + 1 ) [ exp ( t / b ) 10 t ] − ∞ 0 + 1 2 ( b ln ( 10 ) − 1 ) [ exp ( − t / b ) 10 t ] 0 ∞ E[10^t] = \frac{1}{2 (b \ln(10) + 1)} \left[ \exp(t/b) 10^t \right]_{-\infty}^{0} + \frac{1}{2 (b \ln(10) - 1)} \left[\exp(-t/b) 10^t \right]_{0}^{\infty} E [ 1 0 t ] = 2 ( b ln ( 10 ) + 1 ) 1 [ exp ( t / b ) 1 0 t ] − ∞ 0 + 2 ( b ln ( 10 ) − 1 ) 1 [ exp ( − t / b ) 1 0 t ] 0 ∞ Using 10 t = e ln ( 10 ) t 10^t = e^{\ln(10) t} 1 0 t = e l n ( 10 ) t
E [ 10 t ] = 1 2 ( b ln ( 10 ) + 1 ) [ exp ( [ 1 / b + ln ( 10 ) ] t ) ] − ∞ 0 + 1 2 ( b ln ( 10 ) − 1 ) [ exp ( [ − 1 / b + ln ( 10 ) ] t ) ] 0 ∞ E[10^t] = \frac{1}{2 (b \ln(10) + 1)} \left[ \exp([1/b + \ln(10)] t) \right]_{-\infty}^{0} + \frac{1}{2 (b \ln(10) - 1)} \left[ \exp([-1/b + \ln(10)] t) \right]_{0}^{\infty} E [ 1 0 t ] = 2 ( b ln ( 10 ) + 1 ) 1 [ exp ([ 1/ b + ln ( 10 )] t ) ] − ∞ 0 + 2 ( b ln ( 10 ) − 1 ) 1 [ exp ([ − 1/ b + ln ( 10 )] t ) ] 0 ∞ The mean will diverge unless the exponentials are zero at infinity, requiring σ ^ < 2 / ln ( 10 ) ≈ 0.61 \hat{\sigma} < \sqrt{2}/\ln(10) \approx 0.61 σ ^ < 2 / ln ( 10 ) ≈ 0.61 σ ^ \hat{\sigma} σ ^ σ ^ \hat{\sigma} σ ^ N N N
Evaluating the primitive at the limits gives:
E [ 10 t ] = 1 2 ( b ln ( 10 ) + 1 ) − 1 2 ( b ln ( 10 ) − 1 ) = 1 1 − ( b ln ( 10 ) ) 2 E[10^t] = \frac{1}{2 (b \ln(10) + 1)} - \frac{1}{2 (b \ln(10) - 1)} = \frac{1}{1 - (b \ln(10))^2} E [ 1 0 t ] = 2 ( b ln ( 10 ) + 1 ) 1 − 2 ( b ln ( 10 ) − 1 ) 1 = 1 − ( b ln ( 10 ) ) 2 1 If the Laplace shocks are centered at μ ≠ 0 \mu \neq 0 μ = 0 10 μ 10^\mu 1 0 μ
E [ 10 t ] = 10 μ 1 − ( b ln ( 10 ) ) 2 = 10 μ 1 − 1 2 σ 2 ln 2 ( 10 ) E[10^t] = \frac{10^\mu}{1 - (b \ln(10))^2} = \frac{10^\mu}{1 - \frac{1}{2}\sigma^2 \ln^2(10)} E [ 1 0 t ] = 1 − ( b ln ( 10 ) ) 2 1 0 μ = 1 − 2 1 σ 2 ln 2 ( 10 ) 1 0 μ General Expression for Moments
An expression for the moments of a log-Laplace, generalizing to possible asymmetries, is given by Kozubowski (2003):
E [ 10 k t ] = ∫ p L ( t ) 10 k t d t = 10 k μ α β ( α − k ln ( 10 ) ) ( β + k ln ( 10 ) ) E[10^{k t}] = \int p_L(t) 10^{k t} \, dt = 10^{k \mu} \frac{\alpha \beta}{(\alpha - k \ln(10)) (\beta + k \ln(10))} E [ 1 0 k t ] = ∫ p L ( t ) 1 0 k t d t = 1 0 k μ ( α − k ln ( 10 )) ( β + k ln ( 10 )) α β where α \alpha α β \beta β α = β = 1 / b \alpha = \beta = 1/b α = β = 1/ b
E [ 10 k t ] = 10 k μ 1 1 − ( k b ln ( 10 ) ) 2 E[10^{k t}] = 10^{k\mu} \frac{1}{1 - (k b \ln(10))^2} E [ 1 0 k t ] = 1 0 k μ 1 − ( kb ln ( 10 ) ) 2 1 The parameter b b b σ ^ \hat{\sigma} σ ^ b = σ ^ / 2 b = \hat{\sigma}/\sqrt{2} b = σ ^ / 2
The second moment is:
E [ 10 2 t ] = 10 2 μ 1 1 − ( σ ^ ln ( 10 ) ) 2 / 2 E[10^{2 t}] = 10^{2 \mu} \frac{1}{1 - (\hat{\sigma} \ln(10))^2/\sqrt{2}} E [ 1 0 2 t ] = 1 0 2 μ 1 − ( σ ^ ln ( 10 ) ) 2 / 2 1 From here, the variance v a r [ 10 L ( ⋅ ) ] = E [ 10 2 L ( ⋅ ) ] − E 2 [ 10 L ( ⋅ ) ] var[10^{L(\cdot)}] = E[10^{2 L(\cdot)}] - E^2[10^{L(\cdot)}] v a r [ 1 0 L ( ⋅ ) ] = E [ 1 0 2 L ( ⋅ ) ] − E 2 [ 1 0 L ( ⋅ ) ]
v a r [ 10 L ( ⋅ ) ] = 10 2 μ ( 1 1 − 2 σ ^ − 4 ( 4 − σ ^ 2 ) 2 ) var[10^{L(\cdot)}] = 10^{2\mu} \left( \frac{1}{1 - 2 \hat{\sigma}} - \frac{4}{(4 - \hat{\sigma}^2)^2} \right) v a r [ 1 0 L ( ⋅ ) ] = 1 0 2 μ ( 1 − 2 σ ^ 1 − ( 4 − σ ^ 2 ) 2 4 ) In the limit of small micro fluctuations:
v a r [ 10 L ( ⋅ ) ] ≈ 10 2 μ ( σ ^ 2 + 13 4 σ ^ 4 + 15 2 σ ^ 6 + o ( σ ^ 8 ) ) var[10^{L(\cdot)}] \approx 10^{2 \mu} \left( \hat{\sigma}^2 + \frac{13}{4} \hat{\sigma}^4 + \frac{15}{2} \hat{\sigma}^6 + o(\hat{\sigma}^8) \right) v a r [ 1 0 L ( ⋅ ) ] ≈ 1 0 2 μ ( σ ^ 2 + 4 13 σ ^ 4 + 2 15 σ ^ 6 + o ( σ ^ 8 ) )